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1lgvrw4rj | physics | dual-nature-of-radiation | photoelectric-effect | <p>The variation of stopping potential $$\left(\mathrm{V}_{0}\right)$$ as a function of the frequency $$(v)$$ of the incident light for a metal is shown in figure. The work function of the surface is</p>
<p><img src="data:image/png;base64,UklGRh4RAABXRUJQVlA4IBIRAAAQ9gCdASoAA6ECP4HA2GY2L7knITFpGyAwCWlu+/qt3Vl6f8SejJT19... | [{"identifier": "A", "content": "1.36 eV"}, {"identifier": "B", "content": "18.6 eV"}, {"identifier": "C", "content": "2.98 eV"}, {"identifier": "D", "content": "2.07 eV"}] | ["D"] | null | $$
\begin{aligned}
& e V_0=\frac{h c}{\lambda}-\phi \\\\
& V_0=\frac{h c}{e \lambda}-\left(\frac{\phi}{e}\right)=\frac{h v}{e}-\left(\frac{\phi}{e}\right)
\end{aligned}
$$
<br/><br/>When $V_0=0$,
<br/><br/>$$
\begin{aligned}
\frac{\phi}{e} & =\left(\frac{h v}{e}\right) \\\\
\phi & =h v=\frac{6.626 \times 10^{-34} \time... | mcq | jee-main-2023-online-10th-april-evening-shift | 10,160 |
1lgyqc48m | physics | dual-nature-of-radiation | photoelectric-effect | <p>In photo electric effect</p>
<p>A. The photocurrent is proportional to the intensity of the incident radiation</p>
<p>B. Maximum Kinetic energy with which photoelectrons are emitted depends on the intensity of incident light.</p>
<p>C. Max. K.E with which photoelectrons are emitted depends on the frequency of incide... | [{"identifier": "A", "content": "A and E only"}, {"identifier": "B", "content": "A and B only"}, {"identifier": "C", "content": "B and C only"}, {"identifier": "D", "content": "A and C only"}] | ["D"] | null | <p>The photoelectric effect is the phenomenon of emission of electrons (or photoelectrons) from the surface of a metal when it is illuminated by light of sufficient energy. The observations from the photoelectric effect led to the development of quantum theory. </p>
<p>According to the principles of the photoelectric e... | mcq | jee-main-2023-online-8th-april-evening-shift | 10,161 |
1lh3027a8 | physics | dual-nature-of-radiation | photoelectric-effect | <p>The work functions of Aluminium and Gold are $$4.1 ~\mathrm{eV}$$ and and $$5.1 ~\mathrm{eV}$$ respectively. The ratio of the slope of the stopping potential versus frequency plot for Gold to that of Aluminium is</p> | [{"identifier": "A", "content": "1.5"}, {"identifier": "B", "content": "1.24"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2"}] | ["C"] | null | <p>We are given the work functions of Aluminium and Gold as $$4.1 ~\mathrm{eV}$$ and $$5.1 ~\mathrm{eV}$$, respectively.</p>
<p>The stopping potential ($$V_s$$) is related to the frequency ($$f$$) of the incident light by the equation:</p>
<p>$$eV_s = h(f - f_0)$$</p>
<p>Where $$e$$ is the charge of an electron, $$h$$ ... | mcq | jee-main-2023-online-6th-april-evening-shift | 10,162 |
jaoe38c1lscpar9g | physics | dual-nature-of-radiation | photoelectric-effect | <p>The threshold frequency of a metal with work function $$6.63 \mathrm{~eV}$$ is :</p> | [{"identifier": "A", "content": "$$16 \\times 10^{15} \\mathrm{~Hz}$$\n"}, {"identifier": "B", "content": "$$16 \\times 10^{12} \\mathrm{~Hz}$$\n"}, {"identifier": "C", "content": "$$1.6 \\times 10^{15} \\mathrm{~Hz}$$\n"}, {"identifier": "D", "content": "$$1.6 \\times 10^{12} \\mathrm{~Hz}$$"}] | ["C"] | null | <p>The threshold frequency, $$ \nu_0 $$, corresponds to the minimum frequency of light required to eject electrons from the surface of a metal, a phenomenon known as the photoelectric effect. The work function, represented by $$ \phi $$, is the minimum energy needed to remove an electron from the surface of the metal.<... | mcq | jee-main-2024-online-27th-january-evening-shift | 10,163 |
jaoe38c1lsd5j857 | physics | dual-nature-of-radiation | photoelectric-effect | <p>In a photoelectric effect experiment a light of frequency 1.5 times the threshold frequency is made to fall on the surface of photosensitive material. Now if the frequency is halved and intensity is doubled, the number of photo electrons emitted will be:</p> | [{"identifier": "A", "content": "doubled\n"}, {"identifier": "B", "content": "halved\n"}, {"identifier": "C", "content": "Zero\n"}, {"identifier": "D", "content": "quadrupled"}] | ["C"] | null | <p>Since $$\frac{\mathrm{f}}{2}<\mathrm{f}_0$$</p>
<p>i.e. the incident frequency is less than threshold frequency. Hence there will be no emission of photoelectrons.</p>
<p>$$\Rightarrow \text { current }=0$$</p> | mcq | jee-main-2024-online-31st-january-evening-shift | 10,164 |
