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jaoe38c1lsfmego3 | physics | electromagnetic-induction | motional-emf-and-eddy-current | <p>A horizontal straight wire $$5 \mathrm{~m}$$ long extending from east to west falling freely at right angle to horizontal component of earths magnetic field $$0.60 \times 10^{-4} \mathrm{~Wbm}^{-2}$$. The instantaneous value of emf induced in the wire when its velocity is $$10 \mathrm{~ms}^{-1}$$ is _________ $$\tim... | [] | null | 3 | <p>$$\begin{aligned}
& \mathrm{B}_{\mathrm{H}}=0.60 \times 10^{-4} \mathrm{~Wb} / \mathrm{m}^2 \\
& \text { Induced emf e}=\mathrm{B}_{\mathrm{H}} \mathrm{v} \ell \\
&=0.60 \times 10^{-4} \times 10 \times 5 \\
&=3 \times 10^{-3} \mathrm{~V}
\end{aligned}$$</p> | integer | jee-main-2024-online-29th-january-evening-shift | 10,297 |
1lsgxlz5q | physics | electromagnetic-induction | motional-emf-and-eddy-current | <p>A ceiling fan having 3 blades of length $$80 \mathrm{~cm}$$ each is rotating with an angular velocity of 1200 $$\mathrm{rpm}$$. The magnetic field of earth in that region is $$0.5 \mathrm{G}$$ and angle of dip is $$30^{\circ}$$. The emf induced across the blades is $$\mathrm{N} \pi \times 10^{-5} \mathrm{~V}$$. The ... | [] | null | 32 | <p>$$\begin{aligned}
& B_v=B \sin 30=\frac{1}{4} \times 10^{-4} \\
& \omega=2 \pi \times f=\frac{2 \pi}{60} \times 1200 \mathrm{~rad} / \mathrm{s} \\
& \varepsilon=\frac{1}{2} B_V \omega \ell^2 \\
& =32 \pi \times 10^{-5} \mathrm{~V}
\end{aligned}$$</p> | integer | jee-main-2024-online-30th-january-morning-shift | 10,298 |
luxwer3v | physics | electromagnetic-induction | motional-emf-and-eddy-current | <p>A square loop of side $$15 \mathrm{~cm}$$ being moved towards right at a constant speed of $$2\mathrm{~cm} / \mathrm{s}$$ as shown in figure. The front edge enters the $$50 \mathrm{~cm}$$ wide magnetic field at $$t=0$$. The value of induced emf in the loop at $$t=10 \mathrm{~s}$$ will be :</p>
<p><img src="data:imag... | [{"identifier": "A", "content": "zero"}, {"identifier": "B", "content": "4.5 mV"}, {"identifier": "C", "content": "0.3 mV"}, {"identifier": "D", "content": "3 mV"}] | ["A"] | null | <p>Time taken to cross the field region</p>
<p>$$=\frac{50}{2}=25 \mathrm{~s}$$</p>
<p>At $$10 \mathrm{~s}$$ the loop is inside field and flux is not changing.</p>
<p>$$\therefore \quad \varepsilon_{\text {induced }}=0$$</p> | mcq | jee-main-2024-online-9th-april-evening-shift | 10,299 |
xG2EBDunp2W3GzgQ | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | Electromagnetic waves are transverse in nature is evident by | [{"identifier": "A", "content": "polarization "}, {"identifier": "B", "content": "interference "}, {"identifier": "C", "content": "reflection "}, {"identifier": "D", "content": "diffraction "}] | ["A"] | null | The phenomenon of polarisation is shown only by transverse waves. | mcq | aieee-2002 | 10,300 |
jCRQrB6VurpfAStv | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | An electromagnetic wave of frequency $$v=3.0$$ $$MHz$$ passes from vacuum into a dielectric medium with permittivity $$ \in = 4.0.$$ Then | [{"identifier": "A", "content": "wave length is halved and frequency remains unchanged "}, {"identifier": "B", "content": "wave length is doubled and the frequency becomes half "}, {"identifier": "C", "content": "wave length is doubled and the frequency remains unchanged "}, {"identifier": "D", "content": "wave length... | ["A"] | null | Frequency remains constant during refraction
<br><br>$${v_{med}} = {1 \over {\sqrt {{\mu _0}{ \in _0} \times 4} }} = {c \over 2}$$
<br><br>$${{{\lambda _{med}}} \over {{\lambda _{air}}}} = {{{v_{med}}} \over {{v_{air}}}} = {{c/2} \over c} = {1 \over 2}$$
<br><br>$$\therefore$$ wavelength is halved and frequency remain... | mcq | aieee-2004 | 10,301 |
y8OHX2USJUdhvFvW | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | A radiation of energy $$E$$ falls normally on a perfectly reflecting surface. The momentum transferred to the surface is | [{"identifier": "A", "content": "$$Ec$$ "}, {"identifier": "B", "content": "$$2E/c$$ "}, {"identifier": "C", "content": "$$E/c$$ "}, {"identifier": "D", "content": "$$E/{c^2}$$ "}] | ["B"] | null | Momentum of photon $$ = {E \over c}$$
<br><br>Change in momentum $$ = {{2E} \over c}$$
<br><br>$$=$$ momentum transferred to the surface
<br><br>(the photon will reflect with same magnitude of momentum in opposite direction) | mcq | aieee-2004 | 10,302 |
K9vgBPLdLg2CsG9j | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | The $$rms$$ value of the electric field of the light coming from the Sun is $$720$$ $$N/C.$$ The average total energy density of the electromagnetic wave is | [{"identifier": "A", "content": "$$4.58 \\times {10^{ - 6}}\\,J/{m^3}$$ "}, {"identifier": "B", "content": "$$6.37 \\times {10^{ - 9}}\\,J/{m^3}$$ "}, {"identifier": "C", "content": "$$81.35 \\times {10^{ - 12}}\\,J/{m^3}$$ "}, {"identifier": "D", "content": "$$3.3 \\times {10^{ - 3}}\\,J/{m^3}$$ "}] | ["A"] | null | $${E_{rms}} = 720$$
<br><br>The average total energy density
<br><br>$$ = {1 \over 2}{ \in _0}\,E_0^2 = {1 \over 2}{ \in _0}{\left[ {\sqrt 2 {E_{rms}}} \right]^2} = { \in _0}\,E_{rms}^2$$
<br><br>$$ = 8.85 \times {10^{ - 12}} \times {\left( {720} \right)^2}$$
<br><br>$$ = 4.58 \times {10^{ - 6}}\,J/{m^3}$$ | mcq | aieee-2006 | 10,303 |
4c5niHBpA9z8BUuz | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | The magnetic field in a travelling electromagnetic wave has a peak value of $$20$$ $$n$$$$T$$. The peak value of electric field strength is : | [{"identifier": "A", "content": "$$3V/m$$ "}, {"identifier": "B", "content": "$$6V/m$$ "}, {"identifier": "C", "content": "$$9V/m$$ "}, {"identifier": "D", "content": "$$12V/m$$ "}] | ["B"] | null | From question,
<br><br>$${B_0} = 20nT = 20 \times {10^{ - 9}}T$$
<br><br>( as velocity of light in vacuum $$C = 3 \times {10^8}\,\,m{s^{ - 1}}$$ )
<br><br>$${\overrightarrow E _0} = {\overrightarrow B _0} \times \overrightarrow C $$
<br><br>$$\left| {{{\overrightarrow E }_0}} \right| = \left| {\overrightarrow B } \rig... | mcq | jee-main-2013-offline | 10,305 |
282LzDMYWQcD2x5P | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | During the propagation of electromagnetic waves in a medium : | [{"identifier": "A", "content": "Electric energy density is double of the magnetic energy density. "}, {"identifier": "B", "content": "Electric energy density is half of the magnetic energy density. "}, {"identifier": "C", "content": "Electric energy density is equal to the magnetic energy density."}, {"identifier": "D... | ["C"] | null | $${E_0} = C{B_0}$$ and $$C = {1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$$
<br><br>Electric energy density $$ = {1 \over 2}{\varepsilon _0}{E_0}^2 = {\mu _E}$$
<br><br>Magnetic energy density $$ = {1 \over 2}{{{B_0}^2} \over {{\mu _0}}} = {\mu _B}$$
<br><br>Thus, $${\mu _E} = {\mu _B}$$
<br><br>Energy is equally divi... | mcq | jee-main-2014-offline | 10,306 |
MdoFpfEwQALFcvucqsvSB | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | A lamp emits monochromatic green light uniformly in all directions. The lamp is 3% efficient in converting electrical power to electromagnetic waves and consumes 100 W of power. The amplitude of the electric field associated with the electromagnetic radiation at a distance of 5 m from the lamp will be nearly : | [{"identifier": "A", "content": "1.34 V/m"}, {"identifier": "B", "content": "2.68 V/m"}, {"identifier": "C", "content": "4.02 V/m"}, {"identifier": "D", "content": "5.36 V/m"}] | ["B"] | null | Since $\quad U_E \times c=I=$ Intensity
<br/><br/>$$
\begin{aligned}
& \frac{1}{2} \varepsilon_0 E^2 \times c=I=\frac{P}{4 \pi R^2}=\frac{3 / 100 \times P}{4 \pi R^2} \\\\
E & =\sqrt{\frac{6 P}{\varepsilon_0 \times c \times 100 \times 4 \pi R^2}} \\\\
& =\sqrt{\frac{6 \times 100}{8.85 \times 10^{-12} \times 3 \times 10... | mcq | jee-main-2014-online-12th-april-morning-slot | 10,307 |
mtYykazmyYZmdXIUprpn0 | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | Match <b>List I</b> (Wavelength range of electromagnetic spectrum) with<b> List I</b><b>I</b>. (Method of production of these waves) and select the <b>correct</b> option from the options given below the lists.
<br/><br/><img src="data:image/png;base64,UklGRroyAABXRUJQVlA4IK4yAACQywCdASqJAY8BPm0ulEekIiShJfC9KJANiWlu/l... | [{"identifier": "A", "content": "(a)-(iv), (b)-(iii), (c)-(ii), (d)-(i)"}, {"identifier": "B", "content": "(a)-(iii), (b)-(iv), (c)-(i), (d)-(ii)"}, {"identifier": "C", "content": "(a)-(ii), (b)-(iii), (c)-(iv), (d)-(i)"}, {"identifier": "D", "content": "(a)-(i), &n... | ["D"] | null | <p>Based on the given information, let's match the items from List I with their corresponding descriptions in List II :</p>
<p>(a) Range : 700 nm to 1 mm
<br/><br/>Match : (i) Vibration of atoms and molecules</p>
<p>(b) Range : 1 nm to 400 nm
<br/><br/>Match : (ii) Inner shell electrons in atoms moving from one ene... | mcq | jee-main-2014-online-9th-april-morning-slot | 10,309 |
vduIu2V5MwArzdRKU6sWR | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | For plane electromagnetic waves propagating in the z direction, which one of the following combination gives the correct possible direction for $$\overrightarrow E $$ and $$\overrightarrow B $$ field respectively ? | [{"identifier": "A", "content": "$$\\left( {\\widehat i + 2\\widehat j} \\right)\\,\\,$$ and $$\\left( {2\\widehat i - \\widehat j} \\right)$$"}, {"identifier": "B", "content": "$$\\left(-\\, {2\\widehat i - 3\\widehat j} \\right)$$ and $$\\left( {3\\widehat i - 2\\widehat j} \\right)$$"}, {"identifier": "C", "content... | ["B"] | null | Since $\vec{E}$ and $\vec{B}$ are mutually perpendicular. Therefore,
<br/><br/>$$
\begin{gathered}
\vec{E} \times \vec{B}=\vec{c}=c \hat{k} \\\\
\Rightarrow(-2 \hat{i}-3 \hat{j}) \cdot(3 \hat{i}-2 \hat{j})=-6+6=0 \\\\
\Rightarrow(-2 \hat{i}-3 \hat{j}) \times(3 \hat{i}-2 \hat{j})=(6+9) \hat{k}=15 \hat{k}
\end{gathered}
... | mcq | jee-main-2015-online-11th-april-morning-slot | 10,310 |
5vHGPLXaXO7wEvKW1xmXB | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | The electric field component of a monochromatic radiation is given by
<br/><br/>$$\overrightarrow E $$ = 2 E<sub>0</sub> $$\widehat i$$ cos kz cos $$\omega $$t
