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electromagnetism
|
Magnetic force on a magnetic dipole
|
https://physics.stackexchange.com/questions/66203/magnetic-force-on-a-magnetic-dipole
|
<p>Can someone please help me understand and answer the question of "why are the forces on a magnetic dipole different in a uniform magnetic field and non-uniform magnetic field?" I know in a uniform magnetic field the magnetic dipole material will tend to align with the field lines of the magnetic field it is in but what happens in a non-uniform one? Doesn't it just change movement rapidly whilst trying to align it's poles with the field line of the magnetic field.</p>
|
<p>In uniform field, the net force is zero
$$F=\oint ids\times B=(\oint ids)\times B=0\times B=0 $$
but the torque is not zero,$\tau\neq0$, until it aligns with field lines,</p>
<p>in non-uniform field, both are nonzero $F \neq 0$ , $\tau \neq 0$.</p>
| 300
|
electromagnetism
|
Direction of magnetic force when magnetic field and velocity are not in same plane
|
https://physics.stackexchange.com/questions/69413/direction-of-magnetic-force-when-magnetic-field-and-velocity-are-not-in-same-pla
|
<p>We know from Flaming's Right Hand Rule how to calculate the direction of the magnetic force given the magnetic field and the velocity are in the same plane. Now suppose they are not in the same plane. As an example, consider a uniform magnetic field directed perpendicularly onto the screen and a charged particle moving on an inclined plane with angle X with the horizontal. What will be the direction of the magnetic force here?</p>
<p>I think Flaming's Right Hand Rule can only be applied when they are in the same plane because if they are not we cannot adjust our hand so that our fingers point to both of them. But I'm just a beginner in this field, so please correct me if I am wrong. </p>
|
<p>Two vectors (starting or shifted/imagined to start at the same point) always belong to the same plane, usually one plane. If the vectors are $A\to B$ and $A\to C$, just imagine that you connect $B,C$ by a straight line, thus completing a triangle, $ABC$. This triangle is already a clear "seed" of a plane, isn't it? The plane isn't necessary vertical, horizontal, or parallel to any other plane you may have thought about at the beginning but it is a plane nevertheless. There always exists a vector $\vec n$ that is orthogonal to the whole plane or, equivalently, that is orthogonal both to the vectors $A\to B$ and $A\to C$.</p>
<p>This is not really physics but rather geometry for the 10-year-olds.</p>
| 301
|
electromagnetism
|
Is it correct to say "like poles attract, unlike poles repel" while two magnets are placed such that one is inside another?
|
https://physics.stackexchange.com/questions/73953/is-it-correct-to-say-like-poles-attract-unlike-poles-repel-while-two-magnets
|
<p>As we know a solenoid is considered as a electromagnet(magnet) if there's a current flowing through it. if a soft iron core is placed inside the solenoid, the former get magnetised. Consider the solenoid as a hollow bar magnet, and the magnetised soft iron core inside as a bar magnet. The poles of the magnets next to each other is of same polarity. Is there attracting force between the two magnets? If so, does it mean "like poles attract, unlike poles repel" while two magnets are placed such that one is inside another? </p>
|
<p>A solenoid <em>induces</em> a magnetic field in the iron bar kept inside it. This is not the same as inserting a permanent bar magnet inside the solenoid. </p>
<p>The induced magnetic field is in the same direction as the original magnetic field. One way of thinking about this is that the <a href="https://en.wikipedia.org/wiki/Magnetic_domain" rel="nofollow">domains</a> in the iron bar line up with the external field, producing a net magnetic field in the same direction as the original. </p>
<p>However if you consider a permanent bar magnet kept in a solenoid, and the solenoid was large enough such that the bar magnet could rotate in any arbitrary direction, once the solenoid is switched on, the bar magnet will align itself with the solenoid's field. This has to do with minimizing the energy in the magnetic field. The energy is given by
$$ E = - m .\vec{B} $$ </p>
<p>so clearly if $m$ and $B$ are in the same direction, it is a lower energy configuration than if $m$ and $B$ are in opposite directions.</p>
| 302
|
electromagnetism
|
Particle inside a charged shell
|
https://physics.stackexchange.com/questions/76598/particle-inside-a-charged-shell
|
<p>Imagine that I have a particle of charge $q$ at the center of a spherical insulating shell of charge $Q$ and radius $R$.</p>
<p>Both the particle and shell are initially at rest.</p>
<p>Now I apply a force $\mathbf{F}$ to the particle which causes it to have an acceleration $d\mathbf{v}/dt$.</p>
<p>The electric field due to the charge $q$ is given by:</p>
<p>$$\mathbf{E} = - \mathbf{\nabla} \phi - \frac{\partial \mathbf{A}}{\partial t}.$$</p>
<p>Due to the symmetry of the situation the total static electric force on the shell due to the charge is zero.</p>
<p>The $\mathbf{A}$-field due to the charge $q$ at the position of the shell is approximately given by:</p>
<p>$$\mathbf{A} = \frac{q\mathbf{v}}{4 \pi \epsilon_0 c^2R}.$$</p>
<p>Therefore the charge $q$ induces an electric force $\mathbf{f_s}$ on the shell given by:</p>
<p>$$\mathbf{f_s} = -\frac{qQ}{4 \pi \epsilon_0 c^2R}\frac{d\mathbf{v}}{dt}.$$</p>
<p>But the total force on the system as a whole should remain $\mathbf{F}$.</p>
<p>Does this mean that the shell must induce a balancing electric force $\mathbf{f_s}$ back on the accelerating particle?</p>
<p>P.S. As the particle acceleration is constant I believe that there is no radiation reaction force back on the particle (a controversial view I admit, see related link below). Thus there is no change of momentum in the EM field. Therefore the internal forces on the system of charges should balance.</p>
| 303
|
|
electromagnetism
|
How DC and AC relays works?
|
https://physics.stackexchange.com/questions/79438/how-dc-and-ac-relays-works
|
<p>I was told long time ago that DC relay had a coil. There was a switch (2 wires, one is stable, the other one is flexible) inside the coil. The switch was parallel to the axial direction of the coil. </p>
<p>Today, I am thinking how AC relay works. I go back to think about DC relay. If what I was told is right, the magnetic field direction is parallel to the switch direction. Then how the magnetic field makes two wires touch?</p>
<p>Also, how does AC relay work? </p>
|
<p>This question is not about solid state relays.</p>
<p>In a DC solenoid type relay, a permanent magnet (sliding center core) moves and actuates switch contact(s) when energized. These are typically used in automotive applications, to provide current to a starter motor through fusible links.</p>
<p>Although AC solenoid type relays might be possible, it would be much trickier to mechanically engineer one that operated like their DC counterpart because the center iron core (not magnetized) would need to pull against a retaining spring. That would mean it would normally reside outside of the coil. It is far easier to make AC relays which when energized simply attract a chunk of iron riveted to the end of a lever which actuates the switch contacts.</p>
| 304
|
electromagnetism
|
How do I find the electric field above the center of a square plate (rather than circular)?
|
https://physics.stackexchange.com/questions/79858/how-do-i-find-the-electric-field-above-the-center-of-a-square-plate-rather-than
|
<p>I tried to integrate E due to a line of charge sweeping across the plate, but got bogged down. Any suggestions?</p>
|
<p>The trick to this one is knowing how to do the integral. I am going to assume you got an integral that looks like $$\int_{-L/2}^{L/2} \frac{a \lambda dx}{(x^2+d^2)^{3/2}}.$$
The first thing to do to make this easier is non-dimensionalize the integral. Let's start by switching to a non-dimensional integration variable $u=x/L$. Then $dx = L du$. Our integral becomes $$\int_{-1/2}^{1/2} \frac{a \lambda L du}{((Lu)^2+d^2)^{3/2}}.$$
Now let's pull dimensional factors outside of the integral. We get $$\frac{a \lambda L}{d^2}\int_{-1/2}^{1/2} \frac{du}{((Lu/d)^2+1)^{3/2}}.$$
We will forget about the prefactor right now and concentrate on the integral. The denominator is complicated and we want to simplify it. We are going to simplify with the identy $\tan^2 \theta + 1 = sec^2 \theta$. We will need to make the substition $\tan \theta = Lu/d$. Then $du = \frac{d}{L} \sec^2 \theta d\theta$. The integral is now
$$\int_{-\theta_0}^{\theta_0} \frac{\frac{d}{L} \sec^2 \theta d\theta}{(\tan^2 \theta+1)^{3/2}},$$
where $\theta_0 = \tan^{-1} \frac{L}{2d}$. After simplifying the integral, we get
$$\frac{d}{L} \int_{-\theta_0}^{\theta_0}\cos \theta d\theta,$$
so we are good. This technique is called <a href="http://en.wikipedia.org/wiki/Trig_sub" rel="nofollow">trigonometric substitution</a>. </p>
| 305
|
electromagnetism
|
The curl is not zero?
|
https://physics.stackexchange.com/questions/81063/the-curl-is-not-zero
|
<p>In Maxwell's equations, the curl of the electric field for a steady state processes (ie. No changing electric or magnetic field) zero. However, if we take a curl along the wire for some distance and perpendicularly out from the wire and back to its original position (perpendicularly entering the wire), it is not zero for a resiative wire (non ideal wire) if we assume that there is no field outside the wire. This suggests that there could be electric field outside the wire. Is it? And if there indeed is, how to explain the energy conservation (because usually we just treat the energy to be all converted to thermal energy)?</p>
<p>Thank you </p>
| 306
|
|
electromagnetism
|
Electron motion in a wire
|
https://physics.stackexchange.com/questions/81638/electron-motion-in-a-wire
|
<p>For my introductory course to electromagnetism (I'm an undergraduate student, so ELI5), I'm trying to get the right conceptual model of electron movement in a thin wire (with constant but non-zero cross section, like a cylinder) due to a constant current. It seems to me, that there are several contributing factors, and I wonder which dominate, and if some kind of order exists.</p>
<ol>
<li><p>Electrostatic repulsion between electrons. If this was the only force at play, electrons should all travel on the surface of the wire.</p></li>
<li><p>Magnetostatic attraction caused by the magnetic field from charges in motion. If this was the only force then the electrons would all travel in the center of the cylinder.</p></li>
<li><p>Collisions with matter and thermal scattering, which would make the motion chaotic.</p></li>
<li><p>Material properties of the cylinder, and if in particular the cylinder is a metal, then I suspect that the lattice planes of the metal and orientation of these planes is somehow relevant. </p></li>
</ol>
<p>Basically 1 and 2 create order, 3 and 4 seem to destroy it. I can't seem to find anything relevant in my courseware or online. Specifically I'm interested in common DC household wires.</p>
|
<p>1 and 3 are the major players here, and chaotic/brownian/random movement with lots and lots of collisions is what the electrons do even when they are being pushed from one place to another as a current. </p>
<p>The electrostatic repulsions and attractions tend to both bring them close and separate them at the same time, while heating just gives them a kinetic energy which makes them move randomly in any direction.</p>
<p>They just continue to do their random motion in different parts of wires as they are transferred from one location to other under the applied electric field.</p>
| 307
|
electromagnetism
|
What is the magnetic effect on either of the charges moving parallel
|
https://physics.stackexchange.com/questions/83089/what-is-the-magnetic-effect-on-either-of-the-charges-moving-parallel
|
<p>Consider two electrons moving parallel to each other in the same direction with same constant velocity. Will they experience any force due to either of them?</p>
|
<p>Yes, they will feel both electric and magnetic force. If you apply a Lorentz boost and get into the frame where they are at rest, they will simply feel the electric field of each other.</p>
<p>However if we stay in the lab frame we will notice that the force they experience goes down with $1/\beta \gamma^2$ (relativistic factors), approaching zero as they approach the speed of light. You can picture this also noticing that the time is flowing slower in their fast moving frame of reference and so you kinda see their radial movement in slow-motion.</p>
<p>This is a key point in accelerator physics where you want to get the beam of particles travel as fast as possible in the minimum amount of space thus reducing the radial defocusing effect due the so called "space charge".</p>
| 308
|
electromagnetism
|
Is mass of a particle changed when it is charged?
|
https://physics.stackexchange.com/questions/83382/is-mass-of-a-particle-changed-when-it-is-charged
|
<p>If a particle of mass $M$ is given an electric charge $Q$, will its mass change?</p>
|
<p>What do you mean with "particle"? How would you add a charge?</p>
<p>If you add an electron to an atom you will increase the mass by the electron mass (of course) minus a tiny contribution coming from $m=E/c^2$ where E is the work done taking the electron from where it was to its actual position in the Coulomb potential.</p>
| 309
|
electromagnetism
|
Why is th $\hat{r}$ component zero in this integral?
|
https://physics.stackexchange.com/questions/86084/why-is-th-hatr-component-zero-in-this-integral
|
<p>I'm trying to evaluate the magnetic field by calculating the Coloumb integral $\overrightarrow{A}$, and later I will take:
$$\overrightarrow{B}=\nabla \times \overrightarrow{A}$$</p>
<p>However, in the middle of everything, I get to (cylindrical coordinates):</p>
<p>$$\overrightarrow{A}=\frac{\mu_oI}{4\pi} \oint_{C} \frac{a\hat{\phi'}}{[r²+(z-b)^2-2ar\cos(\phi-\phi')]^{1/2}} d\phi'$$</p>
<p>I should show that the only component of this expression is $\hat{\phi}$, where:</p>
<p>$$\hat{r'}=\hat{x}\cos{\phi'}+\hat{y}\sin{\phi}$$
$$\hat{\phi'}=-\hat{x}\sin{\phi'}+\hat{y}\cos{\phi'}$$
$$\hat{r}=\hat{x}\cos{\phi}+\hat{y}\sin{\phi}$$
$$\hat{\phi}=-\hat{x}\sin{\phi}+\hat{y}\cos{\phi}$$</p>
<p>and that it's leading to only this integral:</p>
<p>$$\overrightarrow{A}=\frac{\mu_oI}{4\pi} \oint_{C} \frac{a\cos{\left(\phi-\phi'\right)}\hat{\phi}}{[r²+(z-b)^2-2ar\cos(\phi-\phi')]^{1/2}} d\phi'$$</p>
<p>But when I rewrite how $\hat{\phi'}$ relates to $\hat{\phi}$ and $\hat{r}$, it's not so obvious that the $\hat{r}$ component disappears. The writer refers to symmetry, but I can't still figure this out. So I basically have this integral instead because I can't show how the $\hat{r}$ component is zero:</p>
<p>$$\overrightarrow{A}=\frac{\mu_oI}{4\pi} \oint_{C} \frac{a\sin{\left(\phi-\phi'\right)}\hat{r}+a\cos{\left(\phi-\phi'\right)}\hat{\phi}}{[r²+(z-b)^2-2ar\cos(\phi-\phi')]^{1/2}} d\phi'$$
Any ideas? I've been stuck at this for a while, and I've tried to show it in Cartesian coordinates without no luck.</p>
|
<p>Hint: substitute $u=\cos{(\phi - \phi')}$ and integrate.</p>
| 310
|
electromagnetism
|
Why don't stationary electric charges possess a magnetic field?
|
https://physics.stackexchange.com/questions/87818/why-dont-stationary-electric-charges-possess-a-magnetic-field
|
<p>Why don't stationary electric charges posses magnetic field, while moving charges do?</p>
| 311
|
|
electromagnetism
|
Calculating $dB/dt$ from velocity
|
https://physics.stackexchange.com/questions/88989/calculating-db-dt-from-velocity
|
<p>I have recently carried out an experiment to verify Faradays law for a falling magnet. My starting point was to keep both the area of the coil and the number of turns constant whilst changing the velocity (the different velocities were obtained by dropping from different heights).What would be good is if a graph of emf induced vs. $dB/dt$ could be plotted so that the gradient of the graph will be equal to the product of the area and number of turns. From $e = -nA(dB/dt)$</p>
<p>In short, is there an equation to change velocity to $dB/dt$?</p>
|
<p>The problem is that the magnetic field around a bar magnet aren't uniform, so different parts of your coil will experience different values of $B$. If the magnetic is small enough, you can approximate it's field by a <a href="http://en.wikipedia.org/wiki/Magnetic_dipole" rel="nofollow">dipole</a>,</p>
<p>$\mathbf{B}({\mathbf{r}})=\frac{\mu_{0}}{4\pi}\left(\frac{3\mathbf{r}(\mathbf{m}\cdot\mathbf{r})}{r^{5}}-\frac{{\mathbf{m}}}{r^{3}}\right)$</p>
<p>(here $\mathbf{m}$ is a constant vector equal to the dipole moment).</p>
<p>Then, supposing your coil is very flat (not really a solenoid), you can calculate what $\frac{\partial\mathbf{B}}{\partial t}$ is when $\mathbf{m} \cdot \mathbf{r} = 0$ , i.e., at the moment the magnet passes through the coil like a basketball falling through a hoop. Of course, $\mathbf{m}\cdot \dot{ \mathbf{r}} \neq 0$. The peak EMF should then be somewhat close to what you would predict from experiment.</p>
<p>A lot of early experiments had to be really clever to isolate exactly what was going on. It really depends on what apparatuses are available to you.</p>
| 312
|
electromagnetism
|
Help Understanding Equation for Characteristic Time of Induced Magnetic Field
|
https://physics.stackexchange.com/questions/90371/help-understanding-equation-for-characteristic-time-of-induced-magnetic-field
|
<p>I am reading <a href="http://books.google.co.uk/books?id=Quv8awSN-RYC&printsec=frontcover&dq=magnetic%20techniques%20for%20the%20treatment%20of%20materials%20pdf&hl=en&sa=X&ei=ePKuUu2lH8yThQfs64D4CA&ved=0CDIQ6AEwAA#v=onepage&q&f=false" rel="nofollow">this</a> book, the part in particular about Eddy-current separation starting at Page 246, in it there is an equation for calculating the "characteristic time with which the induced magnetic field decays in the particle".</p>
<p>$$
\tau = \mu_0\sigma_psb^2
$$</p>
<p>Where:</p>
<ul>
<li>$\mu_0$ is the permeability of free space</li>
<li>$\sigma_p$ is the specific electric conductivity of a particle [S/m]</li>
<li>$s$ is the shape factor of the particle</li>
<li>$b$ is the particle radius</li>
</ul>
<p>I am hoping someone can reflect some light on what $s$ is and how to calculate it, "shape factor" isn't very descriptive and I cannot find anywhere in the book describing it in further detail.</p>
<p>This shape factor is also used in the force equations that I am intersted in...</p>
<p>Another part of the equation I am unclear on is $b$, what if the particle is a cube or cuboid, what should the radius be then?</p>
| 313
|
|
electromagnetism
|
Speed of Magnetic Signal over Large Distance
|
https://physics.stackexchange.com/questions/89701/speed-of-magnetic-signal-over-large-distance
|
<p>If I had a very strong magnet on Earth and a very sensitive compass on Mars (just using planets to illustrate large distance), how long would the compass take to notice if I turned the magnet 180deg? I assume it can't beat the speed of light. Do the outer reaches of the magnetic field move as a rigid body with the turning magnet or do they lag in motion?</p>
|
<p>The core of your question (whether the magnetic field moves as a rigid body at long distances) has a fairly straight-forward answer: no, it does not. Changes in a magnetic field propagate at the speed of light. This is a fundamental consequence of <a href="http://en.wikipedia.org/wiki/Maxwell%27s_equations#Vacuum_equations.2C_electromagnetic_waves_and_speed_of_light" rel="nofollow">Maxwell's Equations</a>.</p>
| 314
|
electromagnetism
|
Lenz's Law and Eddy Currents
|
https://physics.stackexchange.com/questions/90739/lenzs-law-and-eddy-currents
|
<p>You can determine the direction of eddy currents according to Lenz's law.