jaoe38c1lse6bk56 | physics | dual-nature-of-radiation | photoelectric-effect | <p>When a metal surface is illuminated by light of wavelength $$\lambda$$, the stopping potential is $$8 \mathrm{~V}$$. When the same surface is illuminated by light of wavelength $$3 \lambda$$, stopping potential is $$2 \mathrm{~V}$$. The threshold wavelength for this surface is:</p> | [{"identifier": "A", "content": "3$$\\lambda$$"}, {"identifier": "B", "content": "9$$\\lambda$$"}, {"identifier": "C", "content": "5$$\\lambda$$"}, {"identifier": "D", "content": "4.5$$\\lambda$$"}] | ["B"] | null | <p>$$\begin{aligned}
& \mathrm{E}=\phi+\mathrm{K}_{\max } \\
& \phi=\frac{\mathrm{hc}}{\lambda_0} \\
& \mathrm{~K}_{\max }=\mathrm{eV}_0 \\
& 8 \mathrm{e}=\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldots . . \text { (i) } \\
& 2 \mathrm{e}=\frac{\mathrm{hc}}{3 \lambda}-\frac{\mathrm{hc}}{\lambda_0} \ldo... | mcq | jee-main-2024-online-31st-january-morning-shift | 10,165 |
1lsgcxy3n | physics | dual-nature-of-radiation | photoelectric-effect | <p>The work function of a substance is $$3.0 \mathrm{~eV}$$. The longest wavelength of light that can cause the emission of photoelectrons from this substance is approximately;</p> | [{"identifier": "A", "content": "215 nm"}, {"identifier": "B", "content": "400 nm"}, {"identifier": "C", "content": "414 nm"}, {"identifier": "D", "content": "200 nm"}] | ["C"] | null | <p>$$\begin{aligned}
& \text { For P.E.E. : } \lambda \leq \frac{h c}{W_e} \\
& \lambda \leq \frac{1240 \mathrm{~nm}-\mathrm{eV}}{3 \mathrm{eV}} \\
& \lambda \leq 413.33 \mathrm{~nm} \\
& \lambda_{\max } \approx 414 \mathrm{~nm} \text { for P.E.E. }
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift | 10,167 |
lv0vyu60 | physics | dual-nature-of-radiation | photoelectric-effect | <p>Which figure shows the correct variation of applied potential difference (V) with photoelectric current (I) at two different intensities of light $$(\mathrm{I}_1<\mathrm{I}_2)$$ of same wavelengths :</p> | [{"identifier": "A", "content": "<img src=\"https://app-content.cdn.examgoal.net/fly/@width/image/1lv0c72ek/0feeb2ae-ef3c-4577-b53f-97b479d39565/04c672c0-fad0-11ee-891c-2b72d88c1441/file-1lv0c72el.png?format=png\" data-orsrc=\"https://app-content.cdn.examgoal.net/image/1lv0c72ek/0feeb2ae-ef3c-4577-b53f-97b479d39565/04c... | ["C"] | null | <p>Stopping potential is independent on intensity but
photocurrent increases non-linearly on increasing
intensity.</p> | mcq | jee-main-2024-online-4th-april-morning-shift | 10,169 |
lv2erh5l | physics | dual-nature-of-radiation | photoelectric-effect | <p>Given below are two statements: one is labelled as Assertion $$\mathbf{A}$$ and the other is labelled as Reason R.</p>
<p>Assertion A: Number of photons increases with increase in frequency of light.</p>
<p>Reason R: Maximum kinetic energy of emitted electrons increases with the frequency of incident radiation.</p>
... | [{"identifier": "A", "content": "$$\\mathbf{A}$$ is not correct but $$\\mathbf{R}$$ is correct.\n"}, {"identifier": "B", "content": "$$\\mathbf{A}$$ is correct but $$\\mathbf{R}$$ is not correct.\n"}, {"identifier": "C", "content": "Both $$\\mathbf{A}$$ and $$\\mathbf{R}$$ are correct and $$\\mathbf{R}$$ is the correct... | ["A"] | null | <p>In order to determine the most appropriate answer to the question, let's analyze the given Assertion A and Reason R in detail.</p>
<p><strong>Assertion A:</strong> Number of photons increases with increase in frequency of light.</p>
<p>This statement is not correct. The number of photons is determined by the inten... | mcq | jee-main-2024-online-4th-april-evening-shift | 10,170 |
lv9s20zs | physics | dual-nature-of-radiation | photoelectric-effect | <p>Which of the following statement is not true about stopping potential $$(\mathrm{V}_0)$$ ?</p> | [{"identifier": "A", "content": "It depends upon frequency of the incident light.\n"}, {"identifier": "B", "content": "It is $$1 / \\mathrm{e}$$ times the maximum kinetic energy of electrons emitted.\n"}, {"identifier": "C", "content": "It increases with increase in intensity of the incident light.\n"}, {"identifier": ... | ["C"] | null | <p>Stopping potential is independent of intensity of
light. It depends on frequency of light.</p> | mcq | jee-main-2024-online-5th-april-evening-shift | 10,172 |
lvb295g9 | physics | dual-nature-of-radiation | photoelectric-effect | <p>When UV light of wavelength $$300 \mathrm{~nm}$$ is incident on the metal surface having work function $$2.13 \mathrm{~eV}$$, electron emission takes place. The stopping potential is :</p>