<br/><br/>Its magnetic field $$\overrightarrow B $$ is then given by : | [{"identifier": "A", "content": "$${{2{E_0}} \\over c}$$ $$\\widehat j$$ sin kz cos $$\\omega $$t"}, {"identifier": "B", "content": "$$-$$ $${{2{E_0}} \\over c}$$ $$\\widehat j$$ sin kz sin $$\\omega $$t"}, {"identifier": "C", "content": "$${{2{E_0}} \\over c}$$ $$\\widehat j$$ sin kz sin $$\\omega $$t"}, {"identifier"... | ["C"] | null | <p>We have</p>
<p>$${{dE} \over {dz}} = {{ - dE} \over {dt}}$$</p>
<p>$${{dE} \over {dz}} = - 2{E_0}k\sin kz\cos \omega t = {{ - dB} \over {dt}}$$</p>
<p>Therefore, $$dB = + 2{E_0}k\sin kz\cos \omega t\,dt$$</p>
<p>That is, $$B = + 20{E_0}k\sin {k_z}\cos \omega t\,dt$$</p>
<p>$$B = + 2{E_0}k\sin {k_2}\int {\cos \om... | mcq | jee-main-2017-online-9th-april-morning-slot | 10,313 |
oMiaCnABqPIHciDAaCaFC | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | Magnetic field in a plane electromagnetic wave is given by
<br/><br/>$$\overrightarrow B $$ = B<sub>0</sub> sin (k x + $$\omega $$t) $$\widehat j\,T$$
<br/><br/>Expression for corresponding electric field will be :
<br/>Where c is speed of light. | [{"identifier": "A", "content": "$$\\overrightarrow E $$ = B<sub>0</sub> c sin (k x + $$\\omega $$t) $$\\widehat k$$ V/m"}, {"identifier": "B", "content": "$$\\overrightarrow E $$ = $${{{B_0}} \\over c}$$ sin (k x + $$\\omega $$t) $$\\widehat k$$ V/m"}, {"identifier": "C", "content": "$$\\overrightarrow E $$ = $$-$$ B... | ["A"] | null | The relation between electric and magnetic field is ,
<br><br>C = $${{\overrightarrow E } \over {\overrightarrow B }}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $$\overrightarrow E $$ = C $$\overrightarrow B $$
<br><br>Electric field component is perpendicular to the direction of magnetic field. Given magnetic field is alo... | mcq | jee-main-2017-online-8th-april-morning-slot | 10,314 |
u4OOC5iKqsLL5PvFiJxx0 | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | A plane electromagnetic wave of wavelength $$\lambda $$ has an intensity I. It is propagatting along the positive Y-direction. The allowed expressions for the electric and magnetic fields are givn by : | [{"identifier": "A", "content": "<img src=\"https://gateclass.cdn.examgoal.net/n2mitUHpDA7NRIoNf/I6ZKAl0CpaicXPDjZyiUhnGrjp4wx/XBDxm3tcRXs8sWR6euosg3/uploadfile.jpg\" style=\"max-width: 100%; height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2018 (Online) 16th April Morning Slot Physics - E... | ["A"] | null | <p>$$\overrightarrow E $$ is electric field vector, $$\overrightarrow B $$ is magnetic field vector perpendicular to $$\overrightarrow E $$. The direction of propagation is $$(\overrightarrow E \times \overrightarrow B )$$. The direction of propagation of wave is along + y axis, then $$\overrightarrow E $$ is along + ... | mcq | jee-main-2018-online-16th-april-morning-slot | 10,315 |
9rtzfmbrvwaNUyK78Dx2u | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | A monochromatic beam of light has a frequency $$v = {3 \over {2\pi }} \times {10^{12}}Hz$$ and is propagating along the direction $${{\widehat i + \widehat j} \over {\sqrt 2 }}.$$
<br/>It is polarized along the $$\widehat k$$ direction. The acceptable form for the magnetic field is : | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264940/exam_images/usmpqjagiqpkvnwqhgfl.webp\" style=\"max-width: 100%; height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2018 (Online) 15th April Morning Slot Physics - Electromagnetic Waves ... | ["A"] | null | <p>Given : Frequency $$f = {3 \over {2\pi }} \times {10^{12}}$$ Hz; direction $$ = {{\widehat i + \widehat j} \over {\sqrt 2 }}$$.</p>
<p>The beam is polarised along $${\widehat k}$$ direction.</p>
<p>Direction of magnetic field $$\overrightarrow B = \left( {{{\widehat i + \widehat j} \over {\sqrt 2 }} \times \widehat... | mcq | jee-main-2018-online-15th-april-morning-slot | 10,317 |
gZFiWtAr9eIMfS1N | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | An EM wave from air enters a medium. The electric fields are<br/><br>
$$\overrightarrow {{E_1}} $$ = $${E_{01}}\widehat x\cos \left[ {2\pi v\left( {{z \over c} - t} \right)} \right]$$ in air and
<br/><br/>$$\overrightarrow {{E_2}} $$ = $${E_{02}}\widehat x\cos \left[ {k\left( {2z - ct} \right)} \right]$$ in medium, ... | [{"identifier": "A", "content": "$${{{\\varepsilon _{{r_1}}}} \\over {{\\varepsilon _{{r_2}}}}} = 4$$ "}, {"identifier": "B", "content": "$${{{\\varepsilon _{{r_1}}}} \\over {{\\varepsilon _{{r_2}}}}} = 2$$"}, {"identifier": "C", "content": "$${{{\\varepsilon _{{r_1}}}} \\over {{\\varepsilon _{{r_2}}}}} = {1 \\over 4}$... | ["C"] | null | Electric field in air,
<br><br>$$\overrightarrow {{E_1}} $$ = E<sub>01</sub> $$\widehat x$$ cos ( $${{2\pi vz} \over c}$$ $$-$$ 2$$\pi $$vt )
<br><br>$$\therefore\,\,\,$$ Velocity in air = $${{2\pi v} \over {{{2\pi v} \over c}}}$$ = c
<br><br>Also, c = $${1 \over {\sqrt {\mu \varepsilon {r_1}... | mcq | jee-main-2018-offline | 10,318 |
V3bZ7JTDqq0H2lEDHj3rsa0w2w9jwzjtoa5 | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | Light is incident normally on a completely absorbing surface with an energy flux of 25 W cm<sup>–2</sup>. If the surface
has an area of 25 cm<sup>2</sup>, the momentum transferred to the surface in 40 min time duration will be : | [{"identifier": "A", "content": "6.3 \u00d7 10<sup>\u20134</sup> Ns"}, {"identifier": "B", "content": "5.0 \u00d7 10<sup>\u20133</sup> Ns"}, {"identifier": "C", "content": "1.4 \u00d7 10<sup>\u20136</sup> Ns"}, {"identifier": "D", "content": "3.5 \u00d7 10<sup>\u20136</sup> Ns"}] | ["B"] | null | $$P = {{\Delta E} \over C}$$<br><br>
$$ = {{\left( {25 \times 25} \right) \times 40 \times 60} \over {3 \times {{10}^8}}}N - s$$<br><br>
$$ = 5 \times {10^{ - 2}}N - s$$ | mcq | jee-main-2019-online-10th-april-evening-slot | 10,319 |
xqMtPR5EhmDNdUJo613rsa0w2w9jx3o70xv | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | An electromagnetic wave is represented by the electric field $$\overrightarrow E = {E_0}\widehat n\sin \left[ {\omega t + \left( {6y - 8z} \right)} \right]$$
. Taking unit
vectors in x, y and z directions to be $$\widehat i,\widehat j,\widehat k$$
, the direction of propagation $$\widehat s$$, is : | [{"identifier": "A", "content": "$$\\widehat s = {{3\\widehat i - 4\\widehat j} \\over 5}$$"}, {"identifier": "B", "content": "$$\\widehat s = {{ - 4\\widehat k + 3\\widehat j} \\over 5}$$"}, {"identifier": "C", "content": "$$\\widehat s = \\left( {{{ - 3\\widehat j + 4\\widehat k} \\over 5}} \\right)$$"}, {"identifier... | ["C"] | null | $$\overrightarrow E = {E_0}\widehat n\sin \left( {\omega t + \left( {6y - 8z} \right)} \right)$$<br><br>
$$ = {E_0}\widehat n\sin \left( {\omega t + \overrightarrow k .\overrightarrow r } \right)$$<br><br>
where $$\overrightarrow r = x\widehat i + y\widehat j + z\widehat k$$ and $$\overrightarrow k .\overrightarrow r... | mcq | jee-main-2019-online-12th-april-morning-slot | 10,321 |
puOt4a2DT6bNbLLn7S18hoxe66ijvznkqu6 | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | The electric field of a plane electromagnetic
wave is given by<br/>
$$\overrightarrow E = {E_0}\widehat i\cos (kz)cos(\omega t)$$ <br/>The corresponding magnetic field $$\overrightarrow B $$ is then given by | [{"identifier": "A", "content": "$$\\overrightarrow B = {{{E_0}} \\over C}\\widehat j\\sin (kz)\\sin (\\omega t)$$"}, {"identifier": "B", "content": "$$\\overrightarrow B = {{{E_0}} \\over C}\\widehat j\\sin (kz)\\cos (\\omega t)$$"}, {"identifier": "C", "content": "$$\\overrightarrow B = {{{E_0}} \\over C}\\widehat... | ["A"] | null | $$ \therefore \overrightarrow E \times \overrightarrow B \parallel \overrightarrow v $$<br><br>
Given that wave is propagating along positive z-axis and $$\overrightarrow E $$ along positive x-axis. Hence $$\overrightarrow B $$ along y-axis.<br><br>
From Maxwell equation
<br>
$$\overrightarrow V \times \overrightarro... | mcq | jee-main-2019-online-10th-april-morning-slot | 10,322 |
H7wcIWefcCVwIWMeIqsoL | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | 50 W/m<sup>2</sup> energy density of sunlight is normally
incident on the surface of a solar panel. Some
part of incident energy (25%) is reflected from
the surface and the rest is absorbed. The force
exerted on 1m<sup>2</sup> surface area will be close to
(c = 3 × 108 m/s) :- | [{"identifier": "A", "content": "20 \u00d7 10<sup>\u20138</sup> N"}, {"identifier": "B", "content": "35 \u00d7 10<sup>\u20138</sup> N"}, {"identifier": "C", "content": "10 \u00d7 10\u20138 N"}, {"identifier": "D", "content": "15 \u00d7 10<sup>\u20138</sup> N"}] | ["A"] | null | Radiation pressure for 100% reflection = $${{2I} \over C}$$<br><br>
Radiation pressure for 0% reflection = $${I \over C}$$<br><br>
Hence, in given case, radiation pressure <br><br/>
= $$\left( {0.25} \right)\left( {{{2I} \over C}} \right) + \left( {0.75} \right)\left( {{I \over C}} \right)$$<br><br/>
$$\left( {1.25} \... | mcq | jee-main-2019-online-9th-april-evening-slot | 10,323 |
eKBoT4kkGKQmAuKU4a1MH | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | The magnetic field of an electromagnetic wave
is given by :-<br/><br/>
$$\mathop B\limits^ \to = 1.6 \times {10^{ - 6}}\cos \left( {2 \times {{10}^7}z + 6 \times {{10}^{15}}t} \right)\left( {2\mathop i\limits^ \wedge + \mathop j\limits^ \wedge } \right){{Wb} \over {{m^2}}}$$<br/><br/>
The associated electric field... | [{"identifier": "A", "content": "$$\\mathop E\\limits^ \\to = 4.8 \\times {10^2}\\cos \\left( {2 \\times {{10}^7}z - 6 \\times {{10}^{15}}t} \\right)\\left( -2{\\mathop i\\limits^ \\wedge + \\mathop {j}\\limits^ \\wedge } \\right){V \\over m}$$"}, {"identifier": "B", "content": "$$\\mathop E\\limits^ \\to = 4.8 ... | ["C"] | null | If we use that direction of light propagation will be along $$\overrightarrow E \times \overrightarrow B $$. Then (A) option is
correct.<br>
Magnitude of E = CB<br><br>
E = 3 × 10<sup>8</sup> × 1.6 × 10<sup>–6</sup> × $$\sqrt 5 $$<br><br>
E = 4.8 × $${10^{2\sqrt 5 }}$$<br><br>
$$\overrightarrow E $$ and $$\overrighta... | mcq | jee-main-2019-online-8th-april-evening-slot | 10,325 |
PsPvuNusKi6XOgeKA4XQ2 | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | A plane electromagnetic wave travels in free
space along the x-direction. The electric field
component of the wave at a particular point of
space and time is E = 6 V m<sup>–1</sup> along y-direction.