E.g. If a metal sheet is losing flux into the page, it will experience induced eddy currents in a clockwise direction to replace it. </p>
<p>However, this doesn't make sense to me logically. :S Consider a straight current-carrying wire. It has equal flux on either side of it's length. Now warp it into a circle (to simulate eddy currents). It should have equal flux on both outside and inside the the circle. Thus, in the above example, the eddy currents ARE NOT replacing the lost flux into the page, as it is also producing EQUAL flux out of the page (which "cancels" the other flux into the page).</p>
<p>Am I misunderstanding something about Lenz's Law/Eddy currents? :S</p>
| 315
|
|
electromagnetism
|
Why doesn't a particle's velocity effect the strength exerted on it by an electric field?
|
https://physics.stackexchange.com/questions/92170/why-doesnt-a-particles-velocity-effect-the-strength-exerted-on-it-by-an-electr
|
<p>Here is what I know:</p>
<p>$F = E q = m a$ </p>
<p>so $a = \frac{E q}{ m}$</p>
<p>and we can increase the acceleration ($a$) of a particle in an electric field ($E$) by either decreasing its mass ($m$) or increasing its charge ($q$).</p>
<p>Here is where I am confused:</p>
<p>$a = \frac{v}{ t}$, thus $E = \frac{m v }{q t}$</p>
<p>Yet, increasing the velocity ($v$) of a particle traveling through an electric field has no effect on the electric field. Why not?</p>
|
<p>Recall that in this scenario the electric field is some given value - probably a constant - established as a parameter of the problem. If you've been given $E$, you can solve for $a$ (and therefore $v$). Or, having been given $a$, you can compute the necessary $E$ to account for such an acceleration. However, they are not independent in the sense that they cannot both be free parameters of the problem.</p>
<p>In other words, when you change $a$ you are not increasing or decreasing the affect of $E$ on the particle, you are merely recalculating the mathematically necessary electric field for this new kinematical problem.</p>
| 316
|
electromagnetism
|
AC Electromagnets
|
https://physics.stackexchange.com/questions/93356/ac-electromagnets
|
<p>Could someone help explain the uses of AC <a href="http://en.wikipedia.org/wiki/Electromagnet" rel="nofollow">electromagnets</a>. Wherever I look it says that DC electromagnets create stronger magnetic fields. I understand why AC electromagnets could be used in transformers but why use them in motors for example? Power stations also use electromagnets in the generator to create the magnetic field, would these be AC or DC? Finally, which type of current would electromagnets in household motors use e.g. food processors, blenders etc?</p>
|
<p>In motors AC electromagnets can be useful as after half the rotation the magnetic field could be reversed and a permanent magnet placed on rotor could then continue to make a complete turn which would have stopped if the magnetic field was constant. But most of the motors rely on a constant magnetic field in which a coil/rotor of electrical winding is rotated.</p>
<p>As most of the genrators/alternators/dynamos depend on constant magnetic field, I would say that they work on DC electromagents which were a replacement over permanent magnets used in earlier designs.</p>
<p>Household motors usually work by rotating windings of electrical wires in constant magnetic field, so i would say that these contain DC electromagnets.</p>
| 317
|
electromagnetism
|
A simple way to calculate the potential electrical output of a magnet.
|
https://physics.stackexchange.com/questions/93557/a-simple-way-to-calculate-the-potential-electrical-output-of-a-magnet
|
<p>I am a software engineer. This I understand very well. I am attempting to build a prototype with arduino circuits. This I know a little less about. Part of my project requires a electro-magnetic generator. This I know nothing about. </p>
<p>I looked up Gauss's law and embarrassingly the math is a bit above my pay grade. All I want is a simple way to calculate the potential electrical output of a magnet. For example if I use a refrigerator magnet to power my generator its going to do much less then a magnet from a subwolfer. I know that there is also a direct relation to the spacing of the copper wiring that it's passing through but for this example please assume that all other external variables are a constant. How do I determine which magnet would best suit my needs?</p>
|
<p>It is not actually <a href="http://en.wikipedia.org/wiki/Gauss%27s_law" rel="nofollow">Guass' Law</a> (which concerns itself with electric fields) you want, but <a href="http://en.wikipedia.org/wiki/Faraday%27s_law_of_induction" rel="nofollow">Faraday's Law of Induction</a>.</p>
<p>Even then, the <em>only</em> way to answer this question is to <em>measure</em> the magnetic flux available to you. If you are going to use a permanent magnet, then you need a magnetometer as these things are not labeled for strength (indeed it would be very hard to design a useful unit of the generic strength of a permanent magnet because the field geometry is highly variable from example to example).</p>
<p>Unless you just happen to have access to a quality magnetometer, it is almost certainly easier to simply prototype the device and <em>measure</em> the output (which is something you can do with an oscilloscope and a multimeter).</p>
<p>If you do have a magnetometer, you can map the field; guestimate the coils design and relative motion; and estimate the current as a function of time. This doesn't let you off prototyping of course, it just means a good chance of few prototype cycles.</p>
| 318
|
electromagnetism
|
The force exerted on a magnetic dipole $m$ in magnetic field $B$
|
https://physics.stackexchange.com/questions/94605/the-force-exerted-on-a-magnetic-dipole-m-in-magnetic-field-b
|
<p>How can one prove that:</p>
<p>The force exerted on a magnetic dipole $m$ in magnetic field $B$, in addition to $F=\nabla(m\cdot B)$, can be expressed by </p>
<p>$$F=(m\times \nabla)\times B.$$</p>
|
<p>In view of a known identity concerning the cross product:
$$(m\times \nabla)\times B = \nabla (m\cdot B)- m (\nabla \cdot B) = \nabla (m\cdot B) = F$$
because $\nabla \cdot B=0$.</p>
| 319
|
electromagnetism
|
Why aren't all conductors always charged?
|
https://physics.stackexchange.com/questions/96169/why-arent-all-conductors-always-charged
|
<p>If you place a conductor beside an insulator, the insulator will become negatively charged and the conductor will become positively charged. Air is an insulator. So why don't all conductors placed in air automatically become positively charged and the air around it become negatively charged?</p>
<p>See, for example, the "Cause" section on the <a href="http://en.wikipedia.org/wiki/Triboelectric_effect#Cause" rel="nofollow">Triboelectric effect Wikipedia page.</a></p>
|
<p>The triboelectric effect isn't a result of placing two materials next to each other, it comes as a result of rubbing them together. This is important because the two materials do not naturally want to become charged, you have to add energy, typically in the form of the friction that comes from rubbing. This overcomes the activation energy, so to speak, that is required for the electrons to jump from one material to the other. Everyday interactions between objects and air molecules are not energetic enough to cause this effect.</p>
| 320
|
electromagnetism
|
Determining the minimum pull force of a magnet required to hold it in place
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https://physics.stackexchange.com/questions/96579/determining-the-minimum-pull-force-of-a-magnet-required-to-hold-it-in-place
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<p>If I place a magnet on the underside of a metal object so that it is held there by its attraction to the metal, how can I determine the minimum pull force the magnet needs to have in order to keep it attracted to the metal and not fall off? Would the required pull force be equivalent to the magnet's weight?</p>
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<p>What makes a magnet attract to a surface? What makes a magnet <I>strongly</I> attract to a surface at a given field energy? Look at the pole structure of a refrigerator magnet, then "divergence" of a field. A simple Fe-Nd-B N45 magnet will pull hundreds of times its weight. But wait!</p>
<p><a href="http://images.tutorvista.com/cms/images/38/magnetic-materials-list.jpg" rel="nofollow">http://images.tutorvista.com/cms/images/38/magnetic-materials-list.jpg</a> <BR>
Composition matters <BR>
<a href="https://www.ameslab.gov/files/graphMagnets.png" rel="nofollow">https://www.ameslab.gov/files/graphMagnets.png</a> <BR>
a lot. <BR>
<a href="http://www.kjmagnetics.com/images/blog/fridgemagnet.greenfilm.jpg" rel="nofollow">http://www.kjmagnetics.com/images/blog/fridgemagnet.greenfilm.jpg</a> <BR>
Refrigerator magnet poles <BR>
<a href="http://www.kjmagnetics.com/images/blog/halbachvsalt.png" rel="nofollow">http://www.kjmagnetics.com/images/blog/halbachvsalt.png</a> <BR>
You can do better.</p>
| 321
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electromagnetism
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induced emf when a wire or coil travel through a magnetic field
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https://physics.stackexchange.com/questions/98463/induced-emf-when-a-wire-or-coil-travel-through-a-magnetic-field
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<p>I've recently learned that If you move a conductor through a magnetic field, an emf is induced across the ends of the conductor: </p>
<p>$E = BLv $ </p>
<p>I've also been told that the same is true for a coil but that the equation is: </p>
<p>$E = BLvN$</p>
<p>And that for a coil, there is only an emf induced when the current is entering or leaving the magnetic field as once the coil is inside, there is no change in magnetic flux linkage. I've been told that this is not the case with just a wire moving through a magnetic field, but I haven't been told why and I can't seem to figure it out. </p>
<p>I've thought about it, and I think it perhaps has something to do with the area covered by the wire moving at speed $v$ per unit time compared to the area covered by a coil moving at the same speed $v$ in the same unit time. </p>
<p>Unfortunately, I can't quite work it out. </p>
<p>Thank you. </p>
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<p>You're almost right. The induced electromotive force due to external magnetic field is proportional to rate of change of magnetic flux through the circuit. In case of the coil moving in a uniform field, the flux through it (proportional to the number of turns) does not change in time. In case of the wire, the flux changes, since the circuit's area grows or shrinks.</p>
| 322
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electromagnetism
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Question about laws of conservation in electrodynamics
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https://physics.stackexchange.com/questions/32335/question-about-laws-of-conservation-in-electrodynamics
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<p>One way of deriving laws of energy, impulse and angular momentum of electromagnetic field conservation is following:</p>
<ol>
<li><p>Introduce two values below:
$$
\mathbf P = \frac{c}{4 \pi}[\mathbf E \times \mathbf B], \quad W = \frac{1}{8 \pi}(\mathbf E^{2} + \mathbf B^{2}).
$$</p></li>
<li><p>Using Maxwell's equations and time derivative of these values, get the laws.</p></li>
</ol>
<p>But how to argue an introduce of values from it. 1? Can Noether's theorem help to argue it?</p>
| 323
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electromagnetism
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How does one prove:$\nabla(\vec{\mu_m}\cdot\vec{B})\cdot\vec{dr}=0$?
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https://physics.stackexchange.com/questions/34056/how-does-one-prove-nabla-vec-mu-m-cdot-vecb-cdot-vecdr-0
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<p>Work done by a magnetic force(even over an infinitesimally short displacement)=0</p>
<p><em>Net Force</em> on a current loop in an external magnetic field is given by: $$\vec{F}=\nabla(\vec{\mu_m } \cdot \vec{B})$$</p>
<p>How does one prove: $$dW=\nabla(\vec{\mu_m} \cdot \vec{B})\cdot\vec{dr}=0$$</p>
<p>$\vec{\mu_m}$: Magnetic Moment of the current loop.</p>
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<p>I had the same problem a few weeks ago. It is clear that an electromagnet does produce work, to solve the paradox you need to take into account the generator that runs the current and balance the energy ( Griffiths p211, introduction to electrodynamics).
In other words, if the magnetic moment is created by a current, the system does produce a net work, but deeper analysis shows that it is not created by the magnetic force (qvxB) but by the generator.</p>
| 324
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electromagnetism
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What law of electro-magnetics explains this?
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https://physics.stackexchange.com/questions/34265/what-law-of-electro-magnetics-explains-this
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<p>I took my son to a science museum where they had a solenoid oriented vertically with a plastic cylinder passing through the solenoid. An employee dropped an aluminum ring over the top of the cylinder when there was no current going through the solenoid. Then they turned on the current going through the solenoid and they aluminum ring went flying up and off the top of the solenoid. What law of electro-magnetics causes the force on the aluminum ring?</p>
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<p>I'll start this with <strong><a href="http://en.wikipedia.org/wiki/Right_hand_grip_rule#Direction_associated_with_a_rotation" rel="nofollow">Right Hand Grip rule</a></strong> for solenoids...</p>
<p>"The coil (solenoid) is held in the right hand so that the fingers point the direction of current through the windings. Then, the extended thumb points the direction of magnetic field". (which would be along the axis of the coil)</p>
<p>The higher the current, the more the magnetic field would be produced... For your example, let us assume the aluminium ring as a circular coil. When the uniform magnetic field is produced, there is a change in magnetic flux (such as this increase in magnetic field) along the axis of the ring, According to <strong><a href="http://en.wikipedia.org/wiki/Faraday%27s_law_of_induction" rel="nofollow">Faraday's law</a></strong>, induced current flows through the ring whose direction is given by <strong><a href="http://en.wikipedia.org/wiki/Lenz_law" rel="nofollow">Lenz's law</a></strong>. This induced current in the ring flows in a direction such that it opposes the magnetic field in the solenoid (the one which <em>actually</em> produces it). (But, the magnitude of induced magnetic field is always lesser than the field in the solenoid). Anyways, there's a repulsion. With the maximum repulsive force produced, the ring is thrown off from the solenoid. This force always depends on the magnitude of $B$ in the solenoid.</p>
| 325
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electromagnetism
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Effects of EMP in superconductor-based devices/equipment?
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https://physics.stackexchange.com/questions/36145/effects-of-emp-in-superconductor-based-devices-equipment
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<p>What are possible effects of electromagnetic pulse / EMP on superconductor-based devices/equimpent/transportation?</p>
<p>Are they resilient or more sensitive to EMP?</p>
| 326
|
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electromagnetism
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Do magnets work in outer space?
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https://physics.stackexchange.com/questions/37920/do-magnets-work-in-outer-space
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<p>Is there any media where magnet lose its property?</p>
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<p>Magnets work perfectly in the vacuum – and in the absence of a gravitational field. They don't depend on any "environment" or "medium". And the electromagnetic force is independent of gravity, too.</p>
<p>In analogy with sound, some 19th century physicists thought that electromagnetic phenomena required a medium called the "luminiferous [=light-carrying] aether" which would play the same role as the "air" for sound. They thought it was a very unusual kind of a material that had to penetrate everything and have other strange properties. That didn't stop them from constructing a mechanical model of the aether out of wheels and gears.</p>
<p>However, it was demonstrated by the Michelson-Morley experiment that there was no aether wind. Also, mostly independently, Einstein's special theory of relativity demonstrated that the "aether" can't be composed of any building blocks that could be "at rest". There isn't any preferred inertial system so there can't be a single aether's system (unlike the case of air that has a preferred frame in which its velocity, the velocity of wind, is zero), either.</p>
<p>From any ambitious point of view, the aether doesn't exist. One may say that the vacuum itself and the fields that always occupy it (including the Higgs field) are a form of the aether except that it is a completely different aether than the luminiferous aether envisioned by the 19th century physicists.</p>
<p>At every point of space and at each moment of time, i.e. for every $(t,x,y,z)$, there exist one electric vector $\vec E$ and one magnetic vector $\vec B$ remembering the field. These fields are inseparable from the vacuum and are responsible for all the electromagnetic phenomena including the electromagnetic waves such as light. According to quantum mechanics, $\vec E$ and $\vec B$ are operators, not just ordinary numbers, that generally don't commute with each other, and that's why the electromagnetic waves' energy comes in packets, the photons.</p>
| 327
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electromagnetism
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Is it possible to see domains in a metal/magnet under microscope
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https://physics.stackexchange.com/questions/39024/is-it-possible-to-see-domains-in-a-metal-magnet-under-microscope
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<p>I was just curious to know that is it possible to see domains of a magent/metal in motion on being magnetized, under microscope. If there is someone who has access to microscopes, can help me out on this. I was thinking of buying one.
Question is lil stupid but I am putting it here because I know no better place !!</p>
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<p>You actually have a couple of options without have to resort to a microscope. One of the most interesting is <a href="http://en.wikipedia.org/wiki/Magnetic_field_viewing_film" rel="nofollow">Magnetic Field Viewing Film</a> which should allow one to view magnetic fields easily over large domains. The film is <a href="http://www.youtube.com/watch?v=AMggixEu0q0&feature=related" rel="nofollow">pretty responsive</a>, so it would likely allow one to view domain formation in larger materials. </p>
<p>An interesting way to viewing magnetic effects is to use an <a href="http://www.youtube.com/watch?v=bliKAQMq5Yc" rel="nofollow">old computer crt and some magnets</a>. If you wanted to get down to the <a href="http://www.psi.ch/media/moving-monopoles-caught-on-camera" rel="nofollow">nano scale, that is possible too,</a> but probably out of most people's budget. The <a href="http://www.arborsci.com/magnetic-field-viewer-film-9in-x-7-2in" rel="nofollow">viewing film is definitely budget friendly</a>.</p>
| 328
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electromagnetism
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Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid. Why?
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https://physics.stackexchange.com/questions/38067/magnetic-field-lines-can-be-entirely-confined-within-the-core-of-a-toroid-but-n
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<p>I need a full explantation for this concept.</p>
<p>Magnetic field lines can be entirely confined within the core of a toroid, but not within a straight solenoid.</p>
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<p>This is a solenoid and its magnetic field lines.</p>
<p><img src="https://i.sstatic.net/R11ij.png" alt="solenoid"></p>
<p>This is a <a href="http://en.wikipedia.org/wiki/Toroidal_inductors_and_transformers" rel="nofollow noreferrer">toroid</a> and its magnetic field lines </p>
<p><img src="https://i.sstatic.net/YGZjF.jpg" alt="toroid"></p>
<p>A solenoid by construction has two magnetic poles at the edges when current is flowing through its windings.</p>
<p>One can think of a toroid as a solenoid that has been curved and joined so no poles are open. A toroid can have magnetic fields outside its geometrical boundary according to the way the currents are flowing, if there is a circumferential current that has not been neutralized. Once neutralized there is magnetic field only inside.( as described in the link given above). A neutralizing design is shown below.</p>
<p><img src="https://i.sstatic.net/3nkTa.jpg" alt="neutralizing circumferential current"> </p>
<p>In contrast a solenoid will always have two open poles.</p>
| 329
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electromagnetism
|
Bend or concentrate magnetic field?
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https://physics.stackexchange.com/questions/41011/bend-or-concentrate-magnetic-field
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<p>What are the ways to modify the form of magnetic field from the permanent magnet? For example I have a permanent neodymium magnet. Its magnetic field is distributed at large volume around the magnet, with decreasing strength at larger distances from the magnet. I'd like to make it concentrated in very small distances around this magnet, and to have no field or largely less strong field beyond some given distance from the magnet.