<p>(Given hc $$=1240 \mathrm{~eV} \mathrm{~nm}$$ )</p> | [{"identifier": "A", "content": "4 V"}, {"identifier": "B", "content": "2 V"}, {"identifier": "C", "content": "4.1 V"}, {"identifier": "D", "content": "1.5 V"}] | ["B"] | null | <p>To find the stopping potential ($V_s$) when UV light of wavelength $300 $ nm is incident on a metal surface with a work function of $2.13$ eV, we can use the photoelectric equation which relates the energy of the incident photons, the work function of the metal, and the kinetic energy of the emitted electrons.</p>
... | mcq | jee-main-2024-online-6th-april-evening-shift | 10,173 |
BtUSKBr9A2PaZ8iQ | physics | electromagnetic-induction | inductance-(self-and-mutual) | The inductance between $$A$$ and $$D$$ is
<img src="data:image/png;base64,UklGRgAMAABXRUJQVlA4IPQLAADQkQCdASoAA4wBP4HA1mY2MCwnIVPpEsAwCWlu4XNBG/P59I9nuQTZ09n8ruFnynlnE8b2vYYcQ6k6u6IdSdXdEOpOruiHUnV3RDqTkXFjJtaVI+C4sZNrSpHwXFjJtaVI+C4sZNsvsZNrSpHwXFjJtaVI+C4sZNrSpHwXFjJtaW2JtaVI+C4sZNrSpHwXFjJtaVI+C4sZNrSpJvrSpHwXFjJta... | [{"identifier": "A", "content": "$$3.66$$ $$H$$ "}, {"identifier": "B", "content": "$$9$$ $$H$$ "}, {"identifier": "C", "content": "$$0.66$$ $$H$$ "}, {"identifier": "D", "content": "$$1$$ $$H$$ "}] | ["D"] | null | These three inductors are connected in parallel. The equivalent inductance $${L_P}$$ is given by
<br><br>$${1 \over {{L_P}}} = {1 \over {{L_1}}} + {1 \over {{L_2}}} + {1 \over {{L_3}}}$$ $$ = {1 \over 3} + {1 \over 3} + {1 \over 3} = {3 \over 3} = 1$$ | mcq | aieee-2002 | 10,175 |
vfWyqHhFmRrnwIx4 | physics | electromagnetic-induction | inductance-(self-and-mutual) | When the current changes from $$ + 2A$$ to $$-2A$$ in $$0.05$$ second, an $$e.m.f.$$ of $$8$$ $$V$$ is inducted in a coil. The coefficient of self- induction of the coil is | [{"identifier": "A", "content": "$$0.2H$$ "}, {"identifier": "B", "content": "$$0.4H$$ "}, {"identifier": "C", "content": "$$0.8$$ $$H$$ "}, {"identifier": "D", "content": "$$0.1$$ $$H$$ "}] | ["D"] | null | $$e = - {{\Delta \phi } \over {\Delta t}} = {{ - \Delta \left( {LI} \right)} \over {\Delta t}} = - L{{\Delta I} \over {\Delta t}}$$
<br><br>$$\therefore$$ $$\left| e \right| = L{{\Delta I} \over {\Delta t}} \Rightarrow 8 = L \times {4 \over {0.05}}$$
<br><br>$$ \Rightarrow L = {{8 \times 0.05} \over 4} = 0.1H$$ | mcq | aieee-2003 | 10,176 |
uKfHsKpQPKa8APcC | physics | electromagnetic-induction | inductance-(self-and-mutual) | Two coils are placed close to each other. The mutual inductance of the pair of coils depends upon | [{"identifier": "A", "content": "the rates at which currents are changing in the two coils "}, {"identifier": "B", "content": "relative position and orientation of the two coils "}, {"identifier": "C", "content": "the currents in the two coils "}, {"identifier": "D", "content": "the materials of the wires of the coils ... | ["B"] | null | Mutual conductance depends on the relative position and orientation of the two coils. | mcq | aieee-2003 | 10,177 |
FxlPB9jcTe9N6KcM | physics | electromagnetic-induction | inductance-(self-and-mutual) | A coil of inductance $$300$$ $$mH$$ and resistance $$2\,\Omega $$ is connected to a source of voltage $$2$$ $$V$$. The current reaches half of its steady state value in | [{"identifier": "A", "content": "$$0.1$$ $$s$$ "}, {"identifier": "B", "content": "$$0.05$$ $$s$$ "}, {"identifier": "C", "content": "$$0.3$$ $$s$$ "}, {"identifier": "D", "content": "$$0.15$$ $$s$$ "}] | ["A"] | null | <b>KEY CONCEPT :</b> The charging of inductance given
<br><br>by, $$i = {i_0}\left( {1 - {e^{ - {{Rt} \over L}}}} \right)$$
<br><br>$${{{i_0}} \over 2} = {i_0}\left( {1 - {e^{ - {{Rt} \over L}}}} \right) \Rightarrow {e^{ - {{Rt} \over L}}} = {1 \over 2}$$
<br><br>Taking log on both the sides,
<br><br>$$ - {{Rt} \over L... | mcq | aieee-2005 | 10,178 |
ksDZGpPdqa85ZXsA | physics | electromagnetic-induction | inductance-(self-and-mutual) | An inductor $$(L=100$$ $$mH)$$, a resistor $$\left( {R = 100\,\Omega } \right)$$ and a battery $$\left( {E = 100V} \right)$$ are initially connected in series as shown in the figure. After a long time the battery is disconnected after short circuiting the points $$A$$ and $$B$$. The current in the circuit $$1$$ $$ms$$ ... | [{"identifier": "A", "content": "$$1/eA$$ "}, {"identifier": "B", "content": "$$eA$$ "}, {"identifier": "C", "content": "$$0.1$$ $$A$$ "}, {"identifier": "D", "content": "$$1$$ $$A$$ "}] | ["A"] | null | Initially, when steady state is achieved,
<br><br>$$i = {E \over R}$$
<br><br>Let $$E$$ is short circuited at a $$t=0.$$ Then
<br><br>At $$t=0,$$ $${i_0} = {E \over R}$$
<br><br>Let during decay of current at any time the current
<br><br>flowing is $$ - L{{di} \over {dt}} - iR = 0$$
<br><br>$$ \Rightarrow {{di} \over... | mcq | aieee-2006 | 10,179 |