Its corresponding magnetic field component,
B would be : | [{"identifier": "A", "content": "2 \u00d7 10<sup>\u20138</sup> T along y-direction"}, {"identifier": "B", "content": "6 \u00d7 10<sup>\u20138</sup> T along z-direction"}, {"identifier": "C", "content": "2 \u00d7 10<sup>\u20138</sup> T along z-direction"}, {"identifier": "D", "content": "6 \u00d7 10<sup>\u20138</sup> T ... | ["C"] | null | The direction of propagation of an EM wave is direction of $$\overrightarrow E \times \overrightarrow B $$<br><br>
$$\widehat i = \widehat j \times \widehat B$$<br><br>
$$ \Rightarrow \widehat B = \widehat k$$<br><br>
$$C = {E \over B} \Rightarrow B = {E \over C} = {6 \over {3 \times {{10}^8}}}$$<br><br>
B = 2 × 10<su... | mcq | jee-main-2019-online-8th-april-morning-slot | 10,326 |
C2BB46HIaHRpRBwZuExwp | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | The electric field of a plane polarized electromagnetic wave in free space at time t = 0 is given by an expression $$\overrightarrow E \left( {x,y} \right) = 10\widehat j\cos \left[ {\left( {6x + 8z} \right)} \right].$$ The magnetic field $$\overrightarrow B $$(x,z, t) is given by $$-$$ (c is the velocity of light) | [{"identifier": "A", "content": "$${1 \\over c}\\left( {6\\hat k + 8\\widehat i} \\right)\\cos \\left[ {\\left( {6x + 8z - 10ct} \\right)} \\right]$$"}, {"identifier": "B", "content": "$${1 \\over c}\\left( {6\\widehat k - 8\\widehat i} \\right)\\cos \\left[ {\\left( {6x + 8z - 10ct} \\right)} \\right]$$"}, {"identifie... | ["B"] | null | $$\overrightarrow E = 10\widehat j\cos \left[ {\left( {6\widehat i + 8\widehat k} \right).\left( {x\widehat i + z\widehat k} \right)} \right]$$
<br><br>$$ = 10\widehat j\,\cos \left[ {\overrightarrow K .\overrightarrow r } \right]$$
<br><br>$$ \therefore $$ $$\overrightarrow K = 6\widehat i + 8\wideh... | mcq | jee-main-2019-online-10th-january-evening-slot | 10,330 |
1LlGm8iaXGIVASHSRiyob | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | If the magnetic field of a plane electromagnetic wave is given by (the speed of light = 3 × 10<sup>8</sup> B = 100 × 10<sup>–6</sup>
sin $$\left[ {2\pi \times 2 \times {{10}^{15}}\left( {t - {x \over c}} \right)} \right]$$ then the maximum electric field associated with it is -
| [{"identifier": "A", "content": "4.5 $$ \\times $$ 10<sup>4</sup> N/C"}, {"identifier": "B", "content": "4 $$ \\times $$ 10<sup>4</sup> N/C"}, {"identifier": "C", "content": "6 $$ \\times $$ 10<sup>4</sup> N/C"}, {"identifier": "D", "content": "3 $$ \\times $$ 10<sup>4</sup> N/C"}] | ["D"] | null | E<sub>0</sub> = B<sub>0</sub> $$ \times $$ C
<br><br>= 100 $$ \times $$ 10<sup>$$-$$</sup><sup>6</sup> $$ \times $$ 3 $$ \times $$ 10<sup>8</sup>
<br><br>= 3 $$ \times $$ 10<sup>4</sup> N/C
<br><br>$$ \therefore $$ correct answer is 3 $$ \times $$ 10<sup>4</sup> N/C | mcq | jee-main-2019-online-10th-january-morning-slot | 10,331 |
WoEjBb6TS1MN0rAZJazT7 | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x-direction. At a particular point in space and time, $$\overrightarrow E = 6.3\widehat j\,V/m.$$ The corresponding magnetic field $$\overrightarrow {B,} $$ at that point will be : | [{"identifier": "A", "content": "18.9 $$ \\times $$ 10<sup>$$-$$8</sup> $$\\widehat k$$T"}, {"identifier": "B", "content": "2.1 $$ \\times $$ 10<sup>$$-$$8</sup> $$\\widehat k$$T"}, {"identifier": "C", "content": "6.3 $$ \\times $$ 10<sup>$$-$$8</sup> $$\\widehat k$$T"}, {"identifier": "D", "content": "18.9 $$ \\times ... | ["B"] | null | Given, $$\overrightarrow E = 6.3\widehat j$$
<br><br>$$ \therefore $$ $$\left| {\overrightarrow E } \right| = \sqrt {{{\left( {6.3} \right)}^2}} = 6.3$$
<br><br>$$ \therefore $$ $$\left| {\overrightarrow B } \right| = {{\left| {\overrightarrow E } \right|} \over C}$$
<br><br>$$ = {{6... | mcq | jee-main-2019-online-9th-january-morning-slot | 10,333 |
gJvYizNTcQVvWKiIF9I8P | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | The mean intensity of radiation on the surface of the Sun is about 10<sup>8</sup> W/m<sup>2</sup> . The rms value of the corresponding magnetic field is closet to : | [{"identifier": "A", "content": "10<sup>2</sup> T"}, {"identifier": "B", "content": "10<sup>$$-$$2</sup> T"}, {"identifier": "C", "content": "10<sup>$$-$$4</sup> T"}, {"identifier": "D", "content": "1 T"}] | ["C"] | null | I = $${\varepsilon _0}\,C\,E_{rms}^2$$
<br><br>& E<sub>rms</sub> = cB<sub>rms</sub>
<br><br>I = $${\varepsilon _0}$$ C<sup>3</sup> B$$_{rms}^2$$
<br><br>B$$_{rms}$$ = $$\sqrt {{{\rm I} \over {{ \in _0}{C^3}}}} $$
<br><br>B<sub>rms</sub> $$ \approx $$ 10<sup>$$-$$4</sup> | mcq | jee-main-2019-online-12th-january-evening-slot | 10,334 |
bP9cEHcrjWMrP7atGljgy2xukfaju4vh | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | The electric field of a plane electromagnetic wave is given by
<br/>$$\overrightarrow E = {E_0}\left( {\widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)$$
<br/>Its magnetic field will be given by : | [{"identifier": "A", "content": "$${{{E_0}} \\over c}\\left( {\\widehat x + \\widehat y} \\right)\\sin \\left( {kz - \\omega t} \\right)$$"}, {"identifier": "B", "content": "$${{{E_0}} \\over c}\\left( {\\widehat x - \\widehat y} \\right)\\sin \\left( {kz - \\omega t} \\right)$$"}, {"identifier": "C", "content": "$${{{... | ["D"] | null | Given, $$\overrightarrow E = {E_0}\left( {\widehat x + \widehat y} \right)\sin \left( {kz - \omega t} \right)$$
<br><br>We know, direction of propagation, $$\overrightarrow C = \overrightarrow E \times \overrightarrow B $$
<br><br>Here direction of propagation = $$\widehat k$$
<br><br>$$ \therefore $$ $$\widehat k$$... | mcq | jee-main-2020-online-4th-september-evening-slot | 10,335 |
KtdVH1xq0D0pZMPsVKjgy2xukfxnxzld | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | For a plane electromagnetic wave, the magnetic field at a point x and time t is
<br/><br/>$$\overrightarrow B \left( {x,t} \right)$$ = $$\left[ {1.2 \times {{10}^{ - 7}}\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat k} \right]$$ T
<br/><br/>The instantaneous electric field $$\overrightarro... | [{"identifier": "A", "content": "$$\\overrightarrow E \\left( {x,t} \\right) = \\left[ {36\\sin \\left( {1 \\times {{10}^3}x + 1.5 \\times {{10}^{11}}t} \\right)\\widehat i} \\right]$$ $${V \\over m}$$"}, {"identifier": "B", "content": "$$\\overrightarrow E \\left( {x,t} \\right) = \\left[ {36\\sin \\left( {0.5 \\times... | ["D"] | null | Given,
<br>$$\overrightarrow B \left( {x,t} \right)$$ = $$\left[ {1.2 \times {{10}^{ - 7}}\sin \left( {0.5 \times {{10}^3}x + 1.5 \times {{10}^{11}}t} \right)\widehat k} \right]$$ T
<br><br>Wave is travelling along (–x) axis and $$\overrightarrow B $$ is along
+z axis.
<br><br>We know, Magnitude of electric field
<br><... | mcq | jee-main-2020-online-6th-september-evening-slot | 10,336 |
B1LCvQE5DtsMKP0cLJjgy2xukf3v4mfs | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | The electric field of a plane electromagnetic wave propagating along the x direction in vacuum is
<br/>$$\overrightarrow E = {E_0}\widehat j\cos \left( {\omega t - kx} \right)$$.
<br/>The magnetic field $$\overrightarrow B $$
, at the moment t = 0 is : | [{"identifier": "A", "content": "$$\\overrightarrow B = {{{E_0}} \\over {\\sqrt {{\\mu _0}{ \\in _0}} }}\\cos \\left( {kx} \\right)\\widehat j$$"}, {"identifier": "B", "content": "$$\\overrightarrow B = {{{E_0}} \\over {\\sqrt {{\\mu _0}{ \\in _0}} }}\\cos \\left( {kx} \\right)\\widehat k$$"}, {"identifier": "C", "co... | ["C"] | null | $$\overrightarrow E = {E_0}\,\cos (\omega t - kx)\widehat j$$<br><br>We know, $$E = BC$$<br><br>$$B_{0} = {E_{0} \over C} = {{{E_0}} \over {{1 \over {\sqrt {{\mu _0}{ \in _0}} }}}}$$<br><br>$$ \Rightarrow $$ $$B_{0} = {E_0}\sqrt {{\mu _0}{ \in _0}} $$
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267... | mcq | jee-main-2020-online-3rd-september-evening-slot | 10,339 |
khhPnOKuzztVxQ3cpjjgy2xukf14w4zh | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | The magnetic field of a plane electromagnetic wave is
<br/>$$\overrightarrow B = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat i$$ T
<br/>where c = 3 $$ \times $$ 10<sup>8</sup>
ms<sup>–1</sup> is the speed of light.
The corresponding electric field is :
| [{"identifier": "A", "content": "$$\\overrightarrow E = - {10^{ - 6}}\\sin \\left[ {200\\pi \\left( {y + ct} \\right)} \\right]\\widehat k$$ V/m"}, {"identifier": "B", "content": "$$\\overrightarrow E = - 9\\sin \\left[ {200\\pi \\left( {y + ct} \\right)} \\right]\\widehat k$$ V/m"}, {"identifier": "C", "content": ... | ["B"] | null | Given, $$\overrightarrow B = 3 \times {10^{ - 8}}\sin \left[ {200\pi \left( {y + ct} \right)} \right]\widehat i$$ T
<br><br>$$ \therefore $$ E<sub>0</sub> = CB<sub>0</sub>
<br><br>= 3 × 10<sup>8</sup>
× 3 × 10<sup>–8</sup> = 9 V/m
<br><br>We know, $$\left( {\overrightarrow E \times \overrightarrow B } \right)||\over... | mcq | jee-main-2020-online-3rd-september-morning-slot | 10,340 |
eOSqxQ3fk1N3k8MNARjgy2xukeuuwq8h | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | A plane electromagnetic wave, has <br/>frequency
of 2.0 $$ \times $$ 10<sup>10</sup> Hz and its energy density is
1.02 $$ \times $$ 10<sup>–8</sup> J/m<sup>3</sup> in vacuum. The amplitude of
the magnetic field of the wave is close
to
<br/>( $${1 \over {4\pi {\varepsilon _0}}} = 9 \times {10^9}{{N{m^2}} \over {{C^2}}}$... | [{"identifier": "A", "content": "190 nT"}, {"identifier": "B", "content": "150 nT"}, {"identifier": "C", "content": "160 nT"}, {"identifier": "D", "content": "180 nT"}] | ["C"] | null | Energy density, $${{dU} \over {dV}} = {{B_0^2} \over {2{\mu _0}}}$$
<br><br>$$ \Rightarrow $$ 1.02 $$ \times $$ 10<sup>–8</sup> = $${{B_0^2} \over {2{\mu _0}}}$$
<br><br>Also, c = $${1 \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$$
<br><br>$$ \Rightarrow $$ $${\mu _0} = {1 \over {{c^2}{\varepsilon _0}}}$$
<br><br>$$ \the... | mcq | jee-main-2020-online-2nd-september-morning-slot | 10,341 |
HyCaeOcNWIDUKna9s97k9k2k5l1tg9x | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | A plane electromagnetic wave is propagating
along the direction
$${{\widehat i + \widehat j} \over {\sqrt 2 }}$$
, with its polarization
along the direction $$\widehat k$$ . The correct form of the
magnetic field of the wave would be (here B<sub>0</sub>
is an appropriate constant) : | [{"identifier": "A", "content": "$${B_0}{{\\widehat i - \\widehat j} \\over {\\sqrt 2 }}\\cos \\left( {\\omega t - k{{\\widehat i + \\widehat j} \\over {\\sqrt 2 }}} \\right)$$"}, {"identifier": "B", "content": "$${B_0}{{\\widehat i + \\widehat j} \\over {\\sqrt 2 }}\\cos \\left( {\\omega t - k{{\\widehat i + \\widehat... | ["A"] | null | Direction of propagation = $${{\widehat i + \widehat j} \over {\sqrt 2 }}$$
<br><br>Electric field is in direction = $$\widehat k$$
<br><br>As $$\overrightarrow E \times \overrightarrow B $$ = $${{\widehat i + \widehat j} \over {\sqrt 2 }}$$
<br><br>Propagation direction of $$\overrightarrow B = {{\widehat i - \wideh... | mcq | jee-main-2020-online-9th-january-evening-slot | 10,342 |
a1MqrQXt4uNz66BZM27k9k2k5i7uvtf | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | The electric fields of two plane electromagnetic
plane waves in vacuum are given by
<br/>$$\overrightarrow {{E_1}} = {E_0}\widehat j\cos \left( {\omega t - kx} \right)$$
and
<br/>$$\overrightarrow {{E_2}} = {E_0}\widehat k\cos \left( {\omega t - ky} \right)$$
<br/>At t = 0, a particle of charge q is at origin with
<b... | [{"identifier": "A", "content": "$${E_0}q\\left( {0.8\\widehat i - \\widehat j + 0.4\\widehat k} \\right)$$"}, {"identifier": "B", "content": "$${E_0}q\\left( { - 0.8\\widehat i + \\widehat j + \\widehat k} \\right)$$"}, {"identifier": "C", "content": "$${E_0}q\\left( {0.8\\widehat i + \\widehat j + 0.2\\widehat k} \\r... | ["C"] | null | $$\overrightarrow {{E_1}} = {E_0}\widehat j\cos \left( {\omega t - kx} \right)$$
<br><br>Its corresponding magnetic field will be
<br><br>$$\overrightarrow {{B_1}} = {{{E_0}} \over c}\widehat k\cos \left( {\omega t - kx} \right)$$
<br><br>$$\overrightarrow {{E_2}} = {E_0}\widehat k\cos \left( {\omega t - ky} \right)... | mcq | jee-main-2020-online-9th-january-morning-slot | 10,343 |
lK7D0NhsPKaMPgSnrh7k9k2k5hgokbk | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | A plane electromagnetic wave of frequency
25 GHz is propagating in vacuum along the
z-direction. At a particular point in space and
time, the magnetic field is given by $$\overrightarrow B = 5 \times {10^{ - 8}}\widehat jT$$. The corresponding electric field $$\overrightarrow E $$ is (speed of light c = 3 × 10<sup>8</... | [{"identifier": "A", "content": "15 $$\\widehat i$$V / m"}, {"identifier": "B", "content": "-15 $$\\widehat i$$V / m"}, {"identifier": "C", "content": "1.66 \u00d7 10<sup>\u201316</sup> $$\\widehat i$$V / m"}, {"identifier": "D", "content": "-1.66 \u00d7 10<sup>\u201316</sup> $$\\widehat i$$V / m"}] | ["A"] | null | $$\overrightarrow E = \overrightarrow B \times \overrightarrow V $$
<br><br>= $$\left( {5 \times {0^{ - 8}}\widehat j} \right) \times \left( {3 \times {{10}^8}\widehat k} \right)$$
<br><br>= $${15\,\widehat i}$$ V/m | mcq | jee-main-2020-online-8th-january-evening-slot | 10,344 |
EbHIkuASeOhQAXqWBI7k9k2k5f970fh | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | The electric field of a plane electromagnetic wave is given by
<br/>$$\overrightarrow E = {E_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos \left( {kz + \omega t} \right)$$
<br/><br/>At t = 0, a positively charged particle is at the point (x, y, z) = $$\left( {0,0,{\pi \over k}} \right)$$.