Is it possible at all, and if yes, how?</p>
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<p>To modify the distribution of magnetic field (irrespectively of the source - PMs or coils), you need some material with non-unit magnetic permeability, such as steel. These materials 'concentrate' the field lines, pulling them in away from the surrounding air (which has a permeability of very nearly 1). So to shield the outside world, you'd need to create a 'circuit' of permeable material (let's say steel), which channels the magnetic flux from the north pole of your magnet to its south pole. If you leave a little gap in one part of your circuit, the field will be enhanced in the gap.</p>
| 330
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electromagnetism
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Proving the existence of the magnetic potential
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https://physics.stackexchange.com/questions/41210/proving-the-existence-of-the-magnetic-potential
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<p>Suppose $\vec{B}$ is a differentiable vector field defined everywhere such that $\nabla\cdot \vec{B}=0$. Define $\vec{A}$ by the integral
$$A_1=\int_0^1 \lambda(xB_2(\lambda x,\lambda y,\lambda z)- yB_3(\lambda x,\lambda y,\lambda z)) d\lambda$$
Together with its two cyclic permutations for $A_2,A_3$</p>
<p>I'm trying to work out two things here:</p>
<p>$1.$ What is $\frac{d}{d\lambda}B_i(\lambda x,\lambda y,\lambda z)$</p>
<p>$2.$ How we can use $1.$ to determine $\frac{\partial A_2}{\partial x}-\frac{\partial A_1}{\partial y}=B_3$</p>
<p>From this we can deduce the existance of the magnetic potential by extending $2.$? This is what I have so far:</p>
<p>Is $\frac{d}{d\lambda}B_i(\lambda x,\lambda y,\lambda z)=(x,y,x) \cdot \nabla B_i$?
And can we bring the partial derivative on $A_i$ inside the integral? I have proceeded along these lines but have not found a way to substitute.</p>
<p>Any help would be greatly appreciated!</p>
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<p>You are right in both your specific questions: your $\lambda$ derivative is right and the partial derivatives can go inside the integral.</p>
<p>You have, however, one crucial mistake in your original formula, which should read
$$A_1=\int_0^1 \lambda(\quad z\quad B_2(\lambda x,\lambda y,\lambda z)- yB_3(\lambda x,\lambda y,\lambda z)) d\lambda$$
- i.e., replacing $x$ by $z$. This is needed to make "permutational sense": it now reads like "(1)=(3)(2)-(2)(3)", instead of "(1)=(1)(2)-(2)(3)", which is clearly wrong.</p>
<p>Plugging this and the cyclically-permuted expression for $A_2$ into the curl then gives (after applying $\nabla\cdot\mathbf{B}=0$, your formula for the $\lambda$ derivative, and an integration by parts) the desired
$$\nabla\times\mathbf{A}=\mathbf{B}.$$</p>
| 331
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electromagnetism
|
how do you destroy magnetic field - demagnetize?
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https://physics.stackexchange.com/questions/43550/how-do-you-destroy-magnetic-field-demagnetize
|
<p>And what happens with the magnetic field of a star that goes supernova? The magnetic radiation is scattered through the cosmos? Each particle will go away with its own magnetic radiation?</p>
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<p>In a star the magnetic field isn't a "thing" in it's own right, it's the result of motion of charged particles in the star. When a supernova goes bang the matter ejected will have some peculiar motion, i.e. motion relative to the overall outward flow, and as a result there will be magnetic fields in the ejected matter. These fields will get gradually weaker as the matter expands and dilutes, and will eventually merge with the magnetic fields in the host galaxy.</p>
<p>I had a quick Google and found <a href="http://arxiv.org/abs/1104.4047" rel="nofollow">this review</a> discussing the magnetic fields in supernova remnants. This would be a good start if you're interested in the area.</p>
| 332
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electromagnetism
|
What happens when a ferromagnetic object encounters a field too strong for it?
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https://physics.stackexchange.com/questions/43584/what-happens-when-a-ferromagnetic-object-encounters-a-field-too-strong-for-it
|
<p>Just thinking aloud ... It is possible for a star to grow so large it collapses under it's own gravity. Along a parallel path (so to speak) when a conductor carries a current too large for it to sustain, it burns out.</p>
<p>Can a ferromagnetic object be imparted a field too strong for it? What happens then? </p>
<p>I would begin to guess the dipoles within the magnet can only be magnetized so far; a stronger field wouldn't have an effect. But surely the same argument could have applied to a wire conductor?</p>
|
<p>In a ferromagnet the field is the result of the aligned elecron spins. Once the spins are all aligned the field cannot be made any stronger. You could put a piece of iron in an external field more powerful than the iron could generate on it's own, but it would no especially surprising effect, and when you withdrew the iron from the field you'd just be left with magnetised iron.</p>
<p>There isn't an analogy with conduction in a wire. The heat generation in a conducting wire is the result of conduction electrons scattering off phonons and defects in the crystal structure of the wire and transferring energy to it. By making the wie a perfect crystal and lowering the temperature to absolute zero you can make the wire conduct arbitrarily large currents. Actually I'm not sure the resistance falls to zero at absolute zero (NB I'm ignoring superconductivity here) but it gets very small.</p>
| 333
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electromagnetism
|
Lorentz model and energy exchange
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https://physics.stackexchange.com/questions/43878/lorentz-model-and-energy-exchange
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<p>The Lorentz model, describing the electron of the atom as an harmonic oscillator forced by an oscillating electric field $\vec{E}$, shows that the dipole moment $\vec{D}$ obeys the following equation in the stationary regime:
$$\vec{D}=\frac{q^2}{2m\omega_0}\frac{\vec{E}}{\omega_0-\omega-i\gamma_d}$$
where $\omega_0$ and $m$ are the natural frequency and mass of the electron, $\gamma_d$ is the dissipation introduced by hand to take account of the energy emitted by the electron when accelerated (and also its collisions).</p>
<p>My trouble is that one finds that the energy transmitted by the electron to the field by unit of time, given by:
$$P=-\vec{E}.\frac{d\vec{D}}{dt}$$
is always negative, implying that the electron cannot "feed" the field, however how much it is radiating. Can it be attributed to the fact that the electron energy is in first place coming from the field, and thus, our electron cannot give more than it receives? In what cases does the electron sources the background field? </p>
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<p>In the driven oscillator approximation you gave above the external field is so strong that the electron motion "follows" $\vec{E}(t)$ (i.e. it moves with the same phase $+\pi$ = the opposite phase). It is possible when the initial (before switching $\vec{E}(t)$ on) oscillations are "absent" or their contribution is small in the solution $\vec{r}_e(t)$. The incident field "feeding" (amplifying) is possible when in the solution $\vec{r}_e(t)$ there is a velocity term <em>along</em> $\vec{E}$ that is not caused by the force $q\vec{E}$.</p>
<p>The radiated field $\vec{E}_{rad}$ is added to the external field $\vec{E}$ - they come in a superposition and get into equations of motion of the <em>other</em> charges.</p>
<p>If you speak of a "self-action", it boils down physically to $\gamma_d$.</p>
| 334
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electromagnetism
|
Will a magnet loose too much strength it cut with a hacksaw?
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https://physics.stackexchange.com/questions/47574/will-a-magnet-loose-too-much-strength-it-cut-with-a-hacksaw
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<p>Can a magnet be cut with a hacksaw, by hand, without losing its essential characteristic of being magnetic? I know i would have to be very careful with heat but I don't know how careful</p>
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<p>This would likely mess up the polarity of the magnet near the area where you are cutting. It would depend on several factors, namely how big the magnet is compared to the hacksaw blade (the bigger the magnet the more likely you are to preserve it) and what the magnet is made out of, for instance if the magnet is steel then it would definitely deteriorate the field significantly, but if it were iron and it would not be as degrading because the atoms hold their polarization better. It would have no effect on an electro-magnet. </p>
| 335
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electromagnetism
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How is the polarity of a magnet decided (before its creation)?
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https://physics.stackexchange.com/questions/48130/how-is-the-polarity-of-a-magnet-decided-before-its-creation
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<p>This is not how to determine a magnet after it's created, but rather before it is created. How is the polarity of a magnet created (why does one side go this way and the other go the opposite)? Is it randomly decided based on forces acting on the soon-to-be magnet, or is it aligned with the Earth’s magnetic fields?</p>
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<p>The alignment is basically random, but it can be tipped in favor of one direction or another based on the ambient magnetic field of the Earth. In fact, when molten ferromagnetic material upwells from deep within the Earth, at, say, the Mid-Atlantic Ridge (see <a href="http://en.wikipedia.org/wiki/Seafloor_spreading#Debate_and_search_for_mechanism" rel="nofollow">this wiki article</a>), it will align with the current magnetic field to the extent that we can measure a net magnetization in the solidified rock. In fact, we can trace the history of the Earth's magnetic field in this way, and it shows that there have been numerous <a href="http://en.wikipedia.org/wiki/Geomagnetic_reversal" rel="nofollow">reversals</a> of North and South over the ages.</p>
| 336
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electromagnetism
|
Does the shielding around a solenoid actuator affect the force on the plunger?
|
https://physics.stackexchange.com/questions/48427/does-the-shielding-around-a-solenoid-actuator-affect-the-force-on-the-plunger
|
<p>I'm looking to minimize the weight of some solenoid actuators I want to build, so I'm trying to better understand the effects of the shielding / frame commonly found on solenoid actuators.</p>
<p>My understanding is the force exerted on the plunger is essentially only dependent on the cross sectional area of the plunger, the permeability of the plunger, and the flux density within the plunger. If that is correct, then it seems that the shielding doesn't really affect the plunger, only the reluctance (and thus the inductance) of the overall system.</p>
<p>If I were to remove the shielding, it seems that I could expect a very low inductance when there was no core, so I would have to be careful not to burn out my coil, but other than that it seems like the system should be largely unaffected. Am I understanding things correctly?</p>
<hr>
<p>It seems like in practice there is a lot of effort taken to minimize the air gap in the system, and random forum posts state that if you don't do this your solenoid won't be as strong - however I have yet to see any math or reason to back this up or to explain how to figure out how much stronger one system is over another.</p>
<p>I'd be very grateful if someone could help clear this up!</p>
|
<p>I spent some time learning <a href="http://femm.info/" rel="nofollow noreferrer">FEMM</a> to figure this out.</p>
<p>The answer is yes, the case does make a notable difference in the pull on your plunger.</p>
<p>Without a case the pull on my sample plunger was 0.036 N</p>
<p><img src="https://i.sstatic.net/Jy9i3.png" alt="No case"></p>
<p>And with a case the pull was 0.047 N</p>
<p><img src="https://i.sstatic.net/QqMSg.png" alt="With case"></p>
<hr>
<p>Edit: Playing with the simulations more, the effects of the case are more drastic as the plunger moves into the cavity. As the plunger approaches the other side, you'll start seeing at least an order of magnitude of difference, so not only is the case important but it is <em>very</em> important.</p>
| 337
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electromagnetism
|
4-velocity and electromagnetic fields
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https://physics.stackexchange.com/questions/49736/4-velocity-and-electromagnetic-fields
|
<p>Can anyone see a reason for $$\left(1+{U_\rho U^\rho\over c^2}\right)\left(U_\nu{d^2 U^\nu\over d\tau^2}\right)=0$$?</p>
<p>Here $U^\rho$ is the 4-velocity for a particle and $\tau$ the proper time. The context is for a particle moving in an electromagnetic field.</p>
<p>I believe it may be useful to introduce the antisymmetric tensor $F_{\mu\nu}$ -- the electromagnetic field tensor. </p>
|
<p>The left parentheses are equal to zero due to $U_{\rho}U^{\rho}=-c^2$. This is true for timelike vectors in the (-1,1,1,1) signature.</p>
| 338
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electromagnetism
|
How to count magnetic repulsion
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https://physics.stackexchange.com/questions/47612/how-to-count-magnetic-repulsion
|
<p>I have two equal flat round magnets. I know amount of force $F$ which attracts iron objects to one of them and geometric characteristics of magnets. I want to fix first of magnet and some additional mass in the air by second magnet. To do so, I am going to orient magnets so that second magnet repulse first and additional mass. But I need to establish dependence between distance between magnets and value of additional mass this construction can held. How to gain this dependence?</p>
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<p>To a good approximation, normal magnets can be treated as dipole magnets, in which case the force between them can be found in <a href="http://en.wikipedia.org/wiki/Magnetic_dipole#Forces_between_two_magnetic_dipoles" rel="nofollow">this wikipedia article</a>. To avoid link-only answers, here it is:
$$\mathbf{F} = \dfrac{3 \mu_0}{4 \pi r^5}\left[(\mathbf{m}_1\cdot\mathbf{r})\mathbf{m}_2 + (\mathbf{m}_2\cdot\mathbf{r})\mathbf{m}_1 + (\mathbf{m}_1\cdot\mathbf{m}_2)\mathbf{r} - \dfrac{5(\mathbf{m}_1\cdot\mathbf{r})(\mathbf{m}_2\cdot\mathbf{r})}{r^2}\mathbf{r}\right].$$</p>
<p>However, be warned that while equilibrium levitation is possible using magnets, <em>stable</em> equilibria are impossible due to <a href="http://en.wikipedia.org/wiki/Earnshaw%27s_theorem" rel="nofollow">Earnshaw's theorem</a>. Magnetic levitation is indeed possible but you need fancier schemes than just two magnets. The <a href="http://en.wikipedia.org/wiki/Magnetic_levitation" rel="nofollow">wikipedia article on it</a> is a good starting point.</p>
| 339
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electromagnetism
|
Relationship between current through a motor and it's load
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https://physics.stackexchange.com/questions/51332/relationship-between-current-through-a-motor-and-its-load
|
<p>When a motor, connected to a battery that has a constant voltage, spins without a load it's speed is higher than with load. I'm told that because of back emfs the current is very small when there's no load because of the higher speed. And so when there is a load the back emf is less as the motor spins slower, and so the current is higher.</p>
<p>What is a back emf and what is the relationship between that and the speed of the motor (well the coils inside the motor)?</p>
<p>Is the output energy of the motor constant, whether it has a load or not? (ignoring friction and electrical resistance) because I don't understand how the current can be higher when the motor is (or seems to be) doing more work due to the load?</p>
|
<p>In a DC motor, when the armature rotates its coils cut the magnetic and induce a voltage in the coils. This voltage is of opposite polarity of the voltage that is powering the motor (the battery) and is called the back emf. It is modeled as a voltage source that is proportional to the speed of the motor times a constant. The faster the motor is rotating the higher the back emf. </p>
<p>The power of the motor is not constant and neither is the energy. The power into a motor is just the voltage measured at the lead wires times the current in the lead wires (P=V*I). The output power of the motor is just the speed of the motor times the torque. The power will be zero when there is zero torque and it will be zero when there is zero speed. In between those two points, the power will increase, peak, and then return to zero. In general, on a DC motor, speed is proportional to voltage and torque is proportional to current. As your load increases and you need more torque, the motor will draw more current. If you need to speed your motor up or slow it down, you need to raise or lower your voltage. </p>
| 340
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electromagnetism
|
Current induced when dropping a magnet through a coil
|
https://physics.stackexchange.com/questions/51341/current-induced-when-dropping-a-magnet-through-a-coil
|
<p>When graphing the induced current in a coil while a magnet is dropped through it why is the total area equal to 0?
The area represents the charge in the coil but why must the resultant flow of charge be 0?</p>
|
<p>The EMF induced in the coil is given <a href="http://en.wikipedia.org/wiki/Faraday%27s_law_of_induction" rel="nofollow">by</a>:</p>
<p>$$
\varepsilon=-N\frac{d\Phi_B}{dt}
$$</p>
<p>where $d\Phi_B$ is the magnetic flux through the coil and $N$ is the number of windings. The current through the coil is given by Ohm's law:</p>
<p>$$
I=\frac{\varepsilon}{R}=-\frac{N}{R}\frac{d\Phi_B}{dt}
$$</p>
<p>where $R$ is the resistance of the coil. The total charge $C$ having gone through a conductor over a period of time is the time integral of the current:</p>
<p>$$
C=\int^{t_2}_{t_1}Idt=-\frac{N}{R}\int^{t_2}_{t_1}\frac{d\Phi_B}{dt}dt=-\frac{N}{R}\left(\Phi_B(t_2)-\Phi_B(t_1)\right)
$$</p>
<p>Assuming $\Phi_B$ has roughly the same value when the magnet has gone through the coil, $t_2$, as when it was dropped, $t_1$, this will be zero.</p>
| 341
|
electromagnetism
|
How do transformers work?
|
https://physics.stackexchange.com/questions/51936/how-do-transformers-work
|
<p>A <a href="http://en.wikipedia.org/wiki/Transformer" rel="nofollow">transformer</a> is basically a primary inductor connected to a voltage $U_P$ which you want to transform. You also have an iron rod and a secondary inductor. So when there is a current $I_P$ the iron rods becomes magnetic. When you connect the primary inductor to AC, that means that you'll have a changing current, which causes a change in flux which causes induction. My question is, is $U_S$ just the induction voltage created by the iron rod?</p>
|
<p>It's always handy to have some background information:</p>
<p>In Europe the mains voltage is 230 Volts, which is too much for a lamp for example, so it needs to be lowered to for example 12 Volts. This is done by using a transformer.</p>
<p>The primary winding is connected to the mains voltage of 230 Volts. The AC in this primary winding causes a varying magnetic flux in the iron rod (core) which on its turn creates a varying magnetic flux through the secondary winding. Because of electromagnetic induction a voltage is induced in the secondary winding. The primary winding has more turns than the secondary winding which causes the secondary voltage to be lower than the primary voltage:</p>
<p>$$ \dfrac{N_P}{N_S} = \dfrac{U_P}{U_S} = \dfrac {I_S}{I_P}$$</p>
<p>You can see that be decreasing/increasing the number of turns in the windings you can control the voltage created by electromagnetic induction. </p>
<p>Here is an illustration with an example ($U_S = 220V, U_P = 110 V$)</p>
<p><img src="https://i.sstatic.net/WNkvn.gif" alt="enter image description here"> You can see how simple it really is. </p>
| 342
|
electromagnetism
|
How to interpret Faraday's Law?
|
https://physics.stackexchange.com/questions/52749/how-to-interpret-faradays-law
|
<p>Faraday's law is given by:</p>
<p>$\nabla \times E = -\frac{\partial{B}}{\partial{t}}$</p>
<p>On the right hand side of the equation, we have a quantity representing how $B$ changes over time. On the left hand side of the equation we have a quantity representing the spatial variation of $E$ (the curl of $E$).</p>
<p>The most common interpretation is that, we can determine the spatial variation of the induced electric field, if we know how $B$ changes over time. That is, the right hand side is producing the results obtained from left hand side of the equation.</p>
<p>Suppose instead however, I set up an electric field such that $\nabla \times E$ is non 0, will this induce a time-varying magnetic field given by the above equation?</p>
|
<p>I think your last question stated in the following way might clear up your confusion. If I setup an electric field $\mathbf E(t, \mathbf x)$ such that $\nabla\times\mathbf E(t_0,\mathbf x_0)\neq 0$ at some spacetime point $(t_0, \mathbf x_0)$, then Faraday's law tells me that whatever $\mathbf B$-field there is around, it must necessarily satisfy
$$
\frac{\partial\mathbf B}{\partial t}(t_0, \mathbf x_0)\neq 0
$$
This does not mean that $\mathbf B(t_0, \mathbf x_0)\neq 0$, the magnetic field could vanish for the instant $t_0$, but since its time derivative must me nonzero, it means that in the next instant the magnetic field must, in fact, be nonzero. So you can be guaranteed that if you have managed to setup an electric field with non-vanishing curl at some instant, then quite soon there will have to be a nonzero magnetic field around.</p>
| 343
|
electromagnetism
|
How to calculate how weak does a magnet get when you get an other magnet closer to it?
|
https://physics.stackexchange.com/questions/55087/how-to-calculate-how-weak-does-a-magnet-get-when-you-get-an-other-magnet-closer
|
<p>I heard that when you take two magnets and get them closer together so they reject each other (north pole to north pole or south pole to south pole) they weakens. Does anybody knows how to calculate how much it weakens depending on their distance? </p>
<p>And does it work the other way around? If you take two magnets the attract each other (north pole to south pole) will they get stronger? If so, how to calculate how stronger they'll get depending on their distance</p>
| 344
|
|
electromagnetism
|
Does a photon have a north and south pole?