yhQHBFLrc6kZBeGI | physics | electromagnetic-induction | inductance-(self-and-mutual) | Which of the following units denotes the dimension $${{M{L^2}} \over {{Q^2}}}$$, where $$Q$$ denotes the electric charge? | [{"identifier": "A", "content": "$$Wb/{m^2}$$ "}, {"identifier": "B", "content": "Henry $$(H)$$ "}, {"identifier": "C", "content": "$$H/{m^2}$$"}, {"identifier": "D", "content": "Weber $$(Wb)$$"}] | ["B"] | null | Mutual inductance $$ = {\phi \over I} = {{BA} \over I}$$
<br><br>[Henry] $$ = {{\left[ {M{T^{ - 1}}{Q^{ - 1}}{L^2}} \right]} \over {\left[ {Q{T^{ - 1}}} \right]}} = M{L^2}{Q^{ - 2}}$$ | mcq | aieee-2006 | 10,180 |
n9Ju0KjDnGigKZZU | physics | electromagnetic-induction | inductance-(self-and-mutual) | An ideal coil of $$10H$$ is connected in series with a resistance of $$5\Omega $$ and a battery of $$5V$$. $$2$$ second after the connection is made, the current flowing in ampere in the circuit is | [{"identifier": "A", "content": "$$\\left( {1 - {e^{ - 1}}} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( {1 - e} \\right)$$ "}, {"identifier": "C", "content": "$$e$$ "}, {"identifier": "D", "content": "$${{e^{ - 1}}}$$ "}] | ["A"] | null | <b>KEY CONCEPT :</b> $$I = {I_0}\left( {1 - {e^{ - {R \over L}t}}} \right)$$
<br><br>(When current is in growth in $$LR$$ circuit)
<br><br>$$ = {E \over R}\left( {1 - {e^{ - {R \over L}t}}} \right)$$
<br><br>$$ = {5 \over 5}\left( {1 - {e^{ - {5 \over {10}} \times 2}}} \right)$$
<br><br>$$ = \left( {1 - {e^{ - 1}}} \r... | mcq | aieee-2007 | 10,181 |
irnszdUGj0LIsKov | physics | electromagnetic-induction | inductance-(self-and-mutual) | Two coaxial solenoids of different radius carry current $$I$$ in the same direction. $$\overrightarrow {{F_1}} $$ be the magnetic force on the inner solenoid due to the outer one and $$\overrightarrow {{F_2}} $$ be the magnetic force on the outer solenoid due to the inner one. Then : | [{"identifier": "A", "content": "$$\\overrightarrow {{F_1}} $$ is radially in wards and $$\\overrightarrow {{F_2}} = 0$$ "}, {"identifier": "B", "content": "$$\\overrightarrow {{F_1}} $$ is radially outwards and $$\\overrightarrow {{F_2}} = 0$$ "}, {"identifier": "C", "content": "$$\\overrightarrow {{F_1}} = \\ove... | ["C"] | null | $$\mathop {F{}_1}\limits^ \to = \mathop {F{}_2}\limits^ \to = 0$$
<br><br> because of action and reaction pair | mcq | jee-main-2015-offline | 10,183 |
3vZjqnMh8d7OkCEtKaCgm | physics | electromagnetic-induction | inductance-(self-and-mutual) | A small circular loop of wire of radius a is located at the centre of a much larger circular wire loop of radius b. The two loops are in the same plane. The outer loop of radius b carries an alternating current I = I<sub>o</sub> cos ($$\omega $$t). The emf induced in the smaller inner loop is nearly : | [{"identifier": "A", "content": "$${{\\pi {\\mu _o}{I_o}} \\over 2}.{{{a^2}} \\over b}\\,\\omega \\sin \\left( {\\omega t} \\right)$$"}, {"identifier": "B", "content": "$${{\\pi {\\mu _o}{I_o}} \\over 2}.{{{a^2}} \\over b}\\,\\omega \\cos \\left( {\\omega t} \\right)$$ "}, {"identifier": "C", "content": "$$\\pi {\\mu _... | ["A"] | null | Mutual inductance,
<br><br>M = $${{{\mu _0}\pi {N_1}{N_2}\,{a^2}} \over {2b}}$$
<br><br>here $${{N_1}}$$ = N<sub>2</sub> = 1
<br><br>$$\therefore\,\,\,$$ M = $${{{\mu _0}\pi {a^2}} \over {2b}}$$
<br><br>Current I = I<sub>0</sub> cos ($$\omega $$t)
<br><br>According to Faraday's law,
<br><br>e = $$-$$ M $${{dI} \over ... | mcq | jee-main-2017-online-8th-april-morning-slot | 10,184 |
cBSvlyrMPPjhU4YyaaDFY | physics | electromagnetic-induction | inductance-(self-and-mutual) | There are two long co-axial solenoids of same length $$l$$. The inner and outer coils have radii r<sub>1</sub> and r<sub>2</sub> and number of turns per unit length n<sub>1</sub> and n<sub>2</sub>, respectively. The ratio of mutual inductance to the self - inductance of the inner-coil is : | [{"identifier": "A", "content": "$${{{n_2}} \\over {{n_1}}}.{{{r_2}^2} \\over {{r_1}^2}}$$"}, {"identifier": "B", "content": "$${{{n_2}} \\over {{n_1}}}$$"}, {"identifier": "C", "content": "$${{{n_1}} \\over {{n_2}}}$$"}, {"identifier": "D", "content": "$${{{n_2}} \\over {{n_1}}}.{{{r_1}} \\over {{r_2}}}$$"}] | ["B"] | null | $$M = {\mu _0}\,{n_1}\,{n_2}\,\pi r_1^2$$
<br><br>$$L = {\mu _0}\,n_1^2\,\pi r_1^2$$
<br><br>$$ \Rightarrow \,\,{M \over L} = {{{n_2}} \over {{n_1}}}$$ | mcq | jee-main-2019-online-11th-january-morning-slot | 10,185 |
IcSQGRlXSXy4MRyzYODvD | physics | electromagnetic-induction | inductance-(self-and-mutual) | Two coils 'P' and 'Q' are separated by some
distance. When a current of 3 A flows through
coil 'P', a magnetic flux of 10<sup>–3</sup> Wb passes
through 'Q'. No current is passed through 'Q'.