<br/> If its instantane... | [{"identifier": "A", "content": "parallel to $$\\widehat k$$"}, {"identifier": "B", "content": "parallel to $${{\\widehat i + \\widehat j} \\over {\\sqrt 2 }}$$"}, {"identifier": "C", "content": "antiparallel to $${{\\widehat i + \\widehat j} \\over {\\sqrt 2 }}$$"}, {"identifier": "D", "content": "zero"}] | ["C"] | null | $$\overrightarrow E = {E_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos \left( {kz + \omega t} \right)$$
<br><br>$$\overrightarrow E $$ at t = 0 at z = $${\pi \over k}$$ is given by
<br><br>$$\overrightarrow E = $$$${E_0}\left( {{{\widehat i + \widehat j} \over {\sqrt 2 }}} \right)\cos \left( {k{\pi \over k} + ... | mcq | jee-main-2020-online-7th-january-evening-slot | 10,345 |
lBP1uqrgpsW4RQjldT7k9k2k5dqw8pe | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | If the magnetic field in a plane electromagnetic wave is given by
<br/>$$\overrightarrow B $$ = 3 $$ \times $$ 10<sup>-8</sup> sin(1.6 $$ \times $$ 10<sup>3</sup>x + 48 $$ \times $$ 10<sup>10</sup>t)$$\widehat j$$ T, then what will be expression for electric field ?
| [{"identifier": "A", "content": "$$\\overrightarrow E $$ = (9sin(1.6 $$ \\times $$ 10<sup>3</sup>x + 48 $$ \\times $$ 10<sup>10</sup>t)$$\\widehat k$$ V/m)"}, {"identifier": "B", "content": "$$\\overrightarrow E $$ = (60sin(1.6 $$ \\times $$ 10<sup>3</sup>x + 48 $$ \\times $$ 10<sup>10</sup>t)$$\\widehat k$$ V/m)"}, {"... | ["A"] | null | Given $$\overrightarrow B $$ = 3 $$ \times $$ 10<sup>-8</sup> sin(1.6 $$ \times $$ 10<sup>3</sup>x + 48 $$ \times $$ 10<sup>10</sup>t)$$\widehat j$$ T
<br><br>We know, $${{{E_0}} \over {{B_0}}} = c$$
<br><br>$$ \Rightarrow $$ E<sub>0</sub> = (3 $$ \times $$ 10<sup>-8</sup>) $$ \times $$ (3 $$ \times $$ 10<sup>-8</sup>)... | mcq | jee-main-2020-online-7th-january-morning-slot | 10,346 |
2fu620cxEZwIT2qKGqjgy2xukexrl14m | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | In a plane electromagnetic wave, the directions
of electric field and magnetic field are
represented by $$\widehat k$$ and $$2\widehat i - 2\widehat j$$, respectively.
What is the unit vector along direction of
propagation of the wave? | [{"identifier": "A", "content": "$${1 \\over {\\sqrt 5 }}\\left( {\\widehat i + 2\\widehat j} \\right)$$"}, {"identifier": "B", "content": "$${1 \\over {\\sqrt 5 }}\\left( {2\\widehat i + \\widehat j} \\right)$$"}, {"identifier": "C", "content": "$${1 \\over {\\sqrt 2 }}\\left( {\\widehat i + \\widehat j} \\right)$$"},... | ["C"] | null | $$\overrightarrow E \times \overrightarrow B = \widehat k \times \left( {2\widehat i - 2\widehat j} \right)$$
<br><br>= $$2\widehat k \times \widehat i - 2\widehat k \times \widehat j$$
<br><br>= $$\left( {2\widehat j + 2\widehat i} \right)$$
<br><br>Unit vector along $$\overrightarrow E \times \overrightarrow B $$ ... | mcq | jee-main-2020-online-2nd-september-evening-slot | 10,347 |
BUn4ZIxEiwIbjPs3uF1klrord1l | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | An electromagnetic wave of frequency 3 GHz enters a dielectric medium of relative electric permittivity 2.25 from vacuum. The wavelength of this wave in that medium will be _________ $$\times$$ 10<sup>$$-$$2</sup> cm. | [] | null | 667 | Given, frequency of wave, f = 3 GHz = 3 $$\times$$ 10<sup>9</sup> Hz<br/><br/>Relative permittivity, $$\varepsilon $$<sub>r</sub> = 2.25<br/><br/>Since, f = C/$$\lambda$$<br/><br/>$$ \Rightarrow \lambda = {c \over f} = {{3 \times {{10}^8}} \over {3 \times {{10}^9}}} = 0.1$$ m<br/><br/>$$\because$$ $$\lambda$$<sub>m</s... | integer | jee-main-2021-online-24th-february-evening-slot | 10,349 |
oq6jfusfmyOTHVVEz31kmhol9dz | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | A plane electromagnetic wave of frequency 500 MHz is travelling in vacuum along y-direction. At a particular point in space and time, <br/>$$\overrightarrow B $$ = 8.0 $$\times$$ 10<sup>$$-$$8</sup> $$\widehat z$$T. The value of electric field at this point is :<br/><br/>(speed of light = 3 $$\times$$ 10<sup>8</sup> ms... | [{"identifier": "A", "content": "2.6 $$\\widehat x$$ V/m"}, {"identifier": "B", "content": "$$-$$24 $$\\widehat x$$ V/m"}, {"identifier": "C", "content": "24 $$\\widehat x$$ V/m"}, {"identifier": "D", "content": "$$-$$2.6 $$\\widehat y$$ V/m"}] | ["B"] | null | $${E_0} = B.C$$<br><br>$${E_0} = (8 \times {10^{ - 8}}) \times (3 \times {10^8})$$<br><br>$$ \Rightarrow {E_0} = 24$$<br><br>Direction of wave travelling is in $$\overrightarrow E \times \overrightarrow B $$<br><br>So, $$( - \widehat x) \times \widehat z = + \widehat y$$<br><br>$$ \therefore $$ $$\widehat E = -24\wid... | mcq | jee-main-2021-online-16th-march-morning-shift | 10,352 |
INnagC7ldNSMlgnAfP1kmhovr2p | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | For an electromagnetic wave travelling in free space, the relation between average energy densities due to electric (U<sub>e</sub>) and magnetic (U<sub>m</sub>) fields is : | [{"identifier": "A", "content": "U<sub>e</sub> = U<sub>m</sub>"}, {"identifier": "B", "content": "U<sub>e</sub> $$\\ne$$ U<sub>m</sub>"}, {"identifier": "C", "content": "U<sub>e</sub> < U<sub>m</sub>"}, {"identifier": "D", "content": "U<sub>e</sub> > U<sub>m</sub>"}] | ["A"] | null | In EMW, average energy density due to electric field (U<sub>e</sub>) and magnetic field (U<sub>m</sub>) is same. | mcq | jee-main-2021-online-16th-march-morning-shift | 10,353 |
3c4eolf0hGdecIPLjE1kmj4bxcs | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | If 2.5 $$\times$$ 10<sup>$$-$$6</sup> N average force is exerted by a light wave on a non-reflecting surface of 30 cm<sup>2</sup> area during 40 minutes of time span, the energy flux of light just before it falls on the surface is ___________ W/cm<sup>2</sup>. (Round off to the Nearest Integer)<br/><br/>(Assume complet... | [] | null | 25 | Pressure = $${{Intensity} \over C}$$ (for absorbing surface)<br><br>I = P $$\times$$ C<br><br>I = $${{2.5 \times {{10}^{ - 6}}} \over {30c{m^2}}}$$ N $$\times$$ 3 $$\times$$ 10<sup>8</sup> m/s<br><br>I = 25 W/cm<sup>2</sup> | integer | jee-main-2021-online-17th-march-morning-shift | 10,354 |
zYlLM2PrkgSuI0D0J61kmkc59oz | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | The electric field intensity produced by the radiation coming from a 100 W bulb at a distance of 3 m is E. The electric field intensity produced by the radiation coming from 60W at the same distance is $$\sqrt {{x \over 5}} $$E. Where the value of x = ____________. | [] | null | 3 | $$I = {1 \over 2}C{ \in _0}{E^2}$$<br><br>$${E^2} \propto I$$<br><br>$$I = {{Power} \over {Area}}$$<br><br>$${E^2} \propto {P \over A}$$<br><br>$$E \propto \sqrt P $$<br><br>$${{E'} \over E} = \sqrt {{{60} \over {100}}} $$<br><br>$$E' = \sqrt {{3 \over 5}} E$$<br><br>So the value of x = 3 | integer | jee-main-2021-online-17th-march-evening-shift | 10,355 |
w6k4Dp7xKRzdYTeAkh1kmkcicua | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | Seawater at a frequency f = 9 $$\times$$ 10<sup>2</sup> Hz, has permittivity $$\varepsilon $$ = 80$$\varepsilon $$<sub>0</sub> and resistivity $$\rho$$ = 0.25 $$\Omega$$m. Imagine a parallel plate capacitor is immersed in seawater and is driven by an alternating voltage source V(t) = V<sub>0</sub> sin(2$$\pi$$ft). Then... | [] | null | 6 | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265979/exam_images/fm55y9ao1bcrdgrr7ts0.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 17th March Evening Shift Physics - Electromagnetic Waves Question 75 English Explanation">
<br>Giv... | integer | jee-main-2021-online-17th-march-evening-shift | 10,356 |
7wGYP8oBZmpaPHqp421kmlvdfre | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | A plane electromagnetic wave propagating along y-direction can have the following pair of electric field $$\left( {\overrightarrow E } \right)$$ and magnetic field $$\left( {\overrightarrow B } \right)$$ components. | [{"identifier": "A", "content": "E<sub>x</sub>, B<sub>z</sub> or E<sub>z</sub>, B<sub>x</sub>"}, {"identifier": "B", "content": "E<sub>x</sub>, B<sub>y</sub> or E<sub>y</sub>, B<sub>x</sub>"}, {"identifier": "C", "content": "E<sub>y</sub>, B<sub>y</sub> or E<sub>z</sub>, B<sub>z</sub>"}, {"identifier": "D", "content": ... | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265694/exam_images/xqklknithqotrk70vj6s.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 18th March Evening Shift Physics - Electromagnetic Waves Question 73 English Explanation">
<br>$$ ... | mcq | jee-main-2021-online-18th-march-evening-shift | 10,358 |
1krsucbx4 | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | Intensity of sunlight is observed as 0.092 Wm<sup>$$-$$2</sup> at a point in free space. What will be the peak value of magnetic field at the point?<br/><br/>($${\varepsilon _0} = 8.85 \times {10^{ - 12}}{C^2}{N^{ - 1}}{m^{ - 2}}$$) | [{"identifier": "A", "content": "2.77 $$\\times$$ 10<sup>$$-$$8</sup> T"}, {"identifier": "B", "content": "1.96 $$\\times$$ 10<sup>$$-$$8</sup> T"}, {"identifier": "C", "content": "8.31 T"}, {"identifier": "D", "content": "5.88 T"}] | ["A"] | null | $${I \over C} = {1 \over 2}{\varepsilon _0}.E_0^2$$<br><br>$$ \Rightarrow {E_0} = \sqrt {{{2I} \over {C{\varepsilon _0}}}} $$<br><br>$${{{E_0}} \over {{B_0}}} = C \Rightarrow {B_0} = {{{E_0}} \over C}$$<br><br>$$ \Rightarrow {B_0} = \sqrt {{{2I} \over {{\varepsilon _0}{C^3}}}} = \sqrt {{{2 \times 0.092} \over {8.85 \t... | mcq | jee-main-2021-online-22th-july-evening-shift | 10,360 |
1kruly8xw | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | A linearly polarized electromagnetic wave in vacuum is <br/><br/>$$E = 3.1\cos \left[ {(1.8)z - (5.4 \times {{10}^6})t} \right]\widehat iN/C$$<br/><br/>is incident normally on a perfectly reflecting wall at z = a. Choose the correct option | [{"identifier": "A", "content": "The wavelength is 5.4 m"}, {"identifier": "B", "content": "The frequency of electromagnetic wave is 54 $$\\times$$ 10<sup>4</sup> Hz."}, {"identifier": "C", "content": "The transmitted wave will be $$3.1\\cos \\left[ {(1.8)z - (5.4 \\times {{10}^6})t} \\right]\\widehat iN/C$$"}, {"ident... | ["D"] | null | Reflected wave will have direction opposite to incident wave. | mcq | jee-main-2021-online-25th-july-morning-shift | 10,361 |
1krytb7sk | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | The relative permittivity of distilled water is 81. The velocity of light in it will be :<br/><br/>(Given $$\mu$$<sub>r</sub> = 1) | [{"identifier": "A", "content": "4.33 $$\\times$$ 10<sup>7</sup> m/s"}, {"identifier": "B", "content": "2.33 $$\\times$$ 10<sup>7</sup> m/s"}, {"identifier": "C", "content": "3.33 $$\\times$$ 10<sup>7</sup> m/s"}, {"identifier": "D", "content": "5.33 $$\\times$$ 10<sup>7</sup> m/s"}] | ["C"] | null | <p>The speed of light in a medium is given by the equation:</p>
<p>$$ v = \frac{c}{\sqrt{\varepsilon_r \mu_r}} $$</p>
<p>where:</p>
<ul>
<li>$c$ is the speed of light in vacuum (approximately $3 \times 10^8$ m/s),</li>
<li>$\varepsilon_r$ is the relative permittivity of the medium (in this case, distilled water, and is... | mcq | jee-main-2021-online-27th-july-morning-shift | 10,362 |