|
https://physics.stackexchange.com/questions/55554/does-a-photon-have-a-north-and-south-pole
|
<p>A photon has an oscillating magnetic and electric field.</p>
<p>Is the magnetic field a dipole?</p>
|
<p>A photon is not the source of an oscillating magnetic and electric field, and it does not have poles. Your confusion stems from the fact that photons can be seen as field quanta for electromagnetic fields. In a sense they are the field, but they do not create it. </p>
| 345
|
electromagnetism
|
Magnetic properties of matter
|
https://physics.stackexchange.com/questions/55991/magnetic-properties-of-matter
|
<p>When a dielectric is placed in an electric field,it gets polarized. The electric field in a polarized material is less than the applied field. Now my query is, when a paramagnetic substance is kept in a magnetic field, the field in the substance is more than the applied fiekd. What is the reason for this opposite behaviour? </p>
|
<p>The external field causes the <a href="http://en.wikipedia.org/wiki/Magnetic_dipole" rel="nofollow">magnetic dipole</a> moments $\mathbf m$ of the atoms in the material to align with the applied field $\mathbf B$. If one now imagines summing up the fields due to all of the tiny little dipole moments that are now aligned with the external field, then one will find that the net effect is that the field inside is augmented.</p>
| 346
|
electromagnetism
|
Conservation, Maxwell tensor
|
https://physics.stackexchange.com/questions/56332/conservation-maxwell-tensor
|
<p>Can someone please explain to me how the conservation of the energy momentum tensor $$\nabla_\beta T^{\alpha \beta}=0$$</p>
<p>imply the conservation of the Maxwell tensor $$\nabla_\beta F^{\alpha \beta}=0$$?</p>
<p>Additional info: </p>
<p>Note that the $\nabla_\beta$ denotes the covariant derivative.</p>
<p>Energy momentum tensor</p>
<p><a href="http://en.wikipedia.org/wiki/Electromagnetic_tensor" rel="nofollow">Maxwell tensor = electromagnetic tensor</a></p>
|
<p>The claim isn't correct and even the way how the second equation is called, "conservation of Maxwell tensor", is deeply misleading. It makes no sense to call it a conservation law because the integral $\int d^3 x\,F^{0\mu}$ isn't any natural conserved quantity.</p>
<p>The second equation is called one of the Maxwell's equations. Moreover, the right form has sources on the right hand side
$$\partial_{\mu} F^{\mu\nu}=j^{\nu}$$
The time component is the charge density and the spatial components make up the current. It is possible to use the definition of the stress-energy tensor in terms of the Maxwell tensor etc. to prove the conservation of the stress-energy tensor of Maxwell's equations (because the stress-energy tensor is bilinear in the Maxwell tensor) but it's not possible to do the opposite, especially because the simplified Maxwell's equation without a $j^\mu$ isn't really true.</p>
| 347
|
electromagnetism
|
Direction of the Area Vector (with regards to magnetic dipole)
|
https://physics.stackexchange.com/questions/59498/direction-of-the-area-vector-with-regards-to-magnetic-dipole
|
<p>I'm learning about torque on a conductive coil in a magnetic field. I have been taught that $\vec\tau = \vec\mu \times \vec{B}$, where $\vec\mu$ is the magnetic dipole moment. Also, $\mu = I\vec{A}$, where $\vec A$ is the area vector of the loop.</p>
<p>To find the direction of the area vector, I am told to use the right hand rule with regards to the current in the loop (curl your fingers in the direction of current, and your thumb points in the direction of the area vector).</p>
<p><strong>My question is:</strong> Why does this give the correct direction for the area vector? Is the area vector just defined to be this way to avoid nasty usage of minus signs, or is there some other reason for this?</p>
<p>My guess is that whoever formalized this law/equation (not sure what correct term is for this instance) started with the direction of torque, and worked backwards defining the direction of $\vec\mu$ and $\vec{A}$ to reduce or eliminate stray minus signs in the equations. However, this is, of course, just a guess; I want to know what the true reason is.</p>
|
<p>As far as I know, the area vector is a purely mathematical object whose definition is related to the orientability of the surface (in this case, a disk). This is a property of surfaces embedded in an Euclidean space that allows to choose surface normal vector to the surface at every point. For an oriented surface, this normal is determined so that we can use the right-hand rule to define a clockwise direction of loops on the surface, which by the way, is needed if we want to apply Stokes' theorem. </p>
| 348
|
electromagnetism
|
Electromagnet, ideal turns depending on ohm
|
https://physics.stackexchange.com/questions/60945/electromagnet-ideal-turns-depending-on-ohm
|
<p>More turns -> stronger field<br>
more turns -> longer copper wire<br>
longer copper wire -> more resistance(ohm)<br>
<br>
at what turn does the resistance make the electromagnet weaker? - I want to make an ideal electromagnet.
<br><br>
(Sorry for being vague but I'm looking for some kind of formula or an example) </p>
|
<p>You need to combine some equations, lets list them:</p>
<p>Whe asume that you want the formula in terms of potential diferential, with Ohm's Law, $\Delta V = I R$, and the magnetic field inside one solenoid:</p>
<p>$$
B = \frac{N}{L}\mu I = \frac{N}{L}\mu \frac{\Delta V}{R} \quad\quad (1)
$$</p>
<p>where $N$ , $L$, $\mu$, stand for the number of turns, lenght of the solenoid and magnetic permeability of you core respectively. Then we also know the dependance of $R$ over the lenght of the wire, $R = \rho\frac{l}{A}$, where $\rho$ is the resistivity, $A$ the cross section of the wire and $l$ the lenght. We can compute the lenght by: </p>
<p>$$
l = 2\pi rN
$$</p>
<p>so if we subtitute the last equations in $(1)$:</p>
<p>$$
B = \mu\frac{N\Delta V}{L}\frac{A}{\rho 2 \pi r N } = \mu\frac{\Delta V A}{2 \pi \rho r L}
$$ </p>
<p>surprisingly it doesn't care the number of turns!! </p>
| 349
|
electromagnetism
|
Attraction and repulsion of Magnetic materials
|
https://physics.stackexchange.com/questions/62653/attraction-and-repulsion-of-magnetic-materials
|
<p>Why are diamagnetic materials repelled when placed in magnetic field and why are paramagnetic materials attracted when placed in magnetic field?</p>
|
<blockquote>
<p>Okay... I write this as an inspiration by Feynman..!</p>
</blockquote>
<p>These "WHY?" and "How's such a thing possible?" are quite related in a way that they make the question <em>ambiguous</em>. The question author wouldn't be satisfied for sure, hearing a physicist who try to answer such <em>hating</em> questions. But, <em>truly speaking</em>, these questions are very very good (In fact, +1'd it). But, the physicist can't <em>really</em> provide a satisfying answer. Well, <a href="http://www.youtube.com/watch?v=fKDnm5F2SmY" rel="nofollow">this guy</a> really hates it. He explains what's the problem with "WHY" to the interviewer..!</p>
<p>The reason I say this is because, here's a similar example.</p>
<ul>
<li>You can ask, why a glass falls down and breaks..? And one says, due to gravity. If someone is satisfied, it's OK. <em>(But, going deeper)</em> into two possibilities -</li>
<li>Why does the glass break? It's brittle (or) Why is this gravity pulling this glass down..? Because it's always attractive. This is also OK. But, if someone <em>goes still deeper</em>...</li>
<li>Why is it brittle? (or) Why is this gravity always attractive..? - The answerer will definitely go nuts trying to put his numbers into you. Because, the physicist knows that it's an observed phenomenon. He can't explain for sure why it's like the way it is..!</li>
</ul>
<p>This is because the physicists always try to explain something on the way it occurs, or <em>roughly</em> how this can be explained with his numbers and <em>Greek</em> symbols. He can't surely telly "WHY". That totally goes to an omnipotent being (if such a thing exists) which has created these things. Now, this goes <em>philosophical</em>.</p>
<hr>
<p>But, we can still correct your question by saying, <em>"How does this phenomenon occur?"</em> or <em>"How it's been theorized by our physics fellas?"</em></p>
<p>To answer this question in a sentence - <em>"All materials are diamagnetic"</em>. It's their magnetic permeability $\mu$ (a number) which determines whether their diamagnetic property is thrown out or their paramagnetic property exceeds it. Thanks to quantum mechanics which helped in relating these <a href="http://en.wikipedia.org/wiki/Paramagnetism#Relation_to_electron_spins" rel="nofollow">paramagnetic</a> and <a href="http://en.wikipedia.org/wiki/Diamagnetism#Theory_of_diamagnetism" rel="nofollow">diamagnetic</a> properties to electron pairing and especially <em>their spins</em>.</p>
<p>Well, there's a lot and lot more than <em>just that</em>...</p>
| 350
|
electromagnetism
|
Is it possible for a charged, fast-moving object to slow down and enter geo-stationary orbit?
|
https://physics.stackexchange.com/questions/62723/is-it-possible-for-a-charged-fast-moving-object-to-slow-down-and-enter-geo-stat
|
<p>I've had a wild idea which I can not discuss at length in this forum, but it comes down to the following problem:</p>
<p>A sphere of radius R=~10μm and mass m=~10-16Kgr is travelling towards the earth at v = ~10^8m/sec. The sphere carries a charge Q and intersects the earth's magnetic field perpendicularly, at a distance r, such that it enters an elliptic orbit. Given an electron wave function of ~5eV, is there any possible configuration of Q and r that would
allow the sphere to end up in a stable geo-stationary orbit?</p>
<p>Would releasing charges at particular points of each rotation make any difference?</p>
<p>Even though I did get a physics BS 13 years ago, I took a different direction and the calculations required are way beyond my current capabilities. If anyone finds the problem interesting, I'll be glad to hear from you.</p>
|
<p>Your sphere is traveling at $10^8 \, \mathrm{m/s}$ (33% the speed of light) but Earth's escape velocity is $11200 \, \mathrm{m/s}$. There is absolutely no way the sphere could ever enter any sort of orbit around the Earth, regardless of the interaction between it and the Earth's magnetic field.</p>
<p>$$KE_{relativistic} = \left( \left( \frac{1}{\sqrt {1 - \frac{v^2}{c^2}}} \right) - 1 \right)mc^2 = 8.7 \cdot 10^{16} \, \mathrm{J}$$</p>
<p>That's <strong>20 megatons of TNT</strong>. Even if the magnetic field could perform some sort of braking, the amount of heat generated to dissipate that much energy would induce nuclear fusion.</p>
| 351
|
electromagnetism
|
Small charged sphere's motion in earth's magnetosphere?
|
https://physics.stackexchange.com/questions/63520/small-charged-spheres-motion-in-earths-magnetosphere
|
<p>Suppose that a spherical metal sphere with mass $m=10^{-16}kgr$ radius $R=10μm$ charge $Q=10^{-9}C$ travels with $v=c/3$ and is trapped in the earth's magnetosphere at a distance around $r = 1000km$. <em>The exact numbers are not that important, I am providing them for order of magnitude considerations</em>. </p>
<p>Assuming a non-zero velocity component parallel to the field lines:</p>
<ul>
<li>Will magnetic mirroring and magnetic drift be the same as with plasma
(i.e. can one use the same equations to calculate the motion of the sphere)? Will the eddy currents in the sphere affect mirroring/drift and how? </li>
<li>If the sphere is already spinning, what effect will the spin have on its motion and vice-versa? </li>
<li>If instead of a solid sphere we had a spherical shell with the same characteristics, would it behave differently and how?</li>
</ul>
<p>I am looking for a qualitative answer, or a pointer to work that has been done along these lines. </p>
|
<blockquote>
<p><em>Will magnetic mirroring and magnetic drift be the same as with plasma</em></p>
</blockquote>
<p><strong>No.</strong> It's behaviour will be vastly different than that for a particle. Large metal objects have free electrons/ions which will attract the charged particles in the magnetosphere. Ultimately the magnetospheric plasma will shield the metal object such that the net electromagnetic force on the sphere is zero. </p>
<p>This phenomena is called Debye shielding. </p>
<p>Now, how much of a force is provided by radiation pressure is another question altogether ...</p>
<blockquote>
<p><em>If the sphere is already spinning, what effect will the spin have on its motion and vice-versa?</em></p>
</blockquote>
<p>From the above answer, we can assume that the electromagnetic forces on the sphere are negligible. Hence, its motion will be determined by laws of motion and gravitation. A spinning sphere will have angular momentum, and details for its motion can be found in any orbital mechanics textbook. </p>
<blockquote>
<p><em>If instead of a solid sphere we had a spherical shell with the same characteristics, would it behave differently and how?</em></p>
</blockquote>
<p>Well yes, but this is a complex question dealing with electrostatics and the principles of spacecraft charging, and also orbital mechanics. In theory, the moment of inertia ($I$) of a sphere and shell are very different. Refer to the textbook mentioned above for an in-depth explanation and relevant equations. </p>
| 352
|
electromagnetism
|
What really is the Magnetic Force on a wire?
|
https://physics.stackexchange.com/questions/65368/what-really-is-the-magnetic-force-on-a-wire
|
<p>I have a doubt regarding the significance of a force on a wire. Well, first of all, I know that if I have a particle and if there are several forces acting over it, then we can compute one total force $F$ that gives the same effect as the combination of the several forces, and this force is just the vector sum.</p>
<p>Well, this is pretty good: a vector <em>depends</em> on the point it's being applied (since it's an element of the tangent space at the point), so that since all the forces are on the same point (the same tangent space) we can take their sum and get another thing acting on the same point. This is pretty clear and simple to understand.</p>
<p>My doubt is when we have a wire for example. For a wire normally we use the relationship:</p>
<p>$$F=\int_\gamma i \ \gamma'(t)\times B(\gamma(t)) \ dt$$</p>
<p>In other words, we parametrize the wire with some curve $\gamma$, and we integrate $i \gamma' \times B$ over the wire to get the "force on the wire". But now this is confusing me, we are summing vectors at different points, and getting a vector that I don't know where's located. What I mean is: while when working with particles it makes sense to add the forces and use the total force on the same particle, with a wire this is kind of confusing, because it has length, so what should really mean "a force on a wire" since it's not just a point?</p>
<p>Thanks in advance for the help.</p>
|
<p>The force law you show gives us the <em>total</em> force on the wire. This force comes from the sum total of forces on all the electrons moving through the wire. So imagine that your wire is supported at either end, and the magnetic field is strictly between the two supports, so the total force on the wire is – in some sense – between the two supports. Then, if you want to keep the wire in place, the supports have to exert a total force that is equal and opposite to the electromagnetic force.</p>
<p>This answer might make you happy, but it's important to note that I've just swept your question under the rug a little. We usually don't think of forces as being "located" at a precise spot. The reason for this is pretty basic.</p>
<p>To simplify physics, we group particles together into collective objects, and frequently just assume that there are forces (that we ignore) within those objects to make sure that they don't get destroyed by the forces we're analyzing. This allows us to just look at the sum of forces. For example, we frequently deal with rigid bodies, in which all the particles are described completely by a single position (the position of the center of mass, for example) and the orientation of the body. You might say that we ignore the fact that a rigid body is made up of smaller particles, and just treat it as one big particle. Or, when talking about a rope, we might talk about the tension in that rope being transmitted between different sections of the rope.</p>
<p>So, in the case of the wire and its supports, those supports are extended objects, and the forces they exert are not located at a precise spot. But we treat them as rigid bodies, and look at the sum of the forces exerted by each infinitesimal part of them. We don't need to keep track of all those infinitesimal parts because we assume that the supports stay together, and maintain their shape. Same thing with your wire. We assume that the wire is continuous and doesn't break, so we can just treat it as a single object, and sum up all the forces.</p>
| 353
|
electromagnetism
|
Energy in magnetic fields
|
https://physics.stackexchange.com/questions/7238/energy-in-magnetic-fields
|
<p>If I calculate the energy contained in the electric field for an electric dipole p in an electric field E, I get (ignoring the terms independent of orientation):</p>
<p>$U = - \vec{p} \cdot \vec{E}$</p>
<p>which is as expected. However, if I do the same for a magnetic dipole m in a magnetic field B, I get (again ignoring the terms independent of orientation):</p>
<p>$U = \vec{m} \cdot \vec{B}$</p>
<p>which clearly has the wrong sign. Does one have to be careful in interpreting energy in the magnetic field?</p>
<p>A colleague pointed out that David Griffiths mentioned this problem in passing in "Dipoles at Rest" (AJP 60, 979, 1992). But it wasn't very satisfying, as he seems to be referring to the terms which are independent of the orientation to try to argue it fixes the sign somehow. It didn't make sense to me, maybe because it was only mentioned briefly in passing. Can anyone expand upon this?</p>
<p>Also, if I look at the Lagrangian for electrodynamics, I notice the energy in the electric fields has an overall negative sign (like a potential energy) while the energy in the magnetic fields has an overall positive sign (like a kinetic energy). So is this a hint I'm not supposed to treat the energy in a magnetic field on the same footing as the energy in an electric field?</p>
|
<p>First of all, energy is energy. It may be converted from any form to any other form and there is absolutely no ambiguity about the sign of any form of energy. If $E$ is correctly defined so that it is conserved, the signs in front of all quantities are well-defined.</p>
<p>Your observations about $E^2$ and $B^2$ having the opposite sign in the Maxwell Lagrangian are correct, too.</p>
<p>However, what's incorrect is your assumption that the energy carried by a configuration of electric charges or dipoles or magnetic dipoles may always be reduced to the energy carried by the electromagnetic fields only, $(E^2+B^2)/2$. </p>
<p>This statement is kind of coincidentally true in the case of the electrostatic energy. The energy between charges, an integral of
$$\frac 12\int \rho \Phi$$
where $1/2$ is included to avoid double-counting from the interaction energy of pairs of charges - may be integrated by parts, using $\rho = \nabla\cdot E$ and $E=-\nabla \Phi$, to get
$$\frac 12 E^2$$
That's nice and if one manages to subtract the divergent self-interaction energy of point-like charges (which has been a big puzzle in classical physics and became an equally big puzzle in QED, before it was solved and superseded by renormalization), everything is stored in the electrostatic field energy $E^2/2$.</p>
<p>However, this is simply not the case for the magnetic fields. The actual energy is given by terms like $-\vec m\cdot \vec B$, as you correctly write. In particular, two equally oriented magnetic dipoles, separated by an interval in the same direction as the direction of these dipoles, have a negative interaction energy, see </p>
<blockquote>
<p><a href="http://en.wikipedia.org/wiki/Magnetic_dipole-dipole_interaction">http://en.wikipedia.org/wiki/Magnetic_dipole-dipole_interaction</a></p>
</blockquote>
<p>That's why the force between two such dipoles is attractive - and the very same thing with the same signs would hold for two electric dipoles, too. The term $-\vec m\cdot \vec B$ may be boiled down to the $\pm\vec j\cdot \vec A$ in the Maxwell Lagrangian with sources. I wrote $\pm$ because I don't want to spend lots of time by fixing this sign because only some relative signs really matter for this question. </p>
<p>Now, mimicking the electrostatic integration by parts here in magnetostatics, $\vec j = {\rm curl} \vec B$ and $\vec B = {\rm curl} \vec A$, won't quite work because
$$ \vec j\cdot \vec A = {\rm curl} \vec B \cdot \vec A \neq - \vec B \cdot {\rm curl} \vec A +{\rm curl} (\dots). $$
Those curl identities don't seem to work in the same way. So the energy of magnetic dipoles can't completely eliminate the non-$B^2$ terms, I think.</p>
<p>Even in the electrostatic case, we have to be careful how we interpret the different energies to avoid double-counting. Because the $\int\rho\Phi/2$ energy could have been rewritten as $\int E^2/2$, we may imagine that charges are some mysteriously compact lumps of electric field. But aside from the $1/2$ factor that was needed to avoid the double-counting from the electrostatic energy of each pair of charges, there is one more dangerous place where we could double-count: we must either imagine that the energy comes from $\rho\Phi/2$, or from $E^2/2$, but not from both! Again, if we wanted to include both terms, they would have to have the same sign, and we would double the energy relatively to the correct value.</p>
<p>The magnetic counterpart is that we may express the energy density as $B^2/2$ if we don't study how the dipoles and sources are affected, but if we do, we must forget $B^2/2$ and correctly calculate terms such as $-\vec m\cdot \vec B$.</p>
<p>By the way, the problem with the sign would occur even for randomly generated electric dipoles. It's because $\int E^2/2$ is positively definite - for any configuration of dipoles - while the interaction energy between two dipoles may be both positive and negative. So it's just not true that the interaction energy of two ordinary dipoles may be rewritten as $\int E^2/2$. In some sense, the problem does reside in the dipoles' self-interactions. </p>
<p>Note that $\int E^2/2$ is always positively definite - so how it can account for the electric charges' interaction energies that can have both signs? The answer is that $\int E^2/2$ should be identified with $\int \rho\Phi$ plus the "internal mass" of the electric sources - which is basically infinite. However, $\int B^2/2$ is also positively definite while there is significant extra mass (interaction self-energy) associated with the magnetic dipoles themselves. So for the magnetic sources, exactly because the currents may be "light", the transcription into $\int B^2/2$ cannot work.</p>
<p>When we have either electric or magnetic dipoles, we must carefully put the right sign in front of the energy, instead of assuming that the energy may always be rewritten as "field-only" energy. It can't.</p>
| 354
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electromagnetism
|
Why is there independent motion of Earth's magnetic poles?