When no current passes through 'P' and a
current of 2 A passes through 'Q', the flux
through 'P' is :- | [{"identifier": "A", "content": "3.67 \u00d7 10<sup>\u20134</sup> Wb"}, {"identifier": "B", "content": "3.67 \u00d7 10<sup>\u20133</sup> Wb"}, {"identifier": "C", "content": "6.67 \u00d7 10<sup>\u20134</sup> Wb"}, {"identifier": "D", "content": "6.67 \u00d7 10<sup>\u20133</sup> Wb"}] | ["C"] | null | Mutual induction<br><br/>
$${\phi _q} = M{I_p}$$<br><br/>
10<sup>–3</sup> = M(3)<br><br/>
$$ \Rightarrow M = {1 \over 3} \times {10^{ - 3}}$$<br><br>
$${\phi _p} = M{I_q}$$<br><br/>
$$ \Rightarrow {\phi _p} = 6.67{\rm{ }} \times {\rm{ }}{10^{-4}}{\rm{ }}Wb$$ | mcq | jee-main-2019-online-9th-april-evening-slot | 10,186 |
G6knEI7jVv54jc1vhHZlm | physics | electromagnetic-induction | inductance-(self-and-mutual) | The total number of turns and cross-section area
in a solenoid is fixed. However, its length L is varied
by adjusting the separation between windings. The
inductance of solenoid will be proportional to : | [{"identifier": "A", "content": "1/L<sup>2</sup>"}, {"identifier": "B", "content": "1/L"}, {"identifier": "C", "content": "L"}, {"identifier": "D", "content": "L<sup>2</sup>"}] | ["B"] | null | $$\phi $$ = NBA = LI<br><br>
N $$\mu $$<sub>0</sub> nI$$\pi $$R<sup>2</sup> = LI<br><br>
$$N{\mu _0}{N \over l}l\pi {R^2} = LI$$<br><br>
N and R constant<br><br>
Self inductance (L) $$ \propto {1 \over l} \propto {1 \over {length}}$$ | mcq | jee-main-2019-online-9th-april-morning-slot | 10,187 |
GX3irRDIdEDhrTOQ00kKh | physics | electromagnetic-induction | inductance-(self-and-mutual) | A 20 Henry inductor coil is connected to a
10 ohm resistance in series as shown in figure.
The time at which rate of dissipation of energy
(joule's heat) across resistance is equal to the
rate at which magnetic energy is stored in the
inductor is :
<img src="data:image/png;base64,UklGRpQHAABXRUJQVlA4IIgHAACwYwCdASoAAyE... | [{"identifier": "A", "content": "$${2 \\over {\\ln 2}}$$"}, {"identifier": "B", "content": "$${\\ln 2}$$"}, {"identifier": "C", "content": "$$2{\\ln 2}$$"}, {"identifier": "D", "content": "$${1 \\over 2}{\\ln 2}$$"}] | ["C"] | null | P<sub>R</sub> = i<sup>2</sup> × R<br><br>
P<sub>B</sub> = V × i<br><br>
$$ \therefore $$ P<sub>L</sub> = Vi – i<sup>2</sup>R<br><br>
$$ \Rightarrow $$ Vi – i<sup>2</sup>R = i<sup>2</sup>R<br><br>
$$ \Rightarrow $$ $$i = {V \over {2R}}\,\,and\,\,i = {V \over R}\left( {1 - {e^{ - t/\tau }}} \right)$$<br><br>
$$ \therefor... | mcq | jee-main-2019-online-8th-april-morning-slot | 10,188 |
qdJckTjSX35vhgIGYf7k9k2k5hh31x9 | physics | electromagnetic-induction | inductance-(self-and-mutual) | A shown in the figure, a battery of emf $$\varepsilon $$ is
connected to an inductor L and resistance R in
series. The switch is closed at t = 0. The total
charge that flows from the battery, between
t = 0 and t = t<sub>c</sub> (t<sub>c</sub> is the time constant of the
circuit) is :
<img src="data:image/png;base64,Ukl... | [{"identifier": "A", "content": "$${{\\varepsilon L} \\over {e{R^2}}}$$"}, {"identifier": "B", "content": "$${{\\varepsilon L} \\over {{R^2}}}$$"}, {"identifier": "C", "content": "$${{\\varepsilon L} \\over {{R^2}}}\\left( {1 - {1 \\over e}} \\right)$$"}, {"identifier": "D", "content": "$${{\\varepsilon R} \\over {e{L^... | ["A"] | null | $$i = {i_0}\left( {1 - {e^{ - t/{t_c}}}} \right)$$
<br><br>q = $$\int\limits_0^{{t_c}} {idt} $$
<br><br>= $$\int\limits_0^{{t_c}} {{ \in \over R}\left( {1 - {e^{ - t/{t_c}}}} \right)dt} $$
<br><br>= $${ \in \over R}\left[ {t - {{{e^{ - t/{t_c}}}} \over {\left( { - {1 \over {{t_c}}}} \right)}}} \right]_0^{{t_c}}$$
<br... | mcq | jee-main-2020-online-8th-january-evening-slot | 10,191 |
ZOxrrAmB8tJWYzEw3Z7k9k2k5intpa5 | physics | electromagnetic-induction | inductance-(self-and-mutual) | In a fluorescent lamp choke (a small
transformer) 100 V of reverse voltage is
produced when the choke current changes
uniformly from 0.25 A to 0 in a duration of
0.025 ms. The self-inductance of the choke
(in mH) is estimated to be ________. | [] | null | 10 | V = $$\left| {L{{di} \over {dt}}} \right|$$
<br><br>$$ \Rightarrow $$ L = $${V \over {\left| {{{di} \over {dt}}} \right|}}$$ = $${{100} \over {{{0.25} \over {0.025 \times {{10}^{ - 3}}}}}}$$ = 10 mH | integer | jee-main-2020-online-9th-january-morning-slot | 10,192 |