1ktag20eg | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | The electric field in a plane electromagnetic wave is given by <br/><br>$$\overrightarrow E = 200\cos \left[ {\left( {{{0.5 \times {{10}^3}} \over m}} \right)x - \left( {1.5 \times {{10}^{11}}{{rad} \over s} \times t} \right)} \right]{V \over m}\widehat j$$. If this wave falls normally on a perfectly reflecting surfac... | [] | null | 354 | E<sub>0</sub> = 200<br><br>$$I = {1 \over 2}{\varepsilon _0}E_0^2.C$$<br><br>Radiation pressure<br><br>$$P = {{2I} \over C}$$<br><br>$$ = \left( {{2 \over C}} \right)\left( {{1 \over 2}{\varepsilon _0}E_0^2C} \right)$$<br><br>$$ = {\varepsilon _0}E_0^2$$<br><br>$$ = 8.85 \times {10^{ - 12}} \times {200^2}$$<br><br>$$ =... | integer | jee-main-2021-online-26th-august-morning-shift | 10,363 |
1ktbnruai | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | A light beam is described by $$E = 800\sin \omega \left( {t - {x \over c}} \right)$$. An electron is allowed to move normal to the propagation of light beam with a speed of 3 $$\times$$ 10<sup>7</sup> ms<sup>$$-$$1</sup>. What is the maximum magnetic force exerted on the electron? | [{"identifier": "A", "content": "1.28 $$\\times$$ 10<sup>$$-$$18</sup> N"}, {"identifier": "B", "content": "1.28 $$\\times$$ 10<sup>$$-$$21</sup> N"}, {"identifier": "C", "content": "12.8 $$\\times$$ 10<sup>$$-$$17</sup> N"}, {"identifier": "D", "content": "12.8 $$\\times$$ 10<sup>$$-$$18</sup> N"}] | ["D"] | null | $${{{E_0}} \over C} = {B_0}$$<br><br>$${F_{\max }} = e{B_0}V$$<br><br>$$ = 1.6 \times {10^{ - 19}} \times {{800} \over {3 \times {{10}^8}}} \times 3 \times {10^7}$$<br><br>$$ = 12.8 \times {10^{ - 18}}$$ N | mcq | jee-main-2021-online-26th-august-evening-shift | 10,364 |
1kte6j88r | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | Electric field in a plane electromagnetic wave is given by E = 50 sin(500x $$-$$ 10 $$\times$$ 10<sup>10</sup> t) V/m The velocity of electromagnetic wave in this medium is :<br/><br/>(Given C = speed of light in vacuum) | [{"identifier": "A", "content": "$${3 \\over 2}$$C"}, {"identifier": "B", "content": "C"}, {"identifier": "C", "content": "$${2 \\over 3}$$C"}, {"identifier": "D", "content": "$${C \\over 2}$$"}] | ["C"] | null | $$V = {\omega \over K} = {{10 \times {{10}^{10}}} \over {500}} = 2 \times {10^8}$$<br><br>$$V = {{2C} \over 3}$$. | mcq | jee-main-2021-online-27th-august-morning-shift | 10,365 |
1ktfojayy | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | A plane electromagnetic wave with frequency of 30 MHz travels in free space. At particular point in space and time, electric field is 6 V/m. The magnetic field at this point will be x $$\times$$ 10<sup>$$-$$8</sup> T. The value of x is ___________. | [] | null | 2 | $$|B|\, = {{|E|} \over C} = {6 \over {3 \times {{10}^8}}}$$<br><br>= 2 $$\times$$ 10<sup>$$-$$8</sup> T<br><br>$$\therefore$$ x = 2 | integer | jee-main-2021-online-27th-august-evening-shift | 10,366 |
1kth6e908 | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | The electric field in an electromagnetic wave is given by E = (50 NC<sup>$$-$$1</sup>) sin$$\omega$$ (t $$-$$ x/c)<br/><br/>The energy contained in a cylinder of volume V is 5.5 $$\times$$ 10<sup>$$-$$12</sup> J. The value of V is _____________ cm<sup>3</sup>. (given $$\in$$<sub>0</sub> = 8.8 $$\times$$ 10<sup>$$-$$12<... | [] | null | 500 | $$E = 50\sin \left( {\omega t - {\omega \over c}.\,x} \right)$$<br><br>Energy density = $${1 \over 2}{ \in _0}E_0^2$$<br><br>Energy of volume $$V = {1 \over 2}{ \in _0}E_0^2.\,V = 5.5 \times {10^{ - 12}}$$<br><br>$${1 \over 2}8.8 \times {10^{ - 12}} \times 2500V = 5.5 \times {10^{ - 12}}$$<br><br>$$V = {{5.5 \times 2}... | integer | jee-main-2021-online-31st-august-morning-shift | 10,367 |
1ktjp2itf | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | The magnetic field vector of an electromagnetic wave is given by $$B = {B_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos (kz - \omega t)$$; where $$\widehat i,\widehat j$$ represents unit vector along x and y-axis respectively. At t = 0s, two electric charges q<sub>1</sub> of 4$$\pi$$ coulomb and q<sub>2</sub> of 2... | [{"identifier": "A", "content": "$$2\\sqrt 2 :1$$"}, {"identifier": "B", "content": "$$1:\\sqrt 2 $$"}, {"identifier": "C", "content": "2 : 1"}, {"identifier": "D", "content": "$$\\sqrt 2 :1$$"}] | ["C"] | null | $$\overrightarrow F = q\left( {\overrightarrow V \times \overrightarrow B } \right)$$<br><br>$${\overrightarrow F _1} = 4\pi \left[ {0.5c\widehat i \times {B_0}\left( {{{\widehat i + \widehat j} \over 2}} \right)\cos \left( {K.{\pi \over K} - 0} \right)} \right]$$<br><br>$${\overrightarrow F _2} = 2\pi \left[ {0.5c\... | mcq | jee-main-2021-online-31st-august-evening-shift | 10,368 |
1ktmpsv5d | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | Electric field of plane electromagnetic wave propagating through a non-magnetic medium is given by <br/><br>E = 20cos(2 $$\times$$ 10<sup>10</sup> t $$-$$ 200x) V/m. The dielectric constant of the medium is equal to : (Take $$\mu$$<sub>r</sub> = 1)</br> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$${1 \\over 3}$$"}, {"identifier": "D", "content": "3"}] | ["A"] | null | Given, electric field,<br/><br/>E = 20 cos(2 $$\times$$ 10<sup>10</sup>t $$-$$ 200 x) V/m<br/><br/>Comparing with the standard equation,<br/><br/>E = E<sub>0</sub> cos($$\omega$$t $$-$$ kx) V/m, we get<br/><br/>Wave constant, k = 200<br/><br/>Angular frequency, $$\omega$$ = 2 $$\times$$ 10<sup>10</sup> rad/s<br/><br/>S... | mcq | jee-main-2021-online-1st-september-evening-shift | 10,369 |
1l547xmzx | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>The intensity of the light from a bulb incident on a surface is 0.22 W/m<sup>2</sup>. The amplitude of the magnetic field in this light-wave is ______________ $$\times$$ 10<sup>$$-$$9</sup> T.</p>
<p>(Given : Permittivity of vacuum $$\in$$<sub>0</sub> = 8.85 $$\times$$ 10<sup>$$-$$12</sup> C<sup>2</sup> N<sup>$$-$$1... | [] | null | 43 | <p>$$I = {1 \over 2}{\varepsilon _0}E_0^2\,.\,c = {1 \over 2}{\varepsilon _0}{(c{B_0})^2}c$$</p>
<p>$$ \Rightarrow I = {1 \over 2}{\varepsilon _0}{c^3}B_0^2$$</p>
<p>$$ \Rightarrow 0.22 = {1 \over 2}\left( {8.85 \times {{10}^{ - 12}}} \right){\left( {3 \times {{10}^8}} \right)^3}B_0^2$$</p>
<p>$$ \Rightarrow {B_0} \sim... | integer | jee-main-2022-online-29th-june-morning-shift | 10,370 |
1l54vu5to | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>The displacement current of 4.425 $$\mu$$A is developed in the space between the plates of parallel plate capacitor when voltage is changing at a rate of 10<sup>6</sup> Vs<sup>$$-$$1</sup>. The area of each plate of the capacitor is 40 cm<sup>2</sup>. The distance between each plate of the capacitor is x $$\times$$ ... | [] | null | 8 | <p>$$4.425\,\mu A = {{{E_0}A} \over d} \times {{dV} \over {dt}}$$</p>
<p>$$ \Rightarrow d = {{8.85 \times {{10}^{ - 12}} \times 40 \times {{10}^{ - 4}}} \over {4.425 \times {{10}^{ - 6}}}} \times {10^6}$$</p>
<p>$$ \Rightarrow d = 8 \times {10^{ - 3}}$$ m</p>
<p>$$ \Rightarrow x = 8$$</p> | integer | jee-main-2022-online-29th-june-evening-shift | 10,371 |
1l55kidqb | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>An EM wave propagating in x-direction has a wavelength of 8 mm. The electric field vibrating y-direction has maximum magnitude of 60 Vm<sup>$$-$$1</sup>. Choose the correct equations for electric and magnetic fields if the EM wave is propagating in vacuum :</p> | [{"identifier": "A", "content": "<p>$${E_y} = 60\\sin \\left[ {{\\pi \\over 4} \\times {{10}^3}(x - 3 \\times {{10}^8}t)} \\right]\\widehat j\\,\\,V{m^{ - 1}}$$</p>\n<p>$${B_z} = 2\\sin \\left[ {{\\pi \\over 4} \\times {{10}^3}(x - 3 \\times {{10}^8}t)} \\right]\\widehat k\\,\\,T$$</p>"}, {"identifier": "B", "content... | ["B"] | null | <p>In first 3 options speed of light is 3 $$\times$$ 10<sup>8</sup> m/sec and in the fourth option it is 4 $$\times$$ 10<sup>8</sup> m/sec.</p>
<p>Using</p>
<p>E = CB</p>
<p>We can check the option is B.</p> | mcq | jee-main-2022-online-28th-june-evening-shift | 10,372 |
1l56v7nwq | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>Given below are two statements :</p>
<p>Statement I : A time varying electric field is a source of changing magnetic field and vice-versa. Thus a disturbance in electric or magnetic field creates EM waves.</p>
<p>Statement II : In a material medium, the EM wave travels with speed $$v = {1 \over {\sqrt {{\mu _0}{ \in... | [{"identifier": "A", "content": "Both Statement I and Statement II are true"}, {"identifier": "B", "content": "Both Statement I and Statement II are false"}, {"identifier": "C", "content": "Statement I is correct but Statement II is false"}, {"identifier": "D", "content": "Statement I is incorrect but Statement II is t... | ["C"] | null | <p>In a material medium speed of light is given by $$v = {1 \over {\sqrt {{\varepsilon _0}{\varepsilon _r}{\mu _0}{\mu _r}} }}$$. So statement 2 is false.</p> | mcq | jee-main-2022-online-27th-june-evening-shift | 10,374 |
1l58c31uc | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>If Electric field intensity of a uniform plane electromagnetic wave is given as $$E = - 301.6\sin (kz - \omega t){\widehat a_x} + 452.4\sin (kz - \omega t){\widehat a_y}{V \over m}$$. Then magnetic intensity 'H' of this wave in Am<sup>$$-$$1</sup> will be :</p>
<p>[Given : Speed of light in vacuum $$c = 3 \times {1... | [{"identifier": "A", "content": "$$ + 0.8\\sin (kz - \\omega t){\\widehat a_y} + 0.8\\sin (kz - \\omega t){\\widehat a_x}$$"}, {"identifier": "B", "content": "$$ + 1.0 \\times {10^{ - 6}}\\sin (kz - \\omega t){\\widehat a_y} + 1.5 \\times {10^{ - 6}}(kz - \\omega t){\\widehat a_x}$$"}, {"identifier": "C", "content": "$... | ["C"] | null | <p>$$\overrightarrow E = - 301.6\sin (kz - \omega t){\widehat a_x} + 452.4\sin (kz - \omega t){\widehat a_y}$$</p>
<p>$${E_{0x}} = 301.6$$</p>
<p>$${E_{0y}} = + 452.4$$</p>
<p>$${E_0} = \sqrt {E_{0x}^2 + E_{0y}^2} $$</p>
<p>Now, $${{{E_0}} \over {{B_0}}} = C \Rightarrow {B_0} = {{{E_0}} \over C} = {{\sqrt {E_{0x}^2 ... | mcq | jee-main-2022-online-26th-june-morning-shift | 10,375 |
1l58c4m2t | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>In free space, an electromagnetic wave of 3 GHz frequency strikes over the edge of an object of size $${\lambda \over {100}}$$, where $$\lambda$$ is the wavelength of the wave in free space. The phenomenon, which happens there will be :</p> | [{"identifier": "A", "content": "Reflection"}, {"identifier": "B", "content": "Refraction"}, {"identifier": "C", "content": "Diffraction"}, {"identifier": "D", "content": "Scattering"}] | ["D"] | null | $$
\frac{\mathrm{a}}{\lambda}=\frac{1}{100}
$$
<br/><br/>For reflection size of obstacle must be much larger than wavelength, for diffraction size should be order of wavelength.
<br/><br/>Since the object is of size $\frac{\lambda}{100}$, much smaller than wavelength, so scattering will occur. | mcq | jee-main-2022-online-26th-june-morning-shift | 10,376 |
1l59p8cqw | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>The electromagnetic waves travel in a medium at a speed of 2.0 $$\times$$ 10<sup>8</sup> m/s. The relative permeability of the medium is 1.0. The relative permittivity of the medium will be :</p> | [{"identifier": "A", "content": "2.25"}, {"identifier": "B", "content": "4.25"}, {"identifier": "C", "content": "6.25"}, {"identifier": "D", "content": "8.25"}] | ["A"] | null | The speed of electromagnetic waves in a medium is given by the formula:
<br/><br/>$$v = \frac{1}{\sqrt{\mu \varepsilon}}$$
<br/><br/>where $\mu$ and $\varepsilon$ are the absolute permeability and absolute permittivity of the medium, respectively.