|
https://physics.stackexchange.com/questions/7264/why-is-there-independent-motion-of-earths-magnetic-poles
|
<p>Earth's N and S magnetic poles "wander independently of each other and are not at exactly opposite positions on the globe" [from WIKI's "Earth's magnetic field"]. Can these independent motions be consistent with the supposed "dynamo effect" from electric currents of a liquid outer core? Can Earth's spheroidal shape be a factor?</p>
|
<p>Well, Earth does not have a giant bar magnet inside, the outer core is an giant mass of conductive fluid in a complex motion guided by magnetohydrodynamics -- the problem is complicated enough to be unsolvable analytically and really gives no hope of simple solution.</p>
<p>So one should rather think of a geomagnetic <strong>field</strong> which currently happens to be close to a dipole field, thus producing an illusion of poles. </p>
| 355
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electromagnetism
|
Need some help interpreting a formula inspired from Coulomb's law
|
https://physics.stackexchange.com/questions/8856/need-some-help-interpreting-a-formula-inspired-from-coulombs-law
|
<p>It has been more than a decade since I did all vector related math and physics so pardon me if my question does not make sense. I am reading some article that says it was inspired from Coulomb's law and gives the following expression to calculate the partial force that one particle is exerting on another at time $n$:</p>
<p>$f^{i,j}_n = (c - |p^{i}_n - p^{j}_n|)\frac{p^{j}_n - p^{i}_n}{|p^{i}_n - p^{j}_n|}$</p>
<p>where $c$ is the range of the particle over which it's force can spread and $p^{i}_n$ is the location of particle $i$ in the form of $(x,y)$.</p>
<p>I checked up the Wikipedia article on Coulomb's law but this formula does not make sense to me. What exactly is the formula trying to achieve? Can someone please help me understand this?</p>
|
<p>It's nothing you can rigorously derive from Coulomb's law, but the idea is probably the following: the rightmost factor
$$\frac{p^j - p^i}{|p^j - p^i|}$$ is just the unit vector pointing from particle $i$ to particle $j$, just as in Coulomb's law or Newton's law. Normally, for Coulomb's law, you'd have a factor of $q_1 q_2/4\pi\varepsilon_0 |p^j - p^i|^2$ in front of that, but here they replaced the $1/r^2-$ behaviour by a linearly decreasing factor
$$c - |p^j - p^i|,$$ which yields c when $p^i = p^j$, and vanishes as $|p^j - p^i| \rightarrow c.$ (Since you call c a range, they <em>might</em> also mean that whenever $|p^j - p^i| > c$, the force should vanish, but it's not clear from the notation alone.)</p>
| 356
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electromagnetism
|
How to estimate inductive coupling between distant air coils
|
https://physics.stackexchange.com/questions/8968/how-to-estimate-inductive-coupling-between-distant-air-coils
|
<p>I have two air coils (assume they are simple, circular wire loops). They both have diameter <strong>d</strong>. There is a distance <strong>D</strong> between their centres.</p>
<p><strong>D</strong> is much greater than <strong>d</strong> (more than 10x greater)</p>
<p>Both coils are at different angles, <strong>a</strong> and <strong>b</strong>, relative to the line between their centers. There is an alternating current in one coil.</p>
<p>Is there some simple function which estimates the current induced in the second coil? I would be happy if the form of the function was:</p>
<p>k * f(D, a, b) (where k had to be measured)</p>
<p>For example: k * cos(a) * cos(b) * D^-2</p>
|
<p>I think it's neater to express the various directions in terms of unit vectors. We can convert to angles at the end. So say that $\hat n_1$ and $\hat n_2$ are unit vectors perpendicular to the axes of the coils, and let $\hat r$ be a unit vector pointing from the center of coil 1 to the center of coil 2. Since the coils are separated by a distance that's large compared to their diameters, I'll assume that each one can be approximated as a magnetic dipole.</p>
<p>The magnetic field produced by coil 1, at the location of coil 2, is
$$
\vec B \propto {1\over D^3}[3(\hat n_1\cdot\hat r)\hat r-\hat n_1]
$$
(This is a standard formula for a dipole field. See a textbook such as Griffiths. Since you graciously allowed an arbitrary constant that had to be measured, I'm writing proportionalities and ignoring boring constants).</p>
<p>The mutual inductance is determined by the flux through coil 2 due to this field. Because of the large separation, we can assume the field is approximately constant at the coil's location, so this is just $\vec B\cdot\vec A$. So the flux is just proportional to the component of $\vec B$ that's normal to the second coil. That is,
$$
\Phi\propto \vec B\cdot\hat n_2\propto {1\over D^3}[3(\hat n_1\cdot\hat r)(\hat n_2\cdot \hat r)-\hat n_1\cdot\hat n_2].
$$</p>
<p>Since the mutual inductance is proportional to the flux, this is the answer you're looking for. To express it in terms of the angles, I guess we have $\hat n_1\cdot\hat r=\cos a$ and $\hat n_2\cdot\hat r=\cos b$. The dot product $\hat n_1\cdot\hat n_2=\cos c$, where $c$ is the angle between the axes of the coils.
So I think the final answer is
$$
k D^{-3}[3\cos a\cos b-\cos c].
$$
If all of the various axes lie in a plane, then I guess $c=a-b$, in which case the angular dependence is something like $2\cos a\cos b-\sin a\sin b$, but if things can swivel around in all three dimensions, then $c$ is not uniquely determined by the other two.</p>
| 357
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electromagnetism
|
Electron gun - electron in cylindrical anode
|
https://physics.stackexchange.com/questions/11175/electron-gun-electron-in-cylindrical-anode
|
<p>In an electron gun, the heating filament heats the cathode, releasing electrons by thermionic emission. I've read that <em>"electrons are negatively charged particles and the positively charged cylindrical anode develops a strong electric field that exerts a force on the electrons, accelerating them along the tube"</em>. However, I don't think that this explanation is very clear, and I was wondering specifically how the "strong electric field" inside the cylindrical anode is able to accelerate the electrons?</p>
|
<p>any electric field will accelerate electrons, according to the Lorentz force law:</p>
<p>$\vec{F} = q \vec{E} + \vec{v} \times \vec{B}$,</p>
<p>where q is the charge of the electron, and $\vec{E}$ is the electric field. I assume there is no magnetic field in this example, so the second term is zero.</p>
| 358
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electromagnetism
|
One point to change electric field
|
https://physics.stackexchange.com/questions/13662/one-point-to-change-electric-field
|
<p>Can there be a charge configuration in space such that at any instant of time I can change the electric field at one and only one point?</p>
|
<p>The electric field obeys Maxwell's equations and in particula the Gauss's law. This means that $\rm div E \sim \rho$. If you change the field at only one point, this will introduce an infinite divergence and consequently infinite density. In other words, you'd have to introduce a point charge. But then the field $E$ itself will diverge at that point. So the answer is no, you can't do this.</p>
<p>In general, the allowed configurations are continuous except at boundaries of objects (where $E$ has to jump to account for charges in the material) and at point charges (where the field diverges).</p>
| 359
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electromagnetism
|
Magnetic field in the Centre of Circular loop wire with Current?
|
https://physics.stackexchange.com/questions/14067/magnetic-field-in-the-centre-of-circular-loop-wire-with-current
|
<p>By biot-savart:</p>
<p>$$\bar{H} = \frac{I}{4\pi} \oint \frac{d\bar{l} \times \bar{r}}{r^{3}}$$</p>
<p>so</p>
<p>$$\bar{H} = \frac{I}{2a} \hat{n}$$</p>
<p>Please, explain the last implication. I cannot find such integral to match the results. The radius of the loop is $a$. The current is $I$. $d\bar{l}$ is a vector along the perimeter.</p>
|
<p>For the path element $d\vec{l}$ around a circle with radius $a$ you can write $d\vec{l}=rd\phi\vec{e_\phi}$ with $\vec{r}=-r\vec{e_r}$ (note the minus sign, since the vector points <em>from</em> the wire to the center) you get </p>
<p>$$\vec{H}=\frac{I}{4\pi}\oint \frac{r^2 (-\vec{e_\phi}\times \vec{e_r})}{r^3}d\phi$$ substituting $a$ for $r$ and integrating $\phi$ from 0 to $2\pi$ and realizing $-\vec{e_\phi}\times\vec{e_r}=\vec{e_z}$</p>
<p>$$\vec{H}=\frac{I}{4\pi}\frac{1}{a}\vec{e_z}\int_0^{2\pi}d\phi = \frac{I}{2a}\vec{e_z}$$</p>
| 360
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electromagnetism
|
Mangnetic Flux summing up like Kirhoff?
|
https://physics.stackexchange.com/questions/14071/mangnetic-flux-summing-up-like-kirhoff
|
<p>You have a coil over an iron hearth. There is a current in coil which creates the flux $\phi_{1}$. The flux then distributes over the wider area in the iron (using wrong word?): $\phi_{2}$ the flux over the middle and the $\phi_{3}$ over the left.</p>
<pre><code> |-----------------|
| | |
\phi_{3} |\phi_{2}| | COIL HERE
| | | the lines are of the iron hearth
|-----------------|
\phi_{1}
</code></pre>
<p>The flux does not disappear so</p>
<p>$$\phi_{1} = \phi_{2} + \phi_{3}$$</p>
<p>could some explain the last statement? The last statement is a bit like Kirhoff's law. But I am unsure how can you play with fluxes. Could someone elaborate on this? On which rule, is the statement based on?</p>
|
<p>A fundamental postulate of electromagnetism is that the flux of the magnetic field through any <a href="http://en.wikipedia.org/wiki/Gaussian_surface" rel="nofollow">closed surface</a> is zero. This is essentially a statement that there are no <a href="http://en.wikipedia.org/wiki/Magnetic_monopole" rel="nofollow">magnetic monopoles</a>. This implies that the divergence of the magnetic field is zero. The corresponding equation is</p>
<p>$$\nabla \cdot \vec{B} = 0$$</p>
<p>This is not very much like Kirchoff's (Voltage) Law, which is based on the vanishing curl of the electric field</p>
<p>$$\nabla \times \vec{E} = 0$$</p>
<p>and is only true in the absence of a time-dependent magnetic field.</p>
<p>To summarize the difference:
The law about magnetic fields is saying that you take a closed plastic bag, contort it into any shape, and however much magnetic flux enters the bag has to flow back out again somewhere else. The law about electric fields is saying that you take loop of wire, contort it into any shape, and however much the electric field pushes a charged particle forward in one part of the loop, it will push the particle backwards by the same amount over the rest of the loop.</p>
| 361
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electromagnetism
|
What Is the Physical Size of a Magnet?
|
https://physics.stackexchange.com/questions/15343/what-is-the-physical-size-of-a-magnet
|
<p>With present day materials and technology what is the physical size of a one milli Tesla magnet? How much "power" it has to attract pieces of iron? Please compare it with the objects we have around. What difference in size and "power" it gets each order of magnitude that I go up? What is a weak magnet what is a powerful magnet in this scale?</p>
|
<p>I would recommend you take a look at the K&J Magnetics website, particularly the magnet properties calculator: <a href="http://www.kjmagnetics.com/calculator.asp" rel="nofollow">http://www.kjmagnetics.com/calculator.asp</a> </p>
<p>Your question has too much ambiguity for a real answer, but I think that perhaps that page will address what you are wondering about.</p>
| 362
|
electromagnetism
|
Magnetic Fields
|
https://physics.stackexchange.com/questions/15856/magnetic-fields
|
<p>If running a current through a magnet can reverse the magnetic field, then how strong does the current have to be and how would it best be run through the magnet.</p>
|
<p>If you had solenoid or another type of electromagnet that produced its magnetic field from a current carrying like a Helmholtz coil then you could switch the direction of the current to reverse the direction of the magnetic field. Simply running a current through a permanent magnet would not reverse the magnetic field of that magnet as far as I know. </p>
| 363
|
electromagnetism
|
Poisson's Equation
|
https://physics.stackexchange.com/questions/16472/poissons-equation
|
<p>First up, I needed to compute the potential at some external point due to charge stuffed inside the region between two concentric cylinders, the volume charge density being given.<br/> Two methods came to mind but they are apparently yielding different answers(maybe I goofed up somewhere)
<br/> The first is to solve the Poisson's equation considering that the potential doesn't vary with $\theta$ and $z$ and setup and solve a differential equation.
<br/> With that said, what if I consider the composite object as two cylinders, the bigger one with a volume charge density $\rho_v$, radius $a$ and a smaller one inside it (along the common axis) with density $-\rho_v$, radius $b$ <br/> (In this way the stuff along the axis cancels out leaving the same resultant object) and calculate the potentials due to them at some point $r$. The resultant potential would then be the sum. Though this doesn't apparently yield the same result.</p>
|
<p>It does give the same result--- it would be good to give your steps, if you want to find the error. The easiest way to see this is to just use Gauss's law: if you make a cylinder of radius R, the electric field outward times the surface area is the charge inside. If the length of the Gaussian cylinder is L, and the inner and outer radius of the physical cylindrical annulus is $R_0$,$R_1$, then for the exterior</p>
<p>$$L 2\pi R E(R) = L \rho (\pi R_1^2 - \pi R_0^2) $$</p>
<p>Which gives you the standard 1/R field, log(R) potential, outside a charged cylinder and by its form is equivlent to the difference-of-two-cylinders solution.</p>
<p>For inside the inner cylinder, the field is zero, and between the two, it is that sum of a 1/R and constant which matches the other two regions. The field in all regions can be found by Gaussian cylinders, and the potential is found by integrating in R. The result satisfies Laplace's equation with source, since you can prove Gauss's law for solutions of this equation directly.</p>
| 364
|
electromagnetism
|
Power Generation from Axial and Transverse Emf
|
https://physics.stackexchange.com/questions/18915/power-generation-from-axial-and-transverse-emf
|
<p>We consider a flat rectangular plate moving horizontally in a vertical magnetic field,motion being in a direction perpendicular to the length of the plate. We have an emf=BLV between the tips,in the lenhgth wise direction[the axial emf]. During the formation of the axial emf a current flows along the length of the conductor. This current should get deflected in the lateral direction due to the existing magnetic field, producing a transverse emf between the lateral edges.If the tips are connected by a wire we have a closed circuit condition--we should simply have the Hall voltage between the lateral edges[in the direction of the motion].</p>
<p>Query: Would it be possible to use these voltages as supplementary power sourses in moving vehicles---in moving trains,cars ,aeroplanes etc?
[We seem to have several ready-made parallel cells in these vehicles]</p>
|
<p>I'm assuming that the magnetic field you're referring to over here is the Earth's field, then we have the field strength of $6.5 * 10^{-5}$. If we assume the vehicle to be a bullet train, then we get the velocity of $300 km / h$ which is $83.33 m/s$ Assuming a scenario where a rectangular plate of $1m$ is being used and the current is being generated along its diagonal of length $\sqrt2 m$ Then as you stated,</p>
<p>$ \varepsilon = BLV $</p>
<p>$ \varepsilon = 6.5 * 10^{-5} * 83.33 * \sqrt2 $</p>
<p>Then this setup will generate $.00766$ $V$ of electricity, which is too low to be applied for most practical purposes.</p>
<p>If you notice over here then you already have an engine powering the vehicle, and in effect you're trying to capture some of its output. Wouldn't it be much better to capture it at the source?</p>
| 365
|
electromagnetism
|
Magnetization of coin on a railway track
|
https://physics.stackexchange.com/questions/19374/magnetization-of-coin-on-a-railway-track
|
<p>The rumor was you could make a magnet by leaving a piece of iron on a train track. The train going over it would magnetize it. </p>
<p>Is it true?</p>
|
<p>This is almost a duplicate to <a href="https://physics.stackexchange.com/questions/18340/can-you-magnetize-iron-with-a-hammer/18341#comment43493_18341">can you magnetize iron with a hammer</a> . Have a look at it. </p>
<p>The only difference is that the rail lines are fixed in their north south direction for years. The iron in the lines themselves become magnetized and so the argument with the hammer should also hold , i.e. the small magnetic domains, momentarily freed by the impact, reorient to the magnetic field direction of North South of the earth's magnetic field. If the coin partially melts under the weight of the train, even better. It should of course be made from a ferromagnetic metal.</p>
| 366
|
electromagnetism
|
Would the north poles of two magnets repel each other if a weaker south pole was inserted between them?
|
https://physics.stackexchange.com/questions/19383/would-the-north-poles-of-two-magnets-repel-each-other-if-a-weaker-south-pole-was
|
<p>My son asked me this question and I was stumped - my intuition says that is the south pole was strong enough the attraction between the north and south poles would outweigh the repulsion between the two north poles - but how strong would the south pole need to be in this case?</p>
|
<p>Assuming long skinny solenoidal magnets, so that their ends look like point magnetic sources, the force between two poles is $M1M2/r^2$, just like electrostatics, with pole density replacing charge density, where $M1$,$M2$ is the strength of the pole, and $r$ is the separation between the poles.</p>
<p>So you have three poles at position -A,0,A with magnitudes in the ratio 1:-1/4:1 respectively, the total repulsive force on the left-most one from the rightmost one is exactly balanced by the force from the middle one. Twice the distance means 1/4 the force.</p>
<p>For real magnets, you will usually not be able to get the other poles far enough away to have what looks like a monopole source. Then the answer depends on the geometry, but the above is a rough guide.</p>
| 367
|
electromagnetism
|
How are magnetic fields transmitted?
|
https://physics.stackexchange.com/questions/20455/how-are-magnetic-fields-transmitted
|
<p>A common analogy for gravity is the ball-on-a-rubber-sheet model. In this model, mass distorts spacetime and creates a 'valley' into which other mass can fall. Is this same principal valid for magnetic fields as well (proton-electron)? If so, then how is the repulsion effect modelled?</p>
<p>I ask because the underlying properties of both forces is very similar (inverse-square law). Thanks.</p>
|
<p>Two things:
First, it is more intuitive to treat gravitational force as equivalent to electrostatic force, due to the existence of monopoles (and gravitational field lines do not form closed loops). There is a magnetic analogue to gravity, known as gravetomagnetism, frame-dragging, or the Lense-Thirring effect.</p>
<p>Ok, now to answer your question:
Yes, gravitational fields are created by the "valley" mechanism. But, their method of transmission is through gravitational waves.</p>
<p>Imagine, for a moment, that someone instantaneously vaporized the sun. By vaporized, I mean that all the mass was just forced to disappear (note that this is hypothetical). Now, by our classical description, the Earth will change its trajectory immediately (as the net force changed). But the light from the vaporization will take some time to reach the Earth (8 minutes). So we managed to send a signal faster than light!</p>
<p>The flaw in the above situation is that gravity is transmitted by vibrations in the "rubber sheet" known as gravitational waves. Imagine the same situation in rubber-sheet mode. To make it a bit clearer, imagine a very large sheet with a small but massive sun at the center. Removing the sun will create a vibration that travels to the edge of the sheet. Till the vibration reaches the Earth, the Earth will feel that the sheet has not changed in its vicinity. Once the vibration reaches the Earth, it will be jerked onto its new "no force" path (after a brief period of stretching and squeezing).</p>
<p>Now, Electromagnetism is transmitted by electromagnetic waves. There aren't any rubber-sheet analogies for this, due to the repulsive-attractive duality. Basically, an EM wave like light propagates in one direction, and carries a mutually perpendicular set of electric and magnetic fields which oscillate (with the same frequency as the wave itself). A moving charge (for that matter, a stationary one, too) emits these waves, which interact with other charges in a similar way as above. The inverse square law here can be said to arise from the fact that the intensity of an EM wave is inversely proportional to the square of the distance (When the wave arises from a point source).