KF0GnfKYFRM3Y3VAPq1klrngqo0 | physics | electromagnetic-induction | inductance-(self-and-mutual) | Figure shows a circuit that contains four identical resistors with resistance R = 2.0$$\Omega$$, two identical inductors with inductance L = 2.0 mH and an ideal battery with emf E = 9V. The current 'i' just after the switch 'S' is closed will be :<br/><br/><img src="data:image/png;base64,UklGRvIMAABXRUJQVlA4IOYMAACwtgC... | [{"identifier": "A", "content": "3.0 A"}, {"identifier": "B", "content": "3.37 A"}, {"identifier": "C", "content": "9 A"}, {"identifier": "D", "content": "2.25 A"}] | ["D"] | null | Given, resistance, R = 2$$\Omega$$,<br><br>Inductance, L = 2 mH,<br><br>emf, E = 9 V<br><br>and i be the current.<br><br>$$\because$$ At t = 0 when switch is closed, inductors behave as open circuit.<br><br>$$\therefore$$ Effective circuit will be <br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/... | mcq | jee-main-2021-online-24th-february-evening-slot | 10,193 |
3LTmhImqV9RuPxxjym1kmlw2i3l | physics | electromagnetic-induction | inductance-(self-and-mutual) | The time taken for the magnetic energy to reach 25% of its maximum value, when a solenoid of resistance R, inductance L is connected to a battery, is : | [{"identifier": "A", "content": "infinite"}, {"identifier": "B", "content": "$${L \\over R}$$ ln10"}, {"identifier": "C", "content": "$${L \\over R}$$ ln2"}, {"identifier": "D", "content": "$${L \\over R}$$ ln5"}] | ["C"] | null | Magnetic energy, U = $${1 \over 2}$$LI$$_0^2$$<br><br>Given : U = 25% of U<sub>0</sub>.<br><br>$$ \Rightarrow $$ $${1 \over 2}L{I^2} = {1 \over 4} \times {1 \over 2}LI_0^2$$<br><br>$$ \Rightarrow {I^2} = {{I_0^2} \over 4} \Rightarrow I = {{{I_0}} \over 2}$$<br><br>$$ \therefore $$ $$I = {I_0}(1 - {e^{ - t/\tau }})$$<br... | mcq | jee-main-2021-online-18th-march-evening-shift | 10,195 |
1kth3zfh0 | physics | electromagnetic-induction | inductance-(self-and-mutual) | A small square loop of side 'a' and one turn is placed inside a larger square loop of side b and one turn (b >> a). The two loops are coplanar with their centres coinciding. If a current I is passed in the square loop of side 'b', then the coefficient of mutual inductance between the two loops is : | [{"identifier": "A", "content": "$${{{\\mu _0}} \\over {4\\pi }}8\\sqrt 2 {{{a^2}} \\over b}$$"}, {"identifier": "B", "content": "$${{{\\mu _0}} \\over {4\\pi }}{{8\\sqrt 2 } \\over a}$$"}, {"identifier": "C", "content": "$${{{\\mu _0}} \\over {4\\pi }}8\\sqrt 2 {{{b^2}} \\over a}$$"}, {"identifier": "D", "content": "$... | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264132/exam_images/h9aj4cl5tbtovut5jkru.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 31st August Morning Shift Physics - Electromagnetic Induction Question 58 English Explanation"><br... | mcq | jee-main-2021-online-31st-august-morning-shift | 10,197 |
1l58hwz9j | physics | electromagnetic-induction | inductance-(self-and-mutual) | <p>Two coils of self inductance L<sub>1</sub> and L<sub>2</sub> are connected in series combination having mutual inductance of the coils as M. The equivalent self inductance of the combination will be :</p>
<p> <img src="data:image/png;base64,UklGRlYGAABXRUJQVlA4IEoGAABwYgCdASoAA+sAP4HA22U2MK2nIjK5UsAwCWlu4WmhG/Pt9H2o... | [{"identifier": "A", "content": "$${1 \\over {{L_1}}} + {1 \\over {{L_2}}} + {1 \\over M}$$"}, {"identifier": "B", "content": "$${L_1} + {L_2} + M$$"}, {"identifier": "C", "content": "$${L_1} + {L_2} + 2M$$"}, {"identifier": "D", "content": "$${L_1} + {L_2} - 2M$$"}] | ["D"] | null | <p>Self inductances are in series but their mutual inductances are linked oppositely so equivalent self inductance</p>
<p>L = L<sub>1</sub> + L<sub>2</sub> $$-$$ M $$-$$ M = L<sub>1</sub> + L<sub>2</sub> $$-$$ 2M</p> | mcq | jee-main-2022-online-26th-june-evening-shift | 10,199 |
1l5algfye | physics | electromagnetic-induction | inductance-(self-and-mutual) | <p>The current in a coil of self inductance 2.0 H is increasing according to I = 2 sin(t<sup>2</sup>) A. The amount of energy spent during the period when current changes from 0 to 2 A is ____________ J.</p> | [] | null | 4 | <p>$$U = {1 \over 2}L{I^2}$$</p>
<p>$$ = {1 \over 2}2 \times {2^2} = 4$$ J</p> | integer | jee-main-2022-online-25th-june-morning-shift | 10,200 |
1l6dytqj5 | physics | electromagnetic-induction | inductance-(self-and-mutual) | <p>A small square loop of wire of side $$l$$ is placed inside a large square loop of wire $$\mathrm{L}(\mathrm{L}>>l)$$. Both loops are coplanar and their centres coincide at point $$\mathrm{O}$$ as shown in figure. The mutual inductance of the system is :</p>