<br/><br/>Given that $\mu = \mu_0 \mu_r$ and $\varepsilon = \varepsil... | mcq | jee-main-2022-online-25th-june-evening-shift | 10,377 |
1l5akrd1g | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>The electric field in an electromagnetic wave is given by E = 56.5 sin $$\omega$$(t $$-$$ x/c) NC<sup>$$-$$1</sup>. Find the intensity of the wave if it is propagating along x-axis in the free space.</p>
<p>(Given : $$\varepsilon $$<sub>0</sub> = 8.85 $$\times$$ 10<sup>$$-$$12</sup>C<sup>2</sup>N<sup>$$-$$1</sup>m<s... | [{"identifier": "A", "content": "5.65 Wm<sup>$$-$$2</sup>"}, {"identifier": "B", "content": "4.24 Wm<sup>$$-$$2</sup>"}, {"identifier": "C", "content": "1.9 $$\\times$$ 10<sup>$$-$$7</sup> Wm<sup>$$-$$2</sup>"}, {"identifier": "D", "content": "56.5 Wm<sup>$$-$$2</sup>"}] | ["B"] | null | <p>$$I = {1 \over 2}{\varepsilon _0}E_0^2c$$</p>
<p>$$ = {1 \over 2} 8.5 \times {10^{ - 12}} \times {(56.5)^2} \times 3 \times {10^8}$$</p>
<p>$$ = 4.24$$ W/m<sup>2</sup></p> | mcq | jee-main-2022-online-25th-june-morning-shift | 10,378 |
1l5bc6h11 | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>An electric bulb is rated as 200 W. What will be the peak magnetic field at 4 m distance produced by the radiations coming from this bulb? Consider this bulb as a point source with 3.5% efficiency.</p> | [{"identifier": "A", "content": "1.19 $$\\times$$ 10<sup>$$-$$8</sup>T"}, {"identifier": "B", "content": "1.71 $$\\times$$ 10<sup>$$-$$8</sup>T"}, {"identifier": "C", "content": "0.84 $$\\times$$ 10<sup>$$-$$8</sup>T"}, {"identifier": "D", "content": "3.36 $$\\times$$ 10<sup>$$-$$8</sup>T"}] | ["B"] | null | The total power (P<sub>T</sub>) of the light bulb is given as 200 W, but only 3.5% of this power is actually emitted as radiation, which we will call P.
<br/><br/>So, Effective power output of the bulb
<br/><br/>$$
\mathrm{P}=\frac{3 \cdot 5}{100} \times 200=7 \mathrm{~W}
$$
<br/><br/>$$
\begin{aligned}
\text { Intens... | mcq | jee-main-2022-online-24th-june-evening-shift | 10,379 |
1l5c3zqv4 | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>A plane electromagnetic wave travels in a medium of relative permeability 1.61 and relative permittivity 6.44. If magnitude of magnetic intensity is 4.5 $$\times$$ 10<sup>$$-$$2</sup> Am<sup>$$-$$1</sup> at a point, what will be the approximate magnitude of electric field intensity at that point?</p>
<p>(Given : Per... | [{"identifier": "A", "content": "16.96 Vm<sup>$$-$$1</sup>"}, {"identifier": "B", "content": "2.25 $$\\times$$ 10<sup>$$-$$2</sup> Vm<sup>$$-$$1</sup>"}, {"identifier": "C", "content": "8.48 Vm<sup>$$-$$1</sup>"}, {"identifier": "D", "content": "6.75 $$\\times$$ 10<sup>6</sup> Vm<sup>$$-$$1</sup>"}] | ["C"] | null | <p>H = 4.5 $$\times$$ 10<sup>$$-$$2</sup></p>
<p>So B = $$\mu$$<sub>0</sub>$$\mu$$H</p>
<p>Thus $$E = {c \over n}B$$ (where n $$\Rightarrow$$ refractive index)</p>
<p>So $$E = {{3 \times {{10}^8} \times 4\pi \times {{10}^{ - 7}} \times 1.61 \times 4.5 \times {{10}^{ - 2}}} \over {\sqrt {1.61 \times 6.44} }}$$</p>
<p>$... | mcq | jee-main-2022-online-24th-june-morning-shift | 10,380 |
1l5w2wlzw | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>An expression for oscillating electric field in a plane electromagnetic wave is given as E<sub>z</sub> = 300 sin(5$$\pi$$ $$\times$$ 10<sup>3</sup>x $$-$$ 3$$\pi$$ $$\times$$ 10<sup>11</sup>t) Vm<sup>$$-$$1</sup></p>
<p>Then, the value of magnetic field amplitude will be :</p>
<p>(Given : speed of light in Vacuum c ... | [{"identifier": "A", "content": "1 $$\\times$$ 10<sup>$$-$$6</sup> T"}, {"identifier": "B", "content": "5 $$\\times$$ 10<sup>$$-$$6</sup> T"}, {"identifier": "C", "content": "18 $$\\times$$ 10<sup>9</sup> T"}, {"identifier": "D", "content": "21 $$\\times$$ 10<sup>9</sup> T"}] | ["B"] | null | <p>Given the electric field expression:</p>
<p>$E_z = 300 \sin(5\pi \times 10^3 x - 3\pi \times 10^{11} t) \, \text{Vm}^{-1}$</p>
<p>The amplitude of the electric field ($E_0$) is $300 \, \text{V/m}$.</p>
<p>The velocity ($v$) of the wave in the medium is given by the ratio of the coefficients of time and displaceme... | mcq | jee-main-2022-online-30th-june-morning-shift | 10,381 |
1l6dywbms | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>The rms value of conduction current in a parallel plate capacitor is $$6.9 \,\mu \mathrm{A}$$. The capacity of this capacitor, if it is connected to $$230 \mathrm{~V}$$ ac supply with an angular frequency of $$600 \,\mathrm{rad} / \mathrm{s}$$, will be :</p> | [{"identifier": "A", "content": "5 pF"}, {"identifier": "B", "content": "50 pF"}, {"identifier": "C", "content": "100 pF"}, {"identifier": "D", "content": "200 pF"}] | ["B"] | null | <p>$${Z_C} = {V \over I}$$</p>
<p>$$ \Rightarrow {1 \over {\omega C}} = {{230} \over {6.9}}\,M\,\Omega $$</p>
<p>$$ \Rightarrow C = {{6.9} \over {230\,\omega }}\,\mu F$$</p>
<p>$$ = {{6.9} \over {230 \times 600}}\,\mu F$$</p>
<p>$$C = 50\,pF$$</p> | mcq | jee-main-2022-online-25th-july-morning-shift | 10,382 |
1l6gk7gfd | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>The magnetic field of a plane electromagnetic wave is given by :</p>
<p>$$
\overrightarrow{\mathrm{B}}=2 \times 10^{-8} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} \mathrm{t}\right) \,\hat{j} \mathrm{~T}$$.</p>
<p>The amplitude of the electric field would be :</p> | [{"identifier": "A", "content": "$$6\\, \\mathrm{Vm}^{-1}$$ along $$x$$-axis"}, {"identifier": "B", "content": "$$3\\, \\mathrm{Vm}^{-1}$$ along $$z$$-axis"}, {"identifier": "C", "content": "$$6\\, \\mathrm{Vm}^{-1}$$ along $$z$$-axis"}, {"identifier": "D", "content": "$$2 \\times 10^{-8} \\,\\mathrm{Vm}^{-1}$$ along $... | ["C"] | null | <p>Speed of light $$c = {\omega \over k} = {{1.5 \times {{10}^{11}}} \over {0.5 \times {{10}^3}}} = 3 \times {10^8}$$ m/sec</p>
<p>So, $${E_0} = {B_0}c$$</p>
<p>$$ = 2 \times {10^{ - 8}} \times 3 \times {10^8}$$</p>
<p>$$ = 6$$ V/m</p>
<p>Direction will be along z-axis.</p> | mcq | jee-main-2022-online-26th-july-morning-shift | 10,384 |
1l6i0kafr | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>The oscillating magnetic field in a plane electromagnetic wave is given by <br/><br/>$$B_{y}=5 \times 10^{-6} \sin 1000 \pi\left(5 x-4 \times 10^{8} t\right) T$$. The amplitude of electric field will be :</p> | [{"identifier": "A", "content": "$$15 \\times 10^{2} \\,\\mathrm{Vm}^{-1}$$"}, {"identifier": "B", "content": "$$5 \\times 10^{-6} \\,\\mathrm{Vm}^{-1}$$"}, {"identifier": "C", "content": "$$16 \\times 10^{12} \\,\\mathrm{Vm}^{-1}$$"}, {"identifier": "D", "content": "$$4 \\times 10^{2} \\,\\mathrm{Vm}^{-1}$$"}] | ["D"] | null | In an electromagnetic wave, the magnitude of the electric field (E) and the magnetic field (B) are related by the speed of the wave (v), which can be represented as :
<br/><br/>$$E = Bv$$
<br/><br/>Here, B is the magnetic field, E is the electric field, and v is the speed of the electromagnetic wave.
<br/><br/>In t... | mcq | jee-main-2022-online-26th-july-evening-shift | 10,385 |
1l6i1kxdj | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>A velocity selector consists of electric field $$\vec{E}=E \,\hat{k}$$ and magnetic field $$\vec{B}=B \,\hat{j}$$ with $$B=12 \,m T$$. The value of $$E$$ required for an electron of energy $$728 \,\mathrm{e} V$$ moving along the positive $$x$$-axis to pass undeflected is :</p>
<p>(Given, mass of electron $$=9.1 \tim... | [{"identifier": "A", "content": "$$192 \\,\\mathrm{kVm}^{-1}$$"}, {"identifier": "B", "content": "$$192 \\,\\mathrm{mVm}^{-1}$$"}, {"identifier": "C", "content": "$$9600 \\,\\mathrm{kVm}^{-1}$$"}, {"identifier": "D", "content": "$$16 \\,\\mathrm{kVm}^{-1}$$"}] | ["A"] | null | <p>$$v = {E \over B}$$ and $$K = {1 \over 2}m{v^2}$$</p>
<p>$$ \Rightarrow \sqrt {{{2K} \over m}} \times B = E$$</p>
<p>$$ \Rightarrow E = \sqrt {{{2 \times 728 \times 1.6 \times {{10}^{ - 19}}} \over {9.1 \times {{10}^{ - 31}}}}} \times 12 \times {10^{ - 3}}$$</p>
<p>$$ = 192000$$ V/m</p> | mcq | jee-main-2022-online-26th-july-evening-shift | 10,386 |
1l6jibjjv | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>A beam of light travelling along $$X$$-axis is described by the electric field $$E_{y}=900 \sin \omega(\mathrm{t}-x / c)$$. The ratio of electric force to magnetic force on a charge $$\mathrm{q}$$ moving along $$Y$$-axis with a speed of $$3 \times 10^{7} \mathrm{~ms}^{-1}$$ will be :</p>
<p>(Given speed of light $$=... | [{"identifier": "A", "content": "1 : 1"}, {"identifier": "B", "content": "1 : 10"}, {"identifier": "C", "content": "10 : 1"}, {"identifier": "D", "content": "1 : 2"}] | ["C"] | null | <p>Ratio $$ = {{|q\overrightarrow E |} \over {|q\overrightarrow v \times \overrightarrow B |}}$$</p>
<p>$$ = {E \over {vB}} = {{{v_{wave}}} \over v}$$</p>
<p>$$\Rightarrow$$ Ratio $$ = {{3 \times {{10}^8}} \over {3 \times {{10}^7}}} = 10$$</p> | mcq | jee-main-2022-online-27th-july-morning-shift | 10,387 |
1l6nsuzar | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>Sun light falls normally on a surface of area $$36 \mathrm{~cm}^{2}$$ and exerts an average force of $$7.2 \times 10^{-9} \mathrm{~N}$$ within a time period of 20 minutes. Considering a case of complete absorption, the energy flux of incident light is</p> | [{"identifier": "A", "content": "$$25.92 \\times 10^{2} \\mathrm{~W} / \\mathrm{cm}^{2}$$"}, {"identifier": "B", "content": "$$8.64 \\times 10^{-6} \\mathrm{~W} / \\mathrm{cm}^{2}$$"}, {"identifier": "C", "content": "$$6.0 \\mathrm{~W} / \\mathrm{cm}^{2}$$"}, {"identifier": "D", "content": "$$0.06\\mathrm{~W} / \\mathr... | ["D"] | null | <p>Pressure $$ = {l \over c}$$</p>
<p>$$ \Rightarrow {F \over A} = {l \over c}$$</p>
<p>$$ \Rightarrow l = {{7.2 \times {{10}^{ - 9}} \times 3 \times {{10}^8}} \over {36 \times {{10}^{ - 4}}}}$$ W/m<sup>2</sup></p>
<p>$$ = 600$$ W/m<sup>2</sup></p>
<p>$$ \Rightarrow l = 0.06$$ W/cm<sup>2</sup></p> | mcq | jee-main-2022-online-28th-july-evening-shift | 10,389 |
1l6riap6v | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>Nearly 10% of the power of a $$110 \mathrm{~W}$$ light bulb is converted to visible radiation. The change in average intensities of visible radiation, at a distance of $$1 \mathrm{~m}$$ from the bulb to a distance of $$5 \mathrm{~m}$$ is $$a \times 10^{-2} \mathrm{~W} / \mathrm{m}^{2}$$. The value of 'a' will be ___... | [] | null | 84 | $\mathbf{P}^{\prime}=10 \%$ of $110 \mathbf{W}$