Note that this emission of EM waves does not violate conservation if energy, as the particle will also absorb EM waves from others.</p>
<p>The actual explanation for all this comes from quantum mechanics. I'll only show how the electrostatic repulsive force works here, as the attractive force is a bit more complicated to explain.
In quantum mechanics, in any region of space, even an empty one, "virtual" particles can be created and shortly destroyed. These particles are "virtual" as they do not behave normally.</p>
<p>Now, lets take two protons near each other (neglecting nuclear forces). Each proton "emits" virtual photons in all directions. Note that photons have momentum in quantum mechanics. Now, the proton would stay at rest if the second proton was not there, as net momentum change is zero (photons are being emitted in all directions. But, the proton is absorbing photons on one side from the other proton. So, these photons "push" the proton away from the other proton. The same thing happens to the other proton, and the net result is that the protons move away from each other.</p>
<p>To explain attractive force, you have to take into account that the photon has a wavefunction, and has no definite position or momentum. Thus, a photon coming from the left proton can actually hit the right proton <em>from the right</em>, creating an attractive force. So we have to balance the probabilities for this, and the equations involve the charges of the two bodies, this giving rise to the "like charges repel, opposites attract"</p>
<p>One more thing about electric and magnetic fields: They are actually the same thing. An observer moving at some speed will disagree as to which part of the field is electric and which part is magnetic (As magnetic forces involve velocities).</p>
| 368
|
electromagnetism
|
Would a metal enclosure (such as a shipping container) protect its contents from the effects an electromagnetic pulse?
|
https://physics.stackexchange.com/questions/21180/would-a-metal-enclosure-such-as-a-shipping-container-protect-its-contents-from
|
<p>I was watching a program about disaster preparedness, and it was suggested that the metal enclosure of a common shipping container (of the <a href="http://en.wikipedia.org/wiki/Intermodal_container" rel="nofollow">intermodal variety</a>) would be sufficient to protect its contents from a large electromagnetic pulse (the kind that could affect an entire region or continent).</p>
<p>I have my doubts that this is true, as it seems like a misunderstanding of how electromagnetic pulses work—but I can't find any reliable resources on the subject.</p>
<p>What does physics have to say about this? <strong>Would a metal enclosure (such as a shipping container) protect its contents from the effects an electromagnetic pulse large enough to affect a large geographic region?</strong></p>
|
<p>Looks like a metal enclosure would be OK, provided its seams and joints are electromagnetically closed , see <a href="http://www.wbdg.org/ccb/FEDMIL/std188_125_1.pdf" rel="nofollow">http://www.wbdg.org/ccb/FEDMIL/std188_125_1.pdf</a> , however, I am not sure this requirement is satisfied in off-the-shelf containers, so some extra electromagnetic hardening of seams and joints may be required.</p>
| 369
|
electromagnetism
|
Does inducing a current in a wire have any effect on the strength of a permanent magnet?
|
https://physics.stackexchange.com/questions/21062/does-inducing-a-current-in-a-wire-have-any-effect-on-the-strength-of-a-permanent
|
<p>Does inducing a current in a wire result in any changes in the strength of a permanent magnet? Specifically, what would the results of the following controlled experiment be? </p>
<p>You set up two alternators that each simply consist of a permanent magnet located inside a stationary loop of wire as depicted <a href="http://en.wikipedia.org/wiki/File%3aAlternator_1.svg" rel="nofollow">here</a>. With the first (control) alternator, you simply rotate the magnet at a certain velocity for a short time, once per day, and measure the amount of induced current on a daily basis. With the second alternator, you hook the magnet up to your favorite (external) source of rotational motion and spin it continuously and measure the amount of induced current at the same intervals as with the control alternator. </p>
<p>Would the amount of current change over time (in either one)? If it would, in which direction and why? And if this is still not well-specified enough for an definite answer, what would it depend on? (how fast the magnets are rotating? what kind of circuit the wire is hooked up to? what the magnet is made out of? etc.)</p>
|
<p>Friction is a process that converts kinetic energy into heat. Its electrical equivalence is restistance; that converts electrical energy into heat. Materials with high magnetic hysteresis can convert magnetic energy into heat. </p>
<p>You may be confused because the usual source of magnetic energy is the physical movement of a magnet, not a loss of magnetization.</p>
| 370
|
electromagnetism
|
Are there any non magnetic materials that attract to each other as if they were magnetic?
|
https://physics.stackexchange.com/questions/24446/are-there-any-non-magnetic-materials-that-attract-to-each-other-as-if-they-were
|
<p>Are there any non magnetic materials that attract to each other as if they were magnetic?</p>
<p>This is an argument I am having with a friend. </p>
<p>Thanks,</p>
| 371
|
|
electromagnetism
|
Magnetized nail lifting another nail
|
https://physics.stackexchange.com/questions/24616/magnetized-nail-lifting-another-nail
|
<p>I have two nails (made of Fe). A and B.</p>
<p>A can not lift B.
If I rub A on the magnets north pole, then it can lift B.
Then if I rub it again but on the south pole, it can not lift B. </p>
<p>why is that? Why cant it lift B anymore?</p>
<p>the title I have chosen for this question is really bad. If someone else got a better idea please edit. sorry.</p>
|
<p>Iron is composed of small <a href="http://www.britannica.com/EBchecked/topic/357033/magnetic-dipole" rel="nofollow">magnetic dipoles</a> all with their north south axis pointing randomly.</p>
<p>By rubbing the nail in a consistent specific direction the small dipoles line up with the magnetic field, so when you are rubbing the north pole, the dipoles align
south-north. Thus you get a small magnet, not permanent because the little dipoles are not frozen in a crystal.</p>
<p>When you take this nail and rub it on the south pole you essentially demagnetize it for a while, if you take the same time you took on the north pole. If you persist in rubbing, the dipoles will align the other way and you will have a small magnet again.</p>
| 372
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electromagnetism
|
Effect of Charged Particles trapped in Magnetic Field on that Field
|
https://physics.stackexchange.com/questions/29744/effect-of-charged-particles-trapped-in-magnetic-field-on-that-field
|
<p>Given a stream of moving charged particles that encounter a uniform magnetic field such that they are trapped in a circular orbit, what effect do these particles have on the net magnetic field over time? Would the magnetic field get stronger or weaker as the number of trapped particles increase?</p>
|
<p>They always reduce the field, and this is the law that magnetic fields induce currents that reduce their strength, a special case of LeChatelier's principle.</p>
| 373
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electromagnetism
|
How to transfer energy from a generator to a storeage battery
|
https://physics.stackexchange.com/questions/29512/how-to-transfer-energy-from-a-generator-to-a-storeage-battery
|
<p>and thank you in advance for taking the time to read my question. To give an idea of my working level, I'm a 21 year old computer science student entering my senior year at college. It's been a few years since my Electricity/Magnetism course, and i'm a bit rusty on the Lorentz Force.
I wanted to create a sort of "Human Powered Generator", ie. something as simple as a stationary bicycle turning a generator as I peddle. Now I know the "right hand rule", and can quite easily make a motor/generator with some wire and magnets. My question is about voltage/current. I was never clear on the effects of the strength of the magnet and its importance in the amount of potential power generated. In other words, </p>
<p>1.) If i'm turning a generator at a constant rate and it is lightning a bulb, if I magically replaced the magnets with ones twice as strong, what would happen? I'm assuming it gets twice as hard to turn, but outputs potentially twice as much power. </p>
<p>In addition, I was never clear on the relationship between voltage and current in the lorentz force.</p>
<p>2.) While turning a generator at constant speed S with magnets of strength B and # of wire coils C, how much voltage/current is created? I know there are many variables involved here, perhaps such as the width of the rod the coils are wrapped around, thickness of coils, etc. </p>
<p>Finally, </p>
<p>3.) If trying to charge a battery of V volts and A ampre-hours, what measures should I take to ensure safe delivery of energy to the battery? In other words, if I peddle the generator very rapidly, I expect lots of either current/voltage/both. I assume I need a voltage regulator of sorts, and I'm not sure if the current matters (I think its just how "much" energy there is, whereas voltage is the "pressure" or "strength" of the energy). </p>
<p>I appologize if any assumptions I made are incorrect, i'm just going off of old knowledge. I tried wikipedia, but its all symbolic and I cant find a hard example (As in, I dont know how to find magnetic strength of a magnet, always denoted as B in the equation). Thanks to anyone who can answer any/all 3 of my questions!</p>
<p>-John</p>
|
<p>First a few lines of basics. If you put a loop into the magnetic field and this loop turns within it, the magnetic flux through loop shall change according to the formula</p>
<p>$$\Phi_B = \vec{B} \cdot \vec{A} = B A \cos\phi = B A \cos\omega t,$$</p>
<p>where $\vec{B}$ is magnetic field strength, $\vec{A}$ is area of the loop and $\phi$ is angle between $\vec{A}$ (<em>perpendicular</em> to loop) and $\vec{B}$, while $\omega$ is angular velocity of the rotation of the loop.</p>
<p>If you use coil with $N$ loops, then induced voltage on the coil shall be</p>
<p>$$\mathcal{E} = N \frac{\text{d}\Phi_B}{\text{d}t} = - N B A \omega \sin\omega t.$$</p>
<p>Therefore, yes, if you use twice larger magnetic field, you get twice larger voltage.</p>
<p>However, twice larger magnetic field <em>does not</em> necessarily mean it is twice as hard to turn. For example, if you have no electric load on the generator, there is practically no current in the coil and there is practically no Lorentz force! However, if you have completely ohmic load, voltage twice larger means current twice larger and yes it becomes twice as hard to turn.</p>
<p>In practice, you should also consider the internal friction of the generator. So even if there is no load, some muscular power will be required in order to overcome friction. When you increase the electrical load, required power in order to turn generator increases.</p>
<p>Of course, it is difficult to keep rotation constant, which means that with larger angular velocity $\omega$ voltage increases. To keep generator's output voltage constant, you need some electronic circuit, of which the simplest possible includes zener diode.</p>
| 374
|
electromagnetism
|
Find wavelength from relative permittivity and frequency
|
https://physics.stackexchange.com/questions/30710/find-wavelength-from-relative-permittivity-and-frequency
|
<p>If i have a EM wave with frequency 1MHZ and εr=9 in a perfect dielectric ,is it possible to find the wavelength λ and wave propagation speed ? It seems impossible to me..</p>
|
<p>(1) $\lambda \nu = c$</p>
<p>(2) $c^2 = \frac{1}{\mu_0 \epsilon_r \epsilon_0}$</p>
| 375
|
electromagnetism
|
EMF in a half-ring shaped conductor around a solenoid
|
https://physics.stackexchange.com/questions/30715/emf-in-a-half-ring-shaped-conductor-around-a-solenoid
|
<p>A half-ring shaped conductor is being placed around a solenoid. This solenoid has a changing magnetic field.</p>
<p>a) There is a current and EMF (Electromagnetic force) in the half-ring shaped conductor
b) There is a current but no EMF in the half-ring shaped conductor
c) There is no current but there is an EMF in the half-ring shaped conductor
d) There is no current but there is no EMF in the half-ring shaped conductor</p>
<p>My thoughts about this:
First of all the field outside a solenoid isn't very big in contrast to the field inside the solenoid. But I assume that it can't be neglected.
I think there won't be a current as we would need a full ring-shaped conductor for this.
I do think that there will be an EMF in the conductor, and this will generate some eddy currents in the conductor. But there will be no "full-on current" so to speak.
So I would say the answer is C. Is this correct?</p>
<p>A Little side-question;
Is it at all possible for a non-conducting material to have an induced EMF. For example when we move a non-conducting material (ring-shaped) into a magnetic field. The magnetic flux in the ring changes, but will there be an induced EMF? I know that there won't be a current in the loop, but I'm not certain about the EMF.</p>
<p>Thank you very much!</p>
|
<p>Since there is a steadily changing magnetic flux, there are static closed loops of electric field lines circling the solenoid.</p>
<p>If the half-ring shaped conductor is placed around the solenoid, the mobile charge carriers in the conductor will redistribute in such a way that there is zero <em>net</em> E field inside the conductor.</p>
<p>Thus, there is no current (except for the more or less initial near instantaneous current required for the redistribution of charge within the conductor). So, that rules out answers a) and b).</p>
<p>If you place a voltmeter across the ends of the conductor (make sure that the leads do not form a loop threaded by the the magnetic flux!), you will measure a voltage across the conductor that is numerically equal to the emf. (In fact, you're measuring the strength of the opposition to the emf due to the charge distribution in the conductor). So, you are correct, the answer is c).</p>
<p>If you had a half-ring of a non-conductor, no charge redistribution occurs (no free carriers) so you would not measure a voltage across the non-conductor.</p>
| 376
|
electromagnetism
|
How to apply Guass Law to Voltages
|
https://physics.stackexchange.com/questions/29950/how-to-apply-guass-law-to-voltages
|
<p>So, I know $\oint E\centerdot dA = 4\pi Q_{enc}$</p>
<p>I'm trying to solve for a TEM mode with two concentric (infinite) cylindrical wave guides of radius a and b, $a<b$. I know that for TEM modes, I can solve by assuming that the outside and inside are at two different potentials, $\pm V$. </p>
<p>I'm told the solution is $\vec E=a\sqrt {\mu/\epsilon}H_0\hat r/r$. It seems to me that the solutions found $\vec H$ first, and then found $\vec E$. I should be able to do the reverse, and end up with the same answer. So, my question is, how can I apply Gauss' Law in this situation? Or, is there simply a better way to solve for $\vec E $ and $\vec H$?</p>
| 377
|
|
electromagnetism
|
Farady's law and div B = 0
|
https://physics.stackexchange.com/questions/30405/faradys-law-and-div-b-0
|
<p>I'm reading a book on electromagnetism and I am a bit confused about some things in Maxwells equations. This is what I don't like about many physics books: they are very wordy, but at the end you don't know what is an experimental fact, what is a "theorem", what is an assumption and so on,...</p>
<p>Anyway the questions are:</p>
<ol>
<li><p>Is Faraday's law a consequence of Coulomb's and Biot-Savart's (or Ampere's) law?</p></li>
<li><p>Is divB = 0 consequence of Biot-Savart's law i.e. if we found the magnetic monopole, would that mean that Biot-Savart's law is not true? (I think so, but I am not 100% sure)</p></li>
</ol>
|
<ol>
<li>No.</li>
<li>No.</li>
</ol>
<p>More precisely, Biot-Savart relates the curl of $B$ to currents, which is independent of specifying the divergence.</p>
| 378
|
electromagnetism
|
Physical interpretation of the Maxwell stress tensor
|
https://physics.stackexchange.com/questions/409568/physical-interpretation-of-the-maxwell-stress-tensor
|
<p>In 'Introduction to Electrodynamics' by D. Griffiths, shortly after introducing the Maxwell stress tensor there is a paragraph concerning the physical interpretation of the stress tensor $\boldsymbol{T}$</p>
<blockquote>
<p>Physically, $\boldsymbol{T}$ is the force per unit area (or stress) acting on the surface. More precisely, $T_{ij}$ is the force (per unit area) in the $i$th direction acting on an element of the surface oriented in the $j$th direction - "diagonal" elements ($T_{xx}$, $T_{yy}$, $T_{zz}$) represent pressures, and "off-diagonal" elements ($T_{xy}$,$T_{xz}$, etc.) are shears.</p>
</blockquote>
<p>I understand where all this comes from mathematically, but I fail to grasp how this translates into an actual force, and specifically what is meant by "an element of the surface <em>oriented</em> in the $j$th direction". Any clarification would be greatly appreciated.</p>
|
<p>It means a surface element whose tangent plane has a normal in the $j$th direction. For a flat surface, we can shorten that to the normal to the surface pointing in the $j$th direction. For a curved surface, each infinitesimal patch has its own tangent plane.</p>
| 379
|
electromagnetism
|
Do magnets wear out?
|
https://physics.stackexchange.com/questions/411253/do-magnets-wear-out
|
<p>Can a magnet ever wear out or lose strength?</p>
<p>If you break a magnet it (seemingly) gets weaker, but what about from normal use?</p>
<p>Or even very heavy use, like placing 2 magnets facing each other, so that they detract from each other, does that strain cause it to wear quicker?</p>
<hr>
<p><sub>(Note, I'm not looking for a merely yes or no answer; If yes, what will cause it to wear out quicker or slower. If no, why?)</sub></p>
|
<p>Yes, a magnet, as time passes, will lose part of his strength. There are two main reasons: </p>
<ol>
<li><strong>Thermal energy</strong>: it causes the disorientation of the atomic magnetic momenta.</li>
<li>If you have a bar magnet free in space it’s easy to see (using <a href="https://en.m.wikipedia.org/wiki/Amp%C3%A8re%27s_circuital_law" rel="noreferrer">Ampère’s law</a>) that there is inside it a magnetic field $H$ opposite to the magnetisation of the magnet. In order to avoid this phenomenon you should anchor it (that is to say “linking north with south pole“) with a ferromagnet. </li>
</ol>
<p>This two phenomena will cause atomic magnetic momenta to disorient, and, in so doing, the magnetic strength of the magnet will decrease.</p>
<p>The demagnetisation happens even if you apply a sufficiently strong magnetic field opposite to the one generated by the magnet.</p>
<p><strong>EDIT</strong>:
The proof of the existence of a field $H$ inside the magnet is now reported: let’s take a bar magnet as shown in figure <a href="https://i.sstatic.net/PUPba.jpg" rel="noreferrer"><img src="https://i.sstatic.net/PUPba.jpg" alt="Magnet bar with magnetisation $M$ that produces magnetic field $B$"></a></p>
<p>The Ampère’s law tells us that
$$
\int_\gamma H ds =0\; .
$$
Now let’s call $\gamma_1$ the piece of curve inside the magnet and $\gamma_2$ the piece outside with length respectively $L_1$ and $L_2$. The Ampère’s law becomes
$$
\int_{\gamma_1}Hds +\int_{\gamma_2} Hds =0\; .
$$
Let be $H_1$ the mean $H$ field inside the magnet and $H_2$ the one outside. The integral turns into
$$
H_1L_1+H_2L_2=0
$$
From here we have
$$
H_1=-\frac{L_2}{L_1}H_2=-\frac{L_2}{L_1}\frac{B}{\mu_0}\;
$$
And here we have what was to be demonstrated.</p>
| 380
|
electromagnetism
|
Force on circular loop
|
https://physics.stackexchange.com/questions/413159/force-on-circular-loop
|
<p>Would the forces between two circular loops, carrying currents in the same direction, be attractive or repulsive?