<p><img src="data:image/png;base64,UklGRqoKAABXRUJQVl... | [{"identifier": "A", "content": "$$\\frac{2 \\sqrt{2} \\mu_{0} \\mathrm{~L}^{2}}{\\pi l}$$"}, {"identifier": "B", "content": "$$\\frac{\\mu_{0} l^{2}}{2 \\sqrt{2} \\pi \\mathrm{L}}$$"}, {"identifier": "C", "content": "$$\\frac{2 \\sqrt{2} \\mu_{0} l^{2}}{\\pi \\mathrm{L}}$$"}, {"identifier": "D", "content": "$$\n\\frac... | ["C"] | null | <p>We know $$\phi = Mi$$</p>
<p>Let i current be flowing in the larger loop</p>
<p>$$ \Rightarrow \phi \simeq \left[ {4 \times {{{\mu _0}i} \over {4\pi (L/2)}}[\sin 45^\circ + \sin 45^\circ ]} \right] \times $$ Area</p>
<p>$$ = {{2\sqrt 2 {\mu _0}i} \over {\pi L}} \times {I^2}$$</p>
<p>$$ \Rightarrow M = {\phi \ove... | mcq | jee-main-2022-online-25th-july-morning-shift | 10,201 |
1l6ntzhym | physics | electromagnetic-induction | inductance-(self-and-mutual) | <p>For the given circuit the current through battery of 6 V just after closing the switch 'S' will be _________ A.</p>
<p><img src="data:image/png;base64,UklGRlwMAABXRUJQVlA4IFAMAAAQrQCdASoAA58BP4HA2mU2MS0nIbLpcsAwCWlu4W8FJmNwvH5+/3trTZ49vMym356GdRT3Lf//p0+/RCLk1pioyfDapyeyTupcJKLs+oIfUEPqCH1BD6gh9QFnuGByJUA3juhbLtnw/m... | [] | null | 1 | <p>Just after closing the switch, $$i = {6 \over {4 + 2}} = 1\,A$$</p> | integer | jee-main-2022-online-28th-july-evening-shift | 10,202 |
1ldr34jcr | physics | electromagnetic-induction | inductance-(self-and-mutual) | <p><img src="data:image/png;base64,UklGRiIKAABXRUJQVlA4IBYKAACQoACdASoAA6YBP4HA2mW2MLqnIZQZM1AwCWlu4XNBG/Pz9C/qnbW2hPbzuyb2AOrt/9sU/y7kOLO/Ptcu5Dizvz7XLuQ4s78+1y2Ob60ZvrRm+tGb60Y9OVURuncxX2fVm2Q4/WkzRm+tGb60ZvrVT4HZpvq4r/jPwP/F/x767TetZqCntdZrbIXqUKSkyiNY3B/Oe9HnqPGaD82eyaitikEdmeohAuqm6t4vTn2IYth3BcG6r8RYTFwhLFphPfhF2... | [] | null | 30 | <p>From the circuit :</p>
<p>$${V_A} - iR - {{Ldi} \over {dt}} - 12 = {V_B}$$</p>
<p>$$ \Rightarrow {V_A} - {V_B} = 2 \times 12 + 6( - 1) + 12$$ volts</p>
<p>$$ = 30$$ volts</p> | integer | jee-main-2023-online-30th-january-morning-shift | 10,203 |
1ldso1b5o | physics | electromagnetic-induction | inductance-(self-and-mutual) | <p>Find the mutual inductance in the arrangement, when a small circular loop of wire of radius '$$R$$' is placed inside a large square loop of wire of side $$L$$ $$(L \gg R)$$. The loops are coplanar and their centres coincide :</p>
<p><img src="data:image/png;base64,UklGRnQLAABXRUJQVlA4IGgLAAAw6ACdASrvAgADP4HA3WW2MT+n... | [{"identifier": "A", "content": "$$M=\\frac{\\sqrt{2} \\mu_{0} R}{L^{2}}$$"}, {"identifier": "B", "content": "$$M=\\frac{2 \\sqrt{2} \\mu_{0} R}{L^{2}}$$"}, {"identifier": "C", "content": "$$M=\\frac{2 \\sqrt{2} \\mu_{0} R^{2}}{L}$$"}, {"identifier": "D", "content": "$$M=\\frac{\\sqrt{2} \\mu_{0} R^{2}}{L}$$"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lei3q9e1/b707d484-0cd7-4eb2-a354-be8e4d9ce9b0/56a16f90-b405-11ed-8518-6f6f1908382a/file-1lei3q9e2.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lei3q9e1/b707d484-0cd7-4eb2-a354-be8e4d9ce9b0/56a16f90-b405-11ed-8518-6f6f1908382a/fi... | mcq | jee-main-2023-online-29th-january-morning-shift | 10,204 |
1ldwsbfqk | physics | electromagnetic-induction | inductance-(self-and-mutual) | <p>Three identical resistors with resistance R = 12$$\Omega$$ and two identical inductors with self inductance L = 5 mH are connected to an ideal battery with emf of 12 V as shown in figure. The current through the battery long after the switch has been closed will be _____________ A.</p>
<p><img src="data:image/png;ba... | [] | null | 3 | <p>After long time, inductors are shorted.</p>
<p>Effective circuit becomes</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le4deh44/b1217ea3-f19b-4a41-8dbf-ff553ce2b6d6/31164730-ac78-11ed-b725-7db87d3646da/file-1le4deh45.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le... | integer | jee-main-2023-online-24th-january-evening-shift | 10,205 |
1lh30yif9 | physics | electromagnetic-induction | inductance-(self-and-mutual) | <p>Two concentric circular coils with radii $$1 \mathrm{~cm}$$ and $$1000 \mathrm{~cm}$$, and number of turns 10 and 200 respectively are placed coaxially with centers coinciding. The mutual inductance of this arrangement will be ___________ $$\times 10^{-8} \mathrm{H}$$. (Take, $$\pi^{2}=10$$ )</p> | [] | null | 4 | <p>The magnetic field $$B_2$$ due to the current $$I_2$$ in the larger coil with 200 turns is given by:</p>
<p>$$B_2 = \frac{N_2 \mu_0 I_2}{2r_2} = \frac{200 \mu_0 I_2}{2 \times 10}$$</p>
<p>The magnetic flux $$\phi_{1,2}$$ through the smaller coil due to this magnetic field is given by:</p>
<p>$$\phi_{1,2} = N_1 \vec{... | integer | jee-main-2023-online-6th-april-evening-shift | 10,207 |