<br/><br/>$=\frac{10}{100} \times 110 \mathrm{~W}$
<br/><br/>$=11 \mathrm{~W}$
<br/><br/>$\mathrm{I}_1-\mathrm{I}_2=\frac{\mathrm{P}^{\prime}}{4 \pi \mathrm{r}_1^2}-\frac{\mathrm{P}^{\prime}}{4 \pi \mathrm{r}_2^2}$
<br/><br/>$=\frac{11}{4 \pi}\left[\frac{1}{1}-\frac{1}{25... | integer | jee-main-2022-online-29th-july-evening-shift | 10,390 |
1ldnw4zqr | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>The ratio of average electric energy density and total average energy density of electromagnetic wave is :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$\\frac{1}{2}$$"}] | ["D"] | null | Avg electric energy density $=\frac{1}{4} \varepsilon_0 \mathrm{E}_0^2$
<br/><br/>Total Avg energy density $=\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2$
<br/><br/>$$ \therefore $$ Ratio of average electric energy density and total Avg energy density
<br/><br/>$$
= \frac{\frac{1}{4} \varepsilon_0 \mathrm{E}_0^2}{\frac{1}{... | mcq | jee-main-2023-online-1st-february-evening-shift | 10,391 |
1ldpncw54 | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>In a medium the speed of light wave decreases to $$0.2$$ times to its speed in free space The ratio of relative permittivity to the refractive index of the medium is $$x: 1$$. The value of $$x$$ is _________.</p>
<p>(Given speed of light in free space $$=3 \times 10^{8} \mathrm{~m} \mathrm{~s}^{-1}$$ and for the giv... | [] | null | 5 | We know that $v=\frac{c}{n}=\frac{c}{\sqrt{\varepsilon_{r}}}$
<br/><br/>Putting the values:
<br/><br/>$0.2 c=\frac{c}{\sqrt{\varepsilon_{r}}}$
<br/><br/>$\Rightarrow \sqrt{\varepsilon_{r}}=5$
<br/><br/>$\Rightarrow$ Required ratio $=\frac{\varepsilon_{r}}{n}=\frac{\varepsilon_{r}}{\sqrt{\varepsilon_{r}}}=\sqrt{\var... | integer | jee-main-2023-online-31st-january-morning-shift | 10,392 |
ldqvfxpf | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | A point source of $100 \mathrm{~W}$ emits light with $5 \%$ efficiency. At a distance of $5 \mathrm{~m}$ from the source, the intensity produced by the electric field component is: | [{"identifier": "A", "content": "$\\frac{1}{40 \\pi} \\frac{W}{m^2}$"}, {"identifier": "B", "content": "$\\frac{1}{10 \\pi} \\frac{W}{m^2}$"}, {"identifier": "C", "content": "$\\frac{1}{20 \\pi} \\frac{W}{m^2}$"}, {"identifier": "D", "content": "$\\frac{1}{2 \\pi} \\frac{W}{m^2}$"}] | ["A"] | null | <p>A point source of $100 \mathrm{~W}$ emits light with $5 \%$ efficiency. At a distance of $5 \mathrm{~m}$ from the source, the intensity produced by the electric field component is evaluated as follows:</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leoi23xo/0eb20d87-a1d6-4001-9b1f-f87c1a57b... | mcq | jee-main-2023-online-30th-january-evening-shift | 10,393 |
1ldr379ya | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>A point source of light is placed at the centre of curvature of a hemispherical surface. The source emits a power of $$24 \mathrm{~W}$$. The radius of curvature of hemisphere is $$10 \mathrm{~cm}$$ and the inner surface is completely reflecting. The force on the hemisphere due to the light falling on it is _________... | [] | null | 4 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lepxxcug/c3ca6ed1-a3c6-4b09-ab77-d8f46565abdd/ea627c80-b854-11ed-8195-4f3c56fa1eb5/file-1lepxxcuh.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lepxxcug/c3ca6ed1-a3c6-4b09-ab77-d8f46565abdd/ea627c80-b854-11ed-8195-4f3c56fa1eb5... | integer | jee-main-2023-online-30th-january-morning-shift | 10,394 |
1ldsa8zxf | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>Given below are two statements :</p>
<p>Statement I : Electromagnetic waves are not deflected by electric and magnetic field.</p>
<p>Statement II : The amplitude of electric field and the magnetic field in electromagnetic waves are related to each other as $${E_0} = \sqrt {{{{\mu _0}} \over {{\varepsilon _0}}}} {B_0... | [{"identifier": "A", "content": "Both Statement I and Statement II are true"}, {"identifier": "B", "content": "Statement I is true and Statement II is false"}, {"identifier": "C", "content": "Both Statement I and Statement II are false"}, {"identifier": "D", "content": "Statement I is false but Statement II is true"}] | ["B"] | null | <p>Statement I is correct as photon do not carry any charge, hence cannot feel force from either fields.</p>
<p>Statement II is wrong as $$E_0=cB_0$$</p>
<p>$${E_0} = {{{B_0}} \over {\sqrt {{\mu _0}{\varepsilon _0}} }}$$</p> | mcq | jee-main-2023-online-29th-january-evening-shift | 10,395 |
1ldtxzrmx | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>Match List I with List II</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;bo... | [{"identifier": "A", "content": "A-I, B-II, C-III, D-IV"}, {"identifier": "B", "content": "A-III, B-IV, C-I, D-II"}, {"identifier": "C", "content": "A-IV, B-I, C-II, D-III"}, {"identifier": "D", "content": "A-II, B-III, C-IV, D-I"}] | ["C"] | null | Gauss's law $\oint \vec{E} \cdot \overrightarrow{d s}=\frac{q}{\epsilon_{0}} \quad(\mathrm{~A} \rightarrow \mathrm{IV})$
<br/><br/>
Faraday's law $\oint \vec{E} \cdot \overrightarrow{d l}=-\frac{d \phi_{B}}{d t} \quad(\mathrm{~B} \rightarrow \mathrm{I})$
<br/><br/>
Gauss's law in magnetism $\oint \vec{B} \cdot \overrig... | mcq | jee-main-2023-online-25th-january-evening-shift | 10,397 |
1lduhw3w4 | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>An electromagnetic wave is transporting energy in the negative $$z$$ direction. At a certain point and certain time the direction of electric field of the wave is along positive $$y$$ direction. What will be the direction of the magnetic field of the wave at that point and instant?</p> | [{"identifier": "A", "content": "Negative direction of $$y$$"}, {"identifier": "B", "content": "Positive direction of $$z$$"}, {"identifier": "C", "content": "Positive direction of $$x$$"}, {"identifier": "D", "content": "Negative direction $$x$$"}] | ["C"] | null | As, poynting vector<br/><br/>
$$
\overrightarrow{\mathrm{S}}=\overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{H}}
$$<br/><br/>
Given energy transport $=$ negative $\mathrm{z}$ direction<br/><br/> Electric field $=$ positive $\mathrm{y}$ direction <br/><br/>$(-\hat{\mathrm{k}})=(+\hat{\mathrm{j}}) \times[\hat{... | mcq | jee-main-2023-online-25th-january-morning-shift | 10,398 |
1ldyelb09 | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>In $$\overrightarrow E $$ and $$\overrightarrow K $$ represent electric field and propagation vectors of the EM waves in vacuum, then magnetic field vector is given by :</p>
<p>($$\omega$$ - angular frequency) :</p> | [{"identifier": "A", "content": "$${1 \\over \\omega }\\left( {\\overline K \\times \\overline E } \\right)$$"}, {"identifier": "B", "content": "$$\\overline K \\times \\overline E $$"}, {"identifier": "C", "content": "$$\\omega \\left( {\\overline K \\times \\overline E } \\right)$$"}, {"identifier": "D", "content"... | ["A"] | null | Magnetic field vector will be in the direction of $\hat{\mathrm{K}} \times \hat{\mathrm{E}}$<br/><br/>
magnitude of $B=\frac{E}{C}=\frac{K}{\omega} E$<br/><br/>
Or $\overrightarrow{\mathrm{B}}=\frac{1}{\omega}(\overrightarrow{\mathrm{K}} \times \overrightarrow{\mathrm{E}})$ | mcq | jee-main-2023-online-24th-january-morning-shift | 10,400 |
1lgq2j714 | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>Which of the following Maxwell's equation is valid for time varying conditions but not valid for static conditions :</p> | [{"identifier": "A", "content": "$$\\oint \\overrightarrow{\\mathrm{E}} \\cdot \\overrightarrow{d l}=0$$"}, {"identifier": "B", "content": "$$\\oint \\vec{B} \\cdot \\overrightarrow{d l}=\\mu_{0} I$$"}, {"identifier": "C", "content": "$$\\oint \\vec{E} \\cdot \\overrightarrow{d l}=-\\frac{\\partial \\phi_{B}}{\\partial... | ["C"] | null | Maxwell's equations describe the behavior of electric and magnetic fields. There are four equations, and each has a specific role. In the given options, Option C refers to Faraday's Law of Electromagnetic Induction, which is the only equation among the options that is not valid for static conditions.
<br/><br/>
Option ... | mcq | jee-main-2023-online-13th-april-morning-shift | 10,402 |
1lgswh6t1 | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>A plane electromagnetic wave of frequency $$20 ~\mathrm{MHz}$$ propagates in free space along $$\mathrm{x}$$-direction. At a particular space and time, $$\overrightarrow{\mathrm{E}}=6.6 \hat{j} \mathrm{~V} / \mathrm{m}$$. What is $$\overrightarrow{\mathrm{B}}$$ at this point?</p> | [{"identifier": "A", "content": "$$-2.2 \\times 10^{-8} \\hat{i} T$$"}, {"identifier": "B", "content": "$$2.2 \\times 10^{-8} \\hat{i} T$$"}, {"identifier": "C", "content": "$$2.2 \\times 10^{-8} \\hat{k} T$$"}, {"identifier": "D", "content": "$$-2.2 \\times 10^{-8} \\hat{k} T$$"}] | ["C"] | null | In free space, the relationship between the electric field (E) and the magnetic field (B) in an electromagnetic wave is given by:
<br/><br/>
$$
B = \frac{E}{c}
$$
<br/><br/>
where c is the speed of light in a vacuum, approximately equal to $$3 \times 10^8 \mathrm{~m} / \mathrm{s}$$. We are given that the electric field... | mcq | jee-main-2023-online-11th-april-evening-shift | 10,403 |
1lgvrnazv | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>The amplitude of magnetic field in an electromagnetic wave propagating along y-axis is $$6.0 \times 10^{-7} \mathrm{~T}$$. The maximum value of electric field in the electromagnetic wave is</p> | [{"identifier": "A", "content": "$$6.0 \\times 10^{-7} ~\\mathrm{Vm}^{-1}$$"}, {"identifier": "B", "content": "$$5 \\times 10^{14} ~\\mathrm{Vm}^{-1}$$"}, {"identifier": "C", "content": "$$180 ~\\mathrm{Vm}^{-1}$$"}, {"identifier": "D", "content": "$$2 \\times 10^{15} ~\\mathrm{Vm}^{-1}$$"}] | ["C"] | null | In an electromagnetic wave, the maximum value of the electric field E is related to the maximum value of the magnetic field B by the equation
<br/><br/>
$E = cB$
<br/><br/>
where c is the speed of light in a vacuum, which is approximately $3 × 10^8 m/s$.