Would the forces between two circular loops, carrying current in the opposite directions, be attractive or repulsive?
How to find a force acting on a circular loop?
Please explain with a picture. </p>
|
<p>This picture sums up the two scenarios pretty well:</p>
<p><a href="https://i.sstatic.net/6NEvN.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/6NEvN.jpg" alt="enter image description here"></a></p>
<p>To give a little bit of context as to what is happening in each scenario:</p>
<h2>1. When the current in each loop is in the same direction</h2>
<p>Using the RH rule, you can determine the direction of the magnetic field. Curl your right hand in the direction of the current along the wire and your thumb points in the direction of the magnetic field. </p>
<p>The field from this loop is similar to the field of a magnet as is shown in the diagram.</p>
<p>When the two currents go the same direction, the magnetic fields point the same direction so when you put them next to one another, the loops will be attracted as if they were two magnets aligned the same direction.</p>
<h2>2. When the current goes in opposite directions</h2>
<p>You apply the same logic as above to find the direction of the induced magnetic field.</p>
<p>This time the magnetic fields point opposite directions, analogous to having two magnets with like ends facing each other.</p>
<p>As with when you put the like ends of magnets together, they repel one another.</p>
| 381
|
electromagnetism
|
Is it correct way of saying that a moving electric field causes magnetic field?
|
https://physics.stackexchange.com/questions/419379/is-it-correct-way-of-saying-that-a-moving-electric-field-causes-magnetic-field
|
<p>$$\mathbf B=\mathbf V \times \mathbf E \mu _0 \epsilon _0$$</p>
<p>$\mathbf V$ is the velocity vector of moving electric field. Rest of the parameters follow usual notations for those.</p>
|
<p>First of all, <strong>fields don't move</strong> because fields are everywhere in space. For example temperature is a field that has a value at every point in space and all the time, but when the sources moves, the field changes. </p>
<p>So, I suppose you mean by "moving electric field" that the electric field change with respect to time in a way that the values become shifted by $\mathbf{v} dt$ for a time interval $dt$. Electric and magnetic fields can change over time according to some situations, like changing sources such as moving charges or alternating currents. However, electric and magnetic fields do not change each other. I will explain how you get the relation in your question at the end of the answer but first I need to clarify some important points.</p>
<h2>Electromagnetic Potentials</h2>
<p>As a matter of fact, <strong>there is no cause-and-effect relation between electric and magnetic fields</strong>. Rather they both caused by more fundamental fields, called electromagnetic potentials, one is scalar and denoted by $\phi (\mathbf{r},t)$, and the other is vector and denoted by $\mathbf{A}(\mathbf{r},t)$. Their derivatives define how the electric and magnetic field will correlate, as follows:
\begin{align}
\tag{1.a}
\mathbf{E} & \equiv - \mathbf{\nabla} \phi - \frac{\partial \mathbf{A}}{\partial t} \\
\tag{1.b}
\mathbf{B} & \equiv \mathbf{\nabla} \times \mathbf{A}
\end{align}
So, counting the degrees of freedom, $\mathbf{E}$ and $\mathbf{B}$ have 6 in total but they can be derived from 4. The reason why Maxwell, Faraday and Ampére didn't start from them is because they were perceived by forces.</p>
<h2>Maxwell's Equations for Potential Fields</h2>
<p>The electric and magnetic fields correlate according to the potentials, and those potentials change with respect to the Maxwell's equations which become as follows, in this <em>potential formulation</em>:
\begin{align}
\tag{2.a}
\Box \varphi + \frac{\partial}{\partial t} \left ( \mathbf{\nabla} \cdot \mathbf{A} + \frac{1}{c^2} \frac{\partial \varphi}{\partial t}\right ) &= - \frac{\rho}{\varepsilon_0}
\\
\tag{2.b}
\Box \mathbf A - \mathbf \nabla \left ( \mathbf \nabla \cdot \mathbf A + \frac{1}{c^2} \frac{\partial \varphi}{\partial t} \right ) &= - \mu_0 \mathbf J
\end{align}
where $\Box = \nabla^2- \frac{1}{c^2} \frac{\partial^2}{\partial t^2}$ is the d'Alembert operator, and $\rho(\mathbf r, t)$ and $\mathbf{J} (\mathbf{r}, t)$ are charge density and current density, respectively. The expression in paranthesis can be eliminated by shifting those potentials, $\mathbf{A} \rightarrow \mathbf{A} - \mathbf{\nabla} \chi$ and $\varphi \rightarrow \varphi + \partial_t \chi$, with respect to a continuous scalar function, $\chi(\mathbf r, t)$, such that
\begin{align}
\tag{Lorenz gauge}
\mathbf {\nabla} \cdot \mathbf{A} + \frac{1}{c^2} \frac{\partial \varphi}{\partial t} = 0
\end{align}
however, due to their definition in (1), electric and magnetic fields will remain unchanged under this shift, i.e. $\mathbf E \rightarrow \mathbf E$ and $\mathbf B \rightarrow \mathbf B$. Therefore,
\begin{align}
\tag{3.a}
\Box \varphi &= - \frac{\rho}{\varepsilon_0}
\\
\tag{3.b}
\Box \mathbf A &= - \mu_0 \mathbf{J}
\end{align}</p>
<h2>Correlation of E&M Fields Without Sources</h2>
<p>So, your question reduces to how these fields correlate in this way even there is no source to change the potentials. The answer becomes obvious when you write Maxwell's equations of (3) without sources, that is:
\begin{align}
\tag{4}
\Box \varphi = 0
&& and &&
\Box \mathbf A = 0
\end{align}
These are just wave equations. So, if you have an antenna producing a changing $\varphi$ and $\mathbf A$, which is basically a rod that moves charges back and forth, then at really far distance, where the effects of the sources become negligible, there would still be a wave traveling to that distance. This kind of wave is called electromagnetic wave since it has both electric and magnetic components. So, EM wave is an energy transferred by the electromagnetic fields because of some source that is no longer effecting it.</p>
<h2>Moving Sources</h2>
<p>One can ask <em>then how moving charges produce magnetic field, if the field isn't moving?</em> The answer is right there: moving charge means a nonzero current density which produces vector potential according to (3.b), which produces magnetic fields due to (1.b).</p>
<p>However, one can also obtain this result by just switching from an observer A who sees the charge is at rest to an observer B that sees the charge is moving at a velocity, $\mathbf{v}$. Observer A experiences $ \Box \varphi = -\rho / \epsilon_0$ and, say, $\mathbf{A} =0$. In order to obtain what Observer B experiences, one needs to use Lorentz transformations, because Electrodynamics is a Lorentz invariant theory. So, one needs to construct the kinematics of classical electrodynamics on the theory of Special Relativity.</p>
<p>Hence, Lorentz transformations of the potential fields are as follows:
\begin{align}
\varphi' &= \gamma (\varphi - \mathbf{v} \cdot \mathbf{A} ) \\
\mathbf {A}'_{||} &= \gamma (\mathbf{A} _{||} - \mathbf{v} \varphi / c^2 ) \\
\mathbf {A}'_\perp & = \mathbf{A} _\perp
\end{align}
where $\mathbf{A}_{||}$ and $\mathbf {A}_\perp$ are parallel and perpendicular components of the vector potential, and $\gamma = 1/\sqrt{1-v^2/c^2}$ is the Lorentz factor.</p>
<p>Therefore, Observer B will have the following field values in terms of Observer A's fields according to Lorentz transformations:
\begin{align}
\varphi^B &= \gamma \varphi^A \\
\mathbf{A}^B &= -\gamma \mathbf{v} \varphi^A / c^2
\end{align}
which gives a nonzero magnetic field as
\begin{align}
\mathbf{B}^B &= \nabla \times \mathbf{A}^B\\
&= - \frac{1}{c^2}\left( \varphi^B \nabla \times \mathbf{v} - \mathbf{v} \times \nabla \varphi^B \right) \\
&= \frac{1}{c^2} \mathbf{v} \times \mathbf{E}^B
\end{align}
where $\mathbf v$ is constant (or at least curl-free since we only switch to inertial observers).</p>
| 382
|
electromagnetism
|
Electromagnetism Ampere's law Application to solenoid
|
https://physics.stackexchange.com/questions/423382/electromagnetism-amperes-law-application-to-solenoid
|
<p>Electromagnetism;Ampere's Law
Application for finding magnetic field strength(B) inside a current carrying solenoid</p>
<p>Question is that why we multiply the current in one loop to the number of turns(enclosed in amperian rectangular loop) ALTHOUGH the current flowing(charges flowing per unit time) is SAME through all the loops???
There is a SINGLE complete circuit
If there would be more than one circuits comprising each loop, then i think we should add all currents in individual loops
BUT in this situation there is a single circuit...
I hope my question is clear</p>
|
<p>Answer by @Frecher is correct! I will try to give physical essence to your question.</p>
<p>Each loop of current generates its own magnetic field regardless of whether they are part of the same circuit or not. Think of two loops which are in the same circuit and have the same current flowing through them. Now separate the two loops very very far away from each other (you have very long wire!). Now they don't interact with each other but they still generate the magnetic field at their own places. This means each loop must be generating magnetic field even though they are part of the same circuit and the same current flows through them!</p>
<p>All this is actually incorporated automatically in Ampere's law.</p>
| 383
|
electromagnetism
|
How is the waveform of an electromagnetic radiation detected and generated by a resonator (like in FM Radios)?
|
https://physics.stackexchange.com/questions/424907/how-is-the-waveform-of-an-electromagnetic-radiation-detected-and-generated-by-a
|
<p>We know that a resonator consists of an inductor and capacitor. And we also know that tuning them in a specific way will get an Electromagnetic radiation detected with similar characteristics. But when the frequency of the original EM changes how do we detect the change(Like In the FM Radios, the frequency changes but still the same wave is detected)? How and Why? </p>
|
<p>A pure sine wave RF signal does not carry any information and an ideal resonant circuit would not be able to receive information.</p>
<p>In order to pass some information with an RF carrier, it has to be modulated somehow.</p>
<p>When a carrier is modulated, regardless of the modulation method (AM, FM, etc.), it is not a pure sine wave any more - it contains a band of frequencies centered around the carrier frequency, corresponding to the band of frequencies present in the modulating signal, which could be a single tone or a continuous range of frequencies, associated with a speech. </p>
<p>So, a single spectral line, corresponding to a carrier sine wave, is transformed into a relatively wide band of frequencies. We can say that a modulated carrier has some bandwidth.</p>
<p>In order to receive a modulated signal, the resonator of a receiver has to have a similar bandwidth or a relatively wide resonant peak. If the peak was too narrow, part (high frequencies) of the useful signal would be cut off. If the peak was too wide, the receiver could pick up frequencies belonging to other radio stations. </p>
| 384
|
electromagnetism
|
How do you measure and distinguish between E and D fields, and B and H fields
|
https://physics.stackexchange.com/questions/429980/how-do-you-measure-and-distinguish-between-e-and-d-fields-and-b-and-h-fields
|
<p>I’m pretty familiar with maxwells equations, light waves, fields and materials etc, but I’m not clear on how the the various fields are actually measured. How is this typically done? Can the fields inside a material be directly measured?</p>
| 385
|
|
electromagnetism
|
forces on moving charges in a magnetic field
|
https://physics.stackexchange.com/questions/430088/forces-on-moving-charges-in-a-magnetic-field
|
<p>When a conductor is moving relative to a magnetic field (for example a magnet falling in copper pipe, or a Eddy current brake in a train) , it is considered that the conductor moves, the conductor contains electrons, therefore the electrons are moving relative to the magnetic field, therefore EMF is generated.</p>
<p>My question is relative to the positive charges in the atom nucleii that are also moving relative to the magnetic field when the metal is moving, are they too generating a force ? </p>
|
<blockquote>
<p>My question is relative to the positive charges in the atom nucleii
that are also moving relative to the magnetic field when the metal is
moving, are they too generating a force?</p>
</blockquote>
<p>The emf generated in a conductor due to its movement in a magnetic field will be acting or exerting some force on both negative and positive changes: free electrons will move, while nuclei with the rest of the electrons will stay put, due to the internal forces that keep the metal crystal together. </p>
<p>In this sense, it is no different than the electric field acting on negative and positive charges inside a conductor attached to a battery. </p>
| 386
|
electromagnetism
|
What if... you had a bowl of electrons?
|
https://physics.stackexchange.com/questions/431751/what-if-you-had-a-bowl-of-electrons
|
<p>My chemistry teacher used to tell us that if you had a soup bowl with only electrons in it, the explosion could make you fly to Pluto. Was he right? Could this happen?</p>
|
<p>The answer would depend how densely the electrons are packed. Let's say we have 1 kg of electrons, meaning we would have about <span class="math-container">$N = 10^{30}$</span> of them. For simplicity, let's approximate by arranging all of these electrons arranged in a spherical shell of radius <span class="math-container">$r=0.1$</span> meters. By symmetry, the voltage at the location of each charge would be constant and would have the following value.</p>
<p><span class="math-container">$$V = -\frac{1}{4\pi \epsilon_0}\frac{Ne}{r}$$</span></p>
<p>We can thus compute the potential energy <span class="math-container">$U = \frac{1}{2}Q V$</span> associated with this configuration.</p>
<p><span class="math-container">$$U = \frac{1}{2} \cdot Ne \cdot \frac{1}{4\pi \epsilon_0}\frac{Ne}{r} = \frac{N^2 e^2}{8 \pi \epsilon_0 r} \approx \boxed{1.15 \times 10^{33} \text{ Joules}}$$</span></p>
<p>Ignoring air friction, the Earth's escape velocity is <span class="math-container">$v =11200$</span> meters / second, which for a <span class="math-container">$m=2\times 10^6$</span> kg space shuttle, would only require <span class="math-container">$E = \frac{1}{2} mv^2 = 1.25 \times 10^{14}$</span> Joules.</p>
<p>So, yeah, you'd have way more than plenty sufficient energy.</p>
| 387
|
electromagnetism
|
How can electromagnetic waves heat non-conducting media?
|
https://physics.stackexchange.com/questions/432949/how-can-electromagnetic-waves-heat-non-conducting-media
|
<p>According to <a href="https://ws680.nist.gov/publication/get_pdf.cfm?pub_id=906602" rel="nofollow noreferrer">this</a> source, the divergence of the Poynting vector is related to the total energy density of an electromagnetic wave, which is (locally) expressed as</p>
<p><span class="math-container">$$-\nabla\cdot S=EJ+(E\frac{\partial D}{\partial t}+H\frac{\partial B}{\partial t})$$</span></p>
<p>I see, that the first term on the R.H.S. corresponds to Joule heating. But it's only present if there are free charge carriers. The 2nd term is called "radiative loss" in the source I cited above. I'm not sure how to interpret it. I believe it's just "locally" a loss. The wave moves along and carries momentum and energy with it, which implys that this "loss" does not correspond to a loss globally.</p>
<p>So how is it then, that radiation heats media with no free charge carriers?
Or does the radiation induce free carriers by photo-ionization first?</p>
<p>EDIT: I'm also wondering if attenuation of intensity, as an EM-wave travels through a medium, is always necessarily caused by absorption of light. Is the 2nd term able to attenuate the EM-wave in absence of the first one on the RHS?</p>
|
<blockquote>
<p>So how is it then, that radiation heats media with no free charge carriers? Or does the radiation induce free carriers by photo-ionization first?</p>
</blockquote>
<p>This is an excellent question, very well reasoned. As other answers have described there are various microscopic processes whereby non-conductors can absorb EM energy. But there should also be a macroscopic equation that reflects the situation where a material is filled with such microscopic processes. Indeed, there is.</p>
<p>As you have ascertained, the expression your source gave is not the most general expression for macroscopic EM. A more general expression is given in this MIT electromagnetism textbook: <a href="https://web.mit.edu/6.013_book/www/book.html" rel="nofollow noreferrer">https://web.mit.edu/6.013_book/www/book.html</a></p>
<p>In section 11.2 they derive the following more general form: <span class="math-container">$$ -\nabla \cdot (E\times H)=\frac{1}{2}\epsilon_0 \frac{\partial}{\partial t} E^2 + \frac{1}{2}\mu_0 \frac{\partial}{\partial t} H^2 + E \cdot J_u + H \cdot \mu_0 \frac{\partial}{\partial t} M + E \cdot \frac{\partial}{\partial t} P$$</span></p>
<p>In particular, note the last term <span class="math-container">$E\cdot \frac{\partial}{\partial t}P$</span>. As an E field acts on a non-conductor there are no free charges to produce free current or charge, but there are bound charges that can be polarized and create bound charge density. Work can be done on non conductive matter in this way. Thereby resolving your dilemma.</p>
<p>Notice also the other new term <span class="math-container">$H \cdot \mu_0 \frac{\partial}{\partial t} M$</span>. This a major macroscopic means of energy transfer for an induction stovetop.</p>
| 388
|
electromagnetism
|
What causes a moving positive point charge moving right in a uniform into the page magnetic field to specifically move upwards? why not downwards?
|
https://physics.stackexchange.com/questions/434268/what-causes-a-moving-positive-point-charge-moving-right-in-a-uniform-into-the-pa
|
<p>What causes a moving positive point charge moving right in a uniform into the page magnetic field to specifically move upwards? why not downwards?</p>
<p>Upwards and downwards (on the plane of the paper) are viable options. <strong>why does the charge only move upwards?</strong></p>
<p>Doesn't this violate the natural laws of symmetry? Are we just supposed to accept this? Why is the right hand rule more special than the left hand rule?</p>
<p>The direction of the magnetic field can be given by a unit north pole easily (uniformly into the page). So isn't the direction of the magnetic field already fixed?</p>
<p>The reason I ask this is because many posts imply that electromagnetism is parity invariant as we use the right hand rule <em>twice</em> in many cases, negating its arbitrariness, but here we use it only once.</p>
|
<blockquote>
<p>The reason I ask this is because many posts imply that electromagnetism is parity invariant as we use the right hand rule twice in many cases, negating its arbitrariness, but here we use it only once.</p>
</blockquote>
<p>That last bit is incorrect. If you want a description for the dynamics which depends strictly on vector objects, then you do use the right-hand rule twice.</p>
<p>The core of the field picture is that the only thing that matters is the local value of the field, and not how it was produced. In many situations, there will be multiple different possible origins for how a given local field configuration was produced, and we're free to choose whichever provides the most convenient analysis.</p>
<p>In your case, where you have a particle confined to the <span class="math-container">$x,y$</span> plane of the page with a uniform magnetic field going into the page, the cleanest analysis is to provide that magnetic field using a pair of Helmholtz coils above and below the page:</p>
<p><a href="https://i.sstatic.net/GK6q4.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/GK6q4.png" width="450"></a></p>
<p>And, once you do that, it's clear that those Helmholtz coils have an intrinsic circularity, and they break the symmetry between 'up' and 'down' for the particle. If you want the particle to go the other way, then change the direction of the current in the coils.</p>
<p>Now, here you'll probably have a perfectly valid observation - that this still doesn't answer the full analysis for the case where the magnetic field is produced by a permanent magnet. Does that require an only-once usage of the right-hand rule? Well, yes and no.</p>
<ul>
<li><p>From one perspective, as far as macroscopic electromagnetism goes, a permanent magnet's effect is basically indistinguishable from the <a href="https://en.wikipedia.org/wiki/Magnetization#Magnetization_current" rel="nofollow noreferrer">magnetization current</a> on its surface. At that level of analysis, a permanent magnet is exactly identical to a coil of wire carrying that current, which brings you back to the case from above.</p></li>
<li><p>On the other hand, the physical analysis that allows us to underpin that magnetization current with intuition generally asks that we picture a magnetized material as if it were a bunch of tiny current loops, but that's not really the case. The magnetization in ferromagnetic materials comes from the macroscopic organization of the intrinsic magnetic dipole moments of the electron spins inside the material, and those cannot be thought of as tiny current loops.</p>
<p>Nevertheless, both the magnetization and the electron spin that underpins it are still pseudovector objects, and they're still associated with a sense of rotation on the plane of the page. "Spin" the electrons the other way (i.e. give them an angular momentum in the opposite sense) and you'll change the direction of the magnetic field.</p></li>
</ul>
<p>So, actually, no and no: you never require only-once applications of the right-hand rule. It's just that permanent magnets are inherently chiral objects, and when we say "north" or "south" regarding a pole of a magnet, we're really specifying a rotation direction about that axis.</p>
| 389
|
electromagnetism
|
What exactly is magnetic flux?
|
https://physics.stackexchange.com/questions/439108/what-exactly-is-magnetic-flux
|
<p>It is intuitive to think of an electric field, which describes the variance in the force acting on a charged particle if it were located in a certain position. However it is not so easy to understand what magnetism is and what flux is.
Apparently magnetism is an effect of special relativity for moving charges, but this is also hard to understand.
What is magnetic flux?</p>
|
<blockquote>
<p>It is intuitive to think of an electric field, which describes the variance in the force acting on a charged particle if it were located in a certain position</p>
</blockquote>
<p>The electric field <span class="math-container">$\mathbf E$</span> is actually just a force per unit charge <span class="math-container">$(q\mathbf E=\mathbf F)$</span>, where <span class="math-container">$q$</span> is the charge the electric force is acting on. It does not describe a variance of a force.</p>
<blockquote>
<p>However it is not so easy to understand what magnetism is and what flux is. Apparently magnetism is an effect of special relativity for moving charges, but this is also hard to understand. What is magnetic flux?</p>
</blockquote>
<p>So it is true that electric and magnetic fields can be viewed as parts of the same thing in SR, but this view is not needed to understand magnetic flux. Similar to our above discussion about <span class="math-container">$\mathbf E$</span>, the magnetic field <span class="math-container">$\mathbf B$</span> can also be described by the force it exerts on a charged particle moving with velocity <span class="math-container">$\mathbf v$</span>.
<span class="math-container">$$\mathbf F=q\mathbf{v}\times\mathbf{B}$$</span></p>
<p>A good way most introductory physics classes explain the direction of the magnetic field is that it points in the direction a compass would point. Of course, we can determine the entire (magnitude and direction) magnetic field from its current sources using the Biot-Savart law in the same way we can determine the electric field from its charge sources using Coulomb's law, but this is not the crux of your question.</p>
<p>Magnetic flux is a specific case of the more general idea of the flux of a vector field <span class="math-container">$\mathbf V$</span>, whose origin comes from fluid dynamics. The flux <span class="math-container">$\Phi$</span> of a vector field across some surface is given by the surface integral
<span class="math-container">$$\Phi=\int \mathbf V\cdot\text{d}\mathbf A$$</span></p>
<p>Where <span class="math-container">$\text{d}\mathbf A$</span> is an infinitesimal area element vector whose direction is normal to the surface.</p>
<p>This can be applied to the magnetic field, where then you would just replace <span class="math-container">$\mathbf V$</span> with <span class="math-container">$\mathbf B$</span>. Flux tells you "how much field is flowing through the surface". Since the integral depends on the dot product <span class="math-container">$\mathbf V\cdot\text{d}\mathbf A$</span>, you can see that if the field is normal to the surface then we get maximum flux (maximum flow), but if the field is parallel to the surface we get no flux (no flow).</p>
| 390
|
electromagnetism
|
Is the amplitude of an EM wave the combination of the electric AND magnetic fields added together?
|
https://physics.stackexchange.com/questions/439761/is-the-amplitude-of-an-em-wave-the-combination-of-the-electric-and-magnetic-fiel
|
<p>For instance, to get the TOTAL energy of an EM wave(s) or intensity you square the amplitude. But do you first add or combine the strengths of the e and m fields?</p>
|
<p>Suppose you have two sources of Electric field, <span class="math-container">$E_1$</span>, and <span class="math-container">$E_2$</span></p>
<p>Then <span class="math-container">$\vec{E_{tot}}=\vec{E_1}+\vec{E_2}.$</span></p>
<p>So the total intensity is
<span class="math-container">$$E^2_{tot}=(\vec{E_1}+\vec{E_2})^2=E^2_1+E_2^2+2\vec{E_1}\cdot\vec{E_2}$$</span></p>
<p>So the intensity of the sum of two electric fields is not the sum of their intensities individually. </p>
<p>The same thing applies to a magnetic field. </p>
<p>Now both the E field and the B field can carry energy. Not only is there the complexity of the above, summing the associated intensities has it's own complications. </p>
<p>Are you familiar with the Poyting Vector? </p>
<p><span class="math-container">$$\vec{S}=\frac{1}{\mu_0}\vec{E}\times\vec{B}$$</span></p>
<p>It satisfies the energy conservation equation of electro magnetic waves. </p>
<p><span class="math-container">$$\nabla\cdot \vec{S}+\frac{\partial u}{\partial t}=0$$</span></p>
<p><span class="math-container">$$u=\frac{1}{2}\epsilon_0E^2+\frac{1}{2\mu_0}B^2$$</span></p>
<p><span class="math-container">$u$</span> is the energy desnity of the fields. The Poyting Vector indicates the direction of the flow of energy of the EM field. </p>
<p>So the energy density is a linear combination of the sum of their intensities, but neither exactly the sum of their intensities nor the intensities of their sums.</p>
| 391
|
electromagnetism
|
Can concentrating light increase the intensity of the electromagnetic field?
|
https://physics.stackexchange.com/questions/441765/can-concentrating-light-increase-the-intensity-of-the-electromagnetic-field
|
<p>(Correct where applicable)</p>
<p>Light is a wave on the electromagnetic field (or the electromagnetic field can be interpreted as a representation of the magnitude and direction of the force photons will have on another charge).</p>
<p>When we concentrate light, we are essentially changing the paths of the electromagnetic waves or photons so they are closer to each other. Since electromagnetic waves can be said to follow field lines, and in the photon interpretation, the path a photon takes is the field line (and yes, in quantum physics a photon takes all possible paths, so the field lines of a single photon are all over the place), and the electromagnetic field intensity increases when field lines are closer together, does concentrating light also increase electromagnetic field intensity?</p>
|
<p>Yes there is a bigger EM field when photons are concentrated. But the field lines of a photon are perpendicular to the line of travel. Note that the field is quantized in energy, so even though it looks like a bigger wave only part of it corresponding to the single photon energy can be absorbed at a time. Also although the photon sees many paths as its EM field radiates out, most of the energy in concentrated in a small space, but yes a photon going through a hula hoop still gets diffracted slightly. </p>
| 392
|
electromagnetism
|
How do virtual photons exist without violating conservation of energy?
|
https://physics.stackexchange.com/questions/441939/how-do-virtual-photons-exist-without-violating-conservation-of-energy
|
<p>If virtual particles cancel out after being created from spacetime flunctuations because they come in matter-antimatter pairs, how do virtual photons cancel out, due to a photon being its own antiparticle?</p>
| 393
|
|
electromagnetism
|
Reason for force in Stern-Gerlach experiment
|
https://physics.stackexchange.com/questions/447457/reason-for-force-in-stern-gerlach-experiment
|
<p>I'm currently working at an assignment, and I'm having some trouble understanding how the magnetic field deflects the silver atoms passing trough it. From what I understood, the atoms are deflected up or down of a specific amount according to their magnetic moment, but I can't understand what is causing this phenomenon. I'm an engineering student and physics isn't my main field, so I would prefer that the explanation is not too specific. Thanks in advance for your help.</p>
|
<p>If an atom has a magnetic <em>moment</em> (not momentum), that means it acts like a tiny dipole magnet -- like a tiny bar magnet. A uniform magnetic field will not exert a net force on a dipole magnet, because (thinking of it as a bar magnet) the magnetic field will push on one pole and pull on the other pole with equal but opposite force.</p>
<p>However, if the magnetic field is nonuniform - if it diverges upward, for example - you can imagine that when the dipole is aligned with the field, the field is stronger at the location of one of the poles than at the other because of the divergence. Let's say the nonuniform field is due to the N pole of an electromagnet. Then the N pole of the dipole is repelled (upward in this scenario) and the S pole of the dipole is attracted downward. But the <em>net</em> force (upward minus downward) depends on whether the dipole is aligned with its North pole up or its S pole up. In the former case, the net force is downward and in the latter case the net force is upward.</p>
<p>The key result of the Stern-Gerlach experiment is not that the atoms are deflected, but that they are deflected by a fixed amount upward or downward - with nothing in between. This is proof of a quantum property of spin, which relates to the magnetic moment: projection of the spin onto any axis (in this case an axis aligned with the diverging magnetic field) is quantized. "Quantized" means the spin angular momentum along an axis can only have certain discrete values.</p>
| 394
|
electromagnetism
|
Why does a moving charge create electricity
|
https://physics.stackexchange.com/questions/449075/why-does-a-moving-charge-create-electricity
|
<p>Now i have been studying a chapter called current electricity and i found out that moving chages can create electricity why is this possible? Is it the holes and the electrons combining together and creeating heat and light and us pecieving it as electricity?
I haven't put much thought into it but i am also very impatient about knowing what it really is.</p>
|
<p>The term “electricity” doesn’t necessarily refer to any one specific thing, but rather to a whole class of phenomena related to charge, current, voltage, etc. However, it isn’t unreasonable to focus on current and say that electricity is current. So in the remainder of this answer I will talk about current. </p>
<p>More specifically, when a charge density, <span class="math-container">$\rho$</span>, moves with a velocity, <span class="math-container">$\mathbf v$</span>, there is a quantity, <span class="math-container">$\mathbf J = \rho \mathbf v$</span>. It turns out that this quantity is very useful, so we give it a name: current density. Once you integrate the current density over the cross section of a wire you get current. So, roughly speaking moving charges don’t “create” current, they are current. It is just a matter of definition, but it is a very useful definition for studying electric circuits. </p>
| 395
|
electromagnetism
|
Magnetic field inside a current carrying wire
|
https://physics.stackexchange.com/questions/454909/magnetic-field-inside-a-current-carrying-wire
|
<p>Assume an infinite wire carrying DC current. According to Ampere's force law， the moving electrons inside the wire are influenced by the Lorenz force pointing to the center. As a result, the current distribution is changed and finally all current flows in the center.</p>
<p>My question is, what is wrong with the above analysis? Are there other forces on the electrons for balancing so the uniform current distribution can be sustained?</p>
|
<p>You're right, the usual picture of a <em>uniform</em> current density is a mathematical idealization. Realistically, electrons will accumulate near the axis of the wire in such a way as to create an outward pointing electric field that negates this effect. The equilibrium current distribution will no longer be uniform.</p>
<p>Alternatively, one might approximate a uniform distribution by replacing the wire by many parallel thin wires, each one insulated from the others.</p>
| 396
|
electromagnetism
|
Can a coil and a magnet moving together produce a voltage?
|
https://physics.stackexchange.com/questions/458916/can-a-coil-and-a-magnet-moving-together-produce-a-voltage
|
<p>In other words, if a coil is wound around a fixed magnet and the combination is rotated or otherwise moved, will the coil produce emf? Must the permanent magnet move relative the the coil to produce emf? I recall seeing a demonstration during a physics presentation back in the 70s where this was the case. The magnet and coil were spinning together, producing a few watts. But perhaps my memory is faulty, or I am mistaken that the coil was in fact attached to the magnet.</p>
|
<p>Have you heard about the <a href="https://commons.m.wikimedia.org/wiki/File:Faraday_disk_generator.jpg" rel="nofollow noreferrer">Faraday disk generator</a>?</p>
<p><a href="https://i.sstatic.net/wihRm.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/wihRm.jpg" alt="Faraday Disc Generator"></a><br>
A is a horseshoe magnet,<br>
B a rotating metallic disc</p>
<p>Such device produces a current not only when the disc is rotated but also in the case, the magnet is rotating together with the disc.</p>
<p>What is the mechanism behind the phenomenon of a <a href="https://en.wikipedia.org/wiki/Homopolar_generator" rel="nofollow noreferrer">homopolar generator</a>?</p>
<p><a href="https://i.sstatic.net/gm9E5m.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/gm9E5m.jpg" alt="https://en.wikipedia.org/wiki/Homopolar_generator"></a></p>
<p>According to Wikipedia </p>
<blockquote>
<blockquote>
<p>The motion is azimuthal and the field is axial, so the electromotive force is radial. ... If the magnetic field is provided by a permanent magnet, <em>the generator works regardless of whether the magnet is fixed to the stator or rotates with the disc</em>.</p>
</blockquote>
</blockquote>
<p>Our conclusion should be that electrons under a centrifugal force and in a constant magnetic field are moving according the <a href="https://en.m.wikipedia.org/wiki/Lorentz_force" rel="nofollow noreferrer">Lorentz force</a>, the current is perpendicular to both the movement and the magnetic field. And we have to add that this movement should be accompanied by the centrifugal acceleration of the electrons. Moving the disc and the magnet along a line does not induce any current.</p>
<p>Following the above said I try to design a rotating coil with an attached magnet in such a way, it will induce a current. I see no way, but maybe you’ll remember it, applicating the above said. </p>
| 397
|
electromagnetism
|
Can a magnetic field be induced without an electric field?
|
https://physics.stackexchange.com/questions/463014/can-a-magnetic-field-be-induced-without-an-electric-field
|
<p>Can a magnetic field be induced without an electric field?
Because, as far as I know, a time varying electric field induces a magnetic field an vice versa.
But in the case of conductors carrying currennt, it doesn't seem that electric field varies with time, then how is a magnetic field induced?</p>
|
<p>One of Maxwell’s four equations for electromagnetism in a vacuum shows how magnetic fields are produced:</p>
<p><span class="math-container">$$\nabla\times\mathbf{B}=\frac{1}{c}\left(4\pi\mathbf{J}+\frac{\partial\mathbf{E}}{\partial t}\right).$$</span></p>
<p>(I’ve written it in Gaussian units.)</p>
<p>From this equation you can see that there are <em>two different</em> sources for magnetic fields: the first is a current density, and the second is a changing electric field.</p>
<p>So to have a magnetic field you do <em>not</em> need to have a time-varying electric field. You can just have moving charge. But when a magnetic field is produced by moving charge, physicists don’t call it “induced”.</p>
| 398
|
electromagnetism
|
What is electric flux density?
|
https://physics.stackexchange.com/questions/464020/what-is-electric-flux-density
|
<p>We say the electric flux is the number of field "lines", thus electric flux density is the number of field "lines" per a given area. However, let's say we had a point charge <span class="math-container">$Q$</span> centered at the origin and we were to enclose this charge with a surface of radius <span class="math-container">$R$</span>. If we wanted to integrate the electric flux density over that surface, why do we get the total charge enclosed instead of the electric flux? I know this mathematically makes sense because we can write Gauss's law and multiply both sides by epsilon to yield this result, but I don't think this makes sense with what an electric flux density "is". Do I have a misunderstanding of what an electric flux density "is"? Please keep the math simple as I am only in first year university with limited knowledge of multivariate calculus.</p>
<p><span class="math-container">\begin{align}
\int \vec{E} \cdot ds &= Q / \varepsilon \rightarrow \varepsilon \int \vec E \cdot ds = Q_\text{enclosed} \\
\int \vec D \cdot ds &= Q_\text{enclosed} \qquad \vec D = \varepsilon \vec E
\end{align}</span></p>
|
<p>It seems that you're likely to be using non-standard terminology. </p>
<p>If I understand your comments correctly, you seem to be associating the name "electric flux density" with the vector field <span class="math-container">$\vec D= \varepsilon \vec E$</span>. If that is the case, then this is definitely non-standard terminology. The <span class="math-container">$\vec D$</span> field certainly does suffer from a lack of a universally-accepted name (though the problem is less bad than in magnetism), but a better name is <a href="https://en.wikipedia.org/wiki/Electric_displacement_field" rel="nofollow noreferrer">the one used in Wikipedia</a>, "electric displacement field". Using the name "electric flux density" for <span class="math-container">$\vec D$</span> is acceptable when writing in a generic context, but it is vital that it be clearly identified as such when it is introduced.</p>
<p>To be clear: the electric flux across a surface <span class="math-container">$S$</span> is defined as
<span class="math-container">$$
\Phi(S) = \iint_S \vec E(\vec r) \cdot\mathrm d\vec S,
$$</span>
in terms of the electric field <span class="math-container">$\vec E$</span> itself. If the surface <span class="math-container">$S$</span> is closed, then the electric flux is still
<span class="math-container">$$
\Phi(S) = \oint_S \vec E(\vec r) \cdot\mathrm d\vec S
$$</span>
and it just happens, because of Gauss's law, to coincide with <span class="math-container">$Q_S/\varepsilon_0$</span>, i.e. the total charge enclosed by <span class="math-container">$S$</span> divided by the vacuum permittivity. Moreover, if you're dealing with the electric field <span class="math-container">$\vec E$</span> produced by a free-charge distribution that's embedded in a homogeneous, isotropic linear dielectric with permittivity <span class="math-container">$\varepsilon$</span>, then <span class="math-container">$\Phi(S)$</span> can also be seen to equal <span class="math-container">$Q_{S,\mathrm{free}}/\varepsilon$</span>, where <span class="math-container">$Q_{S,\mathrm{free}}$</span> is the free charge enclosed by <span class="math-container">$S$</span>.</p>
<p>Because of this, the direct understanding of the term "electric field density" is to assign that to the vector field <span class="math-container">$\vec E$</span> itself, since it is the vector field whose surface integrals give the electric flux.</p>
<p>It is possible to start re-defining those terms so that you have a bit more operational room in how you distinguish between <span class="math-container">$\vec E$</span> and <span class="math-container">$\vec D$</span>, though you run the risk of creating an extremely confusing situation. It looks to me that what's happened is that your lecturer has chosen a confusing choice of terminology and that's led you into some conceptual contradictions which are impossible to clear up within that framework. However, without seeing the notes in full as provided directly by your lecturer, it's impossible to tell that for sure.</p>
| 399
|
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