<br/><br/>
Given that the amplitude of the magnetic field B is $6... | mcq | jee-main-2023-online-10th-april-evening-shift | 10,405 |
1lgxwty7j | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>The energy of an electromagnetic wave contained in a small volume oscillates with</p> | [{"identifier": "A", "content": "double the frequency of the wave"}, {"identifier": "B", "content": "the frequency of the wave"}, {"identifier": "C", "content": "half the frequency of the wave"}, {"identifier": "D", "content": "zero frequency"}] | ["A"] | null | <p>The energy of an electromagnetic wave contained in a small volume oscillates with double the frequency of the wave. </p>
<p><b>Here's why</b>: The electromagnetic wave consists of oscillating electric and magnetic fields. The energy density of the wave is proportional to the square of the amplitude of these fiel... | mcq | jee-main-2023-online-10th-april-morning-shift | 10,406 |
1lh25dl4z | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>For the plane electromagnetic wave given by $$E=E_{0} \sin (\omega t-k x)$$ and $$B=B_{0} \sin (\omega t-k x)$$, the ratio of average electric energy density to average magnetic energy density is</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1/2"}] | ["A"] | null | <p>The average energy density of an electromagnetic wave is equally shared between the electric field and the magnetic field. This means that the average electric energy density is equal to the average magnetic energy density. </p>
<p>The average electric energy density ($u_E$) is given by:</p>
<p>$ u_E = \frac{1}{2} \... | mcq | jee-main-2023-online-6th-april-morning-shift | 10,407 |
1lh2zrxuq | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>The energy density associated with electric field $$\vec{E}$$ and magnetic field $$\vec{B}$$ of an electromagnetic wave in free space is given by $$\left(\epsilon_{0}-\right.$$ permittivity of free space, $$\mu_{0}-$$ permeability of free space)</p> | [{"identifier": "A", "content": "$$U_{E}=\\frac{\\epsilon_{0} E^{2}}{2}, U_{B}=\\frac{B^{2}}{2 \\mu_{0}}$$"}, {"identifier": "B", "content": "$$U_{E}=\\frac{E^{2}}{2 \\epsilon_{0}}, U_{B}=\\frac{\\mu_{0} B^{2}}{2}$$"}, {"identifier": "C", "content": "$$U_{E}=\\frac{\\epsilon_{0} E^{2}}{2}, U_{B}=\\frac{\\mu_{0} B^{2}}{... | ["A"] | null | <p>The energy density associated with the electric field $$\vec{E}$$ and magnetic field $$\vec{B}$$ of an electromagnetic wave in free space is given by:</p>
<p>For the electric field:
$$U_{E} = \frac{1}{2} \epsilon_{0} E^2$$</p>
<p>For the magnetic field:
$$U_{B} = \frac{1}{2} \frac{B^2}{\mu_{0}}$$</p>
<p>These expres... | mcq | jee-main-2023-online-6th-april-evening-shift | 10,408 |
lsan1vgp | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | If frequency of electromagnetic wave is $60 \mathrm{~MHz}$ and it travels in air along $z$ direction then the corresponding electric and magnetic field vectors will be mutually perpendicular to each other and the wavelength of the wave (in $\mathrm{m}$ ) is : | [{"identifier": "A", "content": "2.5"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "2"}] | ["B"] | null | <p>The speed of electromagnetic waves in air (and in a vacuum) is approximately the speed of light, which we denote as $$ c $$. The speed of light $$ c $$ is $$ 3 \times 10^8 $$ meters per second. The relationship between the speed of light $$ c $$, the frequency $$ f $$, and the wavelength $$ \lambda $$ of an electrom... | mcq | jee-main-2024-online-1st-february-evening-shift | 10,409 |
jaoe38c1lsc3rg5e | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>A plane electromagnetic wave propagating in $$\mathrm{x}$$-direction is described by</p>
<p>$$E_y=\left(200 \mathrm{Vm}^{-1}\right) \sin \left[1.5 \times 10^7 t-0.05 x\right] \text {; }$$</p>
<p>The intensity of the wave is :</p>
<p>(Use $$\epsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2... | [{"identifier": "A", "content": "$$35.4 \\mathrm{~Wm}^{-2}$$\n"}, {"identifier": "B", "content": "$$53.1 \\mathrm{~Wm}^{-2}$$\n"}, {"identifier": "C", "content": "$$26.6 \\mathrm{~Wm}^{-2}$$\n"}, {"identifier": "D", "content": "$$106.2 \\mathrm{~Wm}^{-2}$$"}] | ["B"] | null | <p>$$\begin{aligned}
& \mathrm{I}=\frac{1}{2} \varepsilon_0 \mathrm{E}_0^2 \times \mathrm{c} \\
& \mathrm{I}=\frac{1}{2} \times 8.85 \times 10^{-12} \times 4 \times 10^4 \times 3 \times 10^8 \\
& \mathrm{I}=53.1 \mathrm{~W} / \mathrm{m}^2
\end{aligned}$$</p> | mcq | jee-main-2024-online-27th-january-morning-shift | 10,410 |
jaoe38c1lscpkphx | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>An object is placed in a medium of refractive index 3 . An electromagnetic wave of intensity $$6 \times 10^8 \mathrm{~W} / \mathrm{m}^2$$ falls normally on the object and it is absorbed completely. The radiation pressure on the object would be (speed of light in free space $$=3 \times 10^8 \mathrm{~m} / \mathrm{s}$$... | [{"identifier": "A", "content": "$$6 \\mathrm{~Nm}^{-2}$$\n"}, {"identifier": "B", "content": "$$36 \\mathrm{~Nm}^{-2}$$\n"}, {"identifier": "C", "content": "$$18 \\mathrm{~Nm}^{-2}$$\n"}, {"identifier": "D", "content": "$$2 \\mathrm{~Nm}^{-2}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& \text { Radiation pressure }=\frac{I}{\mathrm{~V}} \\
& =\frac{\mathrm{I} \cdot \mu}{\mathrm{c}} \\
& =\frac{6 \times 10^8 \times 3}{3 \times 10^8} \\
& =6 \mathrm{~N} / \mathrm{m}^2
\end{aligned}$$</p> | mcq | jee-main-2024-online-27th-january-evening-shift | 10,411 |
jaoe38c1lsd5dntd | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>Given below are two statements:</p>
<p>Statement I: Electromagnetic waves carry energy as they travel through space and this energy is equally shared by the electric and magnetic fields.</p>
<p>Statement II: When electromagnetic waves strike a surface, a pressure is exerted on the surface.</p>
<p>In the light of the... | [{"identifier": "A", "content": "Statement I is incorrect but Statement II is correct.\n"}, {"identifier": "B", "content": "Both Statement I and Statement II are correct.\n"}, {"identifier": "C", "content": "Statement I is correct but Statement II is incorrect.\n"}, {"identifier": "D", "content": "Both Statement I and ... | ["B"] | null | <p>$$\begin{aligned}
& \frac{1}{2} \varepsilon_0 \mathrm{E}^2=\frac{\mathrm{B}^2}{2 \mu_0} \\
& \because \mathrm{E}=\mathrm{CB} \text { and } \mathrm{C}=\frac{1}{\mu_0 \varepsilon_0}
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-evening-shift | 10,412 |
jaoe38c1lse6nvzl | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>In a plane EM wave, the electric field oscillates sinusoidally at a frequency of $$5 \times 10^{10} \mathrm{~Hz}$$ and an amplitude of $$50 \mathrm{~Vm}^{-1}$$. The total average energy density of the electromagnetic field of the wave is : [Use $$\varepsilon_0=8.85 \times 10^{-12} \mathrm{C}^2 / \mathrm{Nm}^2$$ ]</p... | [{"identifier": "A", "content": "$$4.425 \\times 10^{-8} \\mathrm{Jm}^{-3}$$\n"}, {"identifier": "B", "content": "$$2.212 \\times 10^{-10} \\mathrm{Jm}^{-3}$$\n"}, {"identifier": "C", "content": "$$2.212 \\times 10^{-8} \\mathrm{Jm}^{-3}$$\n"}, {"identifier": "D", "content": "$$1.106 \\times 10^{-8} \\mathrm{Jm}^{-3}$$... | ["D"] | null | <p>The average energy density of an electromagnetic wave is given by the formula:</p>
<p>$$U = \frac{1}{2} \varepsilon_0 E^2 + \frac{1}{2} \frac{B^2}{\mu_0}$$</p>
<p>For a plane electromagnetic wave, the electric field (E) and the magnetic field (B) contribute equally to the energy density. Therefore, we can focus on... | mcq | jee-main-2024-online-31st-january-morning-shift | 10,413 |
1lsgdfsz5 | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>The electric field of an electromagnetic wave in free space is represented as $$\overrightarrow{\mathrm{E}}=\mathrm{E}_0 \cos (\omega \mathrm{t}-\mathrm{kz}) \hat{i}$$. The corresponding magnetic induction vector will be :</p> | [{"identifier": "A", "content": "$$\\overrightarrow{\\mathrm{B}}=\\mathrm{E}_0 \\mathrm{C} \\cos (\\omega \\mathrm{t}+\\mathrm{k} z) \\hat{j}$$\n"}, {"identifier": "B", "content": "$$\\overrightarrow{\\mathrm{B}}=\\frac{\\mathrm{E}_0}{\\mathrm{C}} \\cos (\\omega \\mathrm{t}-\\mathrm{kz}) \\hat{j}$$\n"}, {"identifier": ... | ["B"] | null | <p>$$\begin{aligned}
& \text { Given } \vec{E}=E_0 \cos (\omega t-k z) \hat{i} \\
& \vec{B}=\frac{E_0}{C} \cos (\omega t-k z) \hat{j} \\
& \hat{C}=\hat{E} \times \hat{B}
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift | 10,415 |
luxweb5k | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>The magnetic field in a plane electromagnetic wave is $$\mathrm{B}_{\mathrm{y}}=\left(3.5 \times 10^{-7}\right) \sin \left(1.5 \times 10^3 x+0.5 \times 10^{11} t\right) \mathrm{T}$$. The corresponding electric field will be :</p> | [{"identifier": "A", "content": "$$E_z=105 \\sin \\left(1.5 \\times 10^3 x+0.5 \\times 10^{11} t\\right) \\mathrm{Vm}^{-1}$$\n"}, {"identifier": "B", "content": "$$E_y=10.5 \\sin \\left(1.5 \\times 10^3 x+0.5 \\times 10^{11} t\\right) \\mathrm{Vm}^{-1}$$\n"}, {"identifier": "C", "content": "$$E_y=1.17 \\sin \\left(1.5 ... | ["A"] | null | <p>For an electromagnetic wave propagating in free space, the relationship between the magnitudes of the electric field ($$E$$) and the magnetic field ($$B$$) can be described using the equation:</p>
<p>$E = cB$</p>
<p>where</p>
<ul>
<li>$c$ is the speed of light in vacuum, approximately $3.0 \times 10^8 \, \te... | mcq | jee-main-2024-online-9th-april-evening-shift | 10,416 |
luy9clsu | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>A plane EM wave is propagating along $$x$$ direction. It has a wavelength of $$4 \mathrm{~mm}$$. If electric field is in $$y$$ direction with the maximum magnitude of $$60 \mathrm{~Vm}^{-1}$$, the equation for magnetic field is :</p> | [{"identifier": "A", "content": "$$\\mathrm{B}_z=2 \\times 10^{-7} \\sin \\left[\\frac{\\pi}{2}\\left(x-3 \\times 10^8 \\mathrm{t}\\right)\\right] \\hat{\\mathrm{k}} \\mathrm{T}$$\n"}, {"identifier": "B", "content": "$$\\mathrm{B}_z=2 \\times 10^{-7} \\sin \\left[\\frac{\\pi}{2} \\times 10^3\\left(x-3 \\times 10^8 \\ma... | ["B"] | null | <p>To find the correct equation for the magnetic field of the plane electromagnetic wave given its parameters, we can use a couple of known relationships from electromagnetism.</p>
<p>Firstly, the wavelength ($ \lambda $) of the wave is given as $4 \mathrm{~mm} = 4 \times 10^{-3} \mathrm{~m}$. The speed of light (and ... | mcq | jee-main-2024-online-9th-april-morning-shift | 10,417 |
lv0vxsz4 | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>The electric field in an electromagnetic wave is given by $$\overrightarrow{\mathrm{E}}=\hat{i} 40 \cos \omega(\mathrm{t}-z / \mathrm{c}) \mathrm{NC}^{-1}$$. The magnetic field induction of this wave is (in SI unit) :</p> | [{"identifier": "A", "content": "$$\\overrightarrow{\\mathrm{B}}=\\hat{j} \\frac{40}{\\mathrm{c}} \\cos \\omega(\\mathrm{t}-z / \\mathrm{c})$$\n"}, {"identifier": "B", "content": "$$\\overrightarrow{\\mathrm{B}}=\\hat{i} \\frac{40}{\\mathrm{c}} \\cos \\omega(\\mathrm{t}-z / \\mathrm{c})$$\n"}, {"identifier": "C", "cont... | ["A"] | null | <p>To determine the magnetic field induction of the given electromagnetic wave, we need to use the relationship between the electric field $$\overrightarrow{\mathrm{E}}$$ and the magnetic field $$\overrightarrow{\mathrm{B}}$$ in an electromagnetic wave. For an electromagnetic wave propagating in vacuum, the following r... | mcq | jee-main-2024-online-4th-april-morning-shift | 10,418 |
lv5gsw9i | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>Average force exerted on a non-reflecting surface at normal incidence is $$2.4 \times 10^{-4} \mathrm{~N}$$. If $$360 \mathrm{~W} / \mathrm{cm}^2$$ is the light energy flux during span of 1 hour 30 minutes, Then the area of the surface is:</p> | [{"identifier": "A", "content": "$$20 \\mathrm{~m}^2$$\n"}, {"identifier": "B", "content": "$$0.2 \\mathrm{~m}^2$$\n"}, {"identifier": "C", "content": "$$0.1 \\mathrm{~m}^2$$\n"}, {"identifier": "D", "content": "$$0.02 \\mathrm{~m}^2$$"}] | ["D"] | null | <p>To solve for the area of the surface, we need to understand the relationship between the force exerted by the light, the light energy flux, and the area of the surface. The pressure exerted by the light on a non-reflecting surface is given by the formula:</p>
<p>
<p>$$ P = \frac{F}{A} $$</p>
</p>
<p>where $$P$$ ... | mcq | jee-main-2024-online-8th-april-morning-shift | 10,419 |
lvb29elu | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>In the given electromagnetic wave $$\mathrm{E}_{\mathrm{y}}=600 \sin (\omega t-\mathrm{kx}) \mathrm{Vm}^{-1}$$, intensity of the associated light beam is (in $$\mathrm{W} / \mathrm{m}^2$$ : (Given $$\epsilon_0=9 \times 10^{-12} \mathrm{C}^2 \mathrm{~N}^{-1} \mathrm{~m}^{-2}$$ )</p> | [{"identifier": "A", "content": "486"}, {"identifier": "B", "content": "729"}, {"identifier": "C", "content": "243"}, {"identifier": "D", "content": "972"}] | ["A"] | null | <p>To find the intensity of the given electromagnetic wave, we need to use the formula for the intensity of an electromagnetic wave:</p>
<p>$$ I = \frac{1}{2} \epsilon_0 c E_0^2 $$</p>
<p>where:</p>
<ul>
<li>$$I$$ is the intensity in $$\mathrm{W} / \mathrm{m}^2$$</li>
<li>$$\epsilon_0$$ is the permittivity of free... | mcq | jee-main-2024-online-6th-april-evening-shift | 10,420 |
lvc57nqe | physics | electromagnetic-waves | displacement-current-and-properties-of-em-waves | <p>Electromagnetic waves travel in a medium with speed of $$1.5 \times 10^8 \mathrm{~m} \mathrm{~s}^{-1}$$. The relative permeability of the medium is 2.0. The relative permittivity will be:</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "5"}] | ["C"] | null | <p>$$\begin{aligned}
& v=\frac{1}{\sqrt{\mu_0 \mu_r \varepsilon_0 \cdot \varepsilon_r}} \\
& \Rightarrow 1.5 \times 10^8=\frac{3 \times 10^8}{\sqrt{2 \cdot \varepsilon_r}} \Rightarrow \varepsilon_r=2
\end{aligned}$$</p> | mcq | jee-main-2024-online-6th-april-morning-shift | 10,421 |
9TyLi4Ad66ficdgT | physics | electromagnetic-waves | em-spectrum | Which of the following are not electromagnetic waves? | [{"identifier": "A", "content": "cosmic rays "}, {"identifier": "B", "content": "gamma rays "}, {"identifier": "C", "content": "$$\\beta $$-rays "}, {"identifier": "D", "content": "$$X$$-rays "}] | ["C"] | null | $$\beta $$ -rays are fast moving beam of electrons. | mcq | aieee-2002 | 10,422 |
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