lsan61ku | physics | electromagnetic-induction | inductance-(self-and-mutual) | A transformer has an efficiency of $80 \%$ and works at $10 \mathrm{~V}$ and $4 \mathrm{~kW}$. If the secondary voltage is $240 \mathrm{~V}$, then the current in the secondary coil is : | [{"identifier": "A", "content": "$1.33 \\mathrm{~A}$"}, {"identifier": "B", "content": "$13.33 \\mathrm{~A}$"}, {"identifier": "C", "content": "$1.59 \\mathrm{~A}$"}, {"identifier": "D", "content": "$15.1 \\mathrm{~A}$"}] | ["B"] | null | <p>To find the current in the secondary coil of the transformer, we first need to calculate the output power, taking into account the efficiency. The efficiency ($$ \eta $$) of the transformer is given by the ratio of the output power ($$ P_{\text{out}} $$) to the input power ($$ P_{\text{in}} $$) times 100%.</p>
<p>Th... | mcq | jee-main-2024-online-1st-february-evening-shift | 10,208 |
jaoe38c1lsc48f9n | physics | electromagnetic-induction | inductance-(self-and-mutual) | <p>Two coils have mutual inductance $$0.002 \mathrm{~H}$$. The current changes in the first coil according to the relation $$\mathrm{i}=\mathrm{i}_0 \sin \omega \mathrm{t}$$, where $$\mathrm{i}_0=5 \mathrm{~A}$$ and $$\omega=50 \pi$$ rad/s. The maximum value of emf in the second coil is $$\frac{\pi}{\alpha} \mathrm{~V}... | [] | null | 2 | <p>$$\begin{aligned}
& \phi=\mathrm{Mi}=\mathrm{Mi}_0 \sin \omega \mathrm{t} \\
& \mathrm{EMF}=-\mathrm{M} \frac{\mathrm{di}}{\mathrm{dt}}=-0.002\left(\mathrm{i}_0 \omega \cos \omega \mathrm{t}\right) \\
& \mathrm{EMF}_{\max }=\mathrm{i}_0 \omega(0.002)=(5)(50 \pi)(0.002) \\
& \mathrm{EMF}_{\max }=\frac{\pi}{2} \mathrm... | integer | jee-main-2024-online-27th-january-morning-shift | 10,209 |
jaoe38c1lse6wdsx | physics | electromagnetic-induction | inductance-(self-and-mutual) | <p>A small square loop of wire of side $$l$$ is placed inside a large square loop of wire of side $$L\left(L=l^2\right)$$. The loops are coplanar and their centers coincide. The value of the mutual inductance of the system is $$\sqrt{x} \times 10^{-7} \mathrm{H}$$, where $$x=$$ _________.</p> | [] | null | 128 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lslumatj/08a420dc-d051-43e9-90c2-11fdf1acba69/e03d5c80-cb3f-11ee-ad47-a16d1086e690/file-6y3zli1lslumatk.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lslumatj/08a420dc-d051-43e9-90c2-11fdf1acba69/e03d5c80-cb3f-11ee... | integer | jee-main-2024-online-31st-january-morning-shift | 10,210 |
lv7v4ron | physics | electromagnetic-induction | inductance-(self-and-mutual) | <p>Two conducting circular loops A and B are placed in the same plane with their centres coinciding as shown in figure. The mutual inductance between them is:</p>
<p><img src="data:image/png;base64,UklGRggXAABXRUJQVlA4IPwWAADQYAGdASoAA/cCP4G+2GU2L7+nIhG5a/AwCWlu/F+3nWinZ18/tx/zP7p/bvfRfTsXcJ/yv/5v+PpP8x/23+D7vn3VYkLgLz... | [{"identifier": "A", "content": "$$\\frac{\\mu_o}{2 \\pi} \\cdot \\frac{b^2}{a}$$\n"}, {"identifier": "B", "content": "$$\\frac{\\mu_{\\mathrm{o}} \\pi \\mathrm{a}^2}{2 \\mathrm{~b}}$$\n"}, {"identifier": "C", "content": "$$\\frac{\\mu_0 \\pi b^2}{2 a}$$\n"}, {"identifier": "D", "content": "$$\\frac{\\mu_0}{2 \\pi} \\c... | ["B"] | null | <p>$$\begin{aligned}
& \frac{\phi_{A / B}}{I_B}=M \\
& \Rightarrow \quad M=\frac{\mu_0 I_B \times \pi a^2}{2 b \times I_B} \\
& =\frac{\mu_0 \pi a^2}{2 b}
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-morning-shift | 10,211 |
1jpI7M7Egal8X9QL | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | In a uniform magnetic field of induction $$B$$ a wire in the form of a semicircle of radius $$r$$ rotates about the diameter of the circle with an angular frequency $$\omega .$$ The axis of rotation is perpendicular to the field. If the total resistance of the circuit is $$R,$$ the mean power generated per period of ro... | [{"identifier": "A", "content": "$${{{{\\left( {B\\pi r\\omega } \\right)}^2}} \\over {2R}}$$ "}, {"identifier": "B", "content": "$${{{{\\left( {B\\pi {r^2}\\omega } \\right)}^2}} \\over {8R}}$$ "}, {"identifier": "C", "content": "$${{B\\pi {r^2}\\omega } \\over {2R}}$$ "}, {"identifier": "D", "content": "$${{{{\\left(... | ["B"] | null | $$\phi = \overrightarrow B .\overrightarrow A ;\phi = BA\cos \,\omega t$$
<br><br>$$\varepsilon = - {{d\phi } \over {dt}} = \omega BA\sin \,\omega t;\,\,$$
<br><br>$$i = {{\omega BA} \over R}\sin \,\omega t$$
<br><br>$${P_{inst}} = {i^2}R = {\left( {{{\omega BA} \over R}} \right)^2} \times R{\sin ^2}\omega t$$
<br>... | mcq | aieee-2004 | 10,214 |
U0L4kw1ReNaC9PQF | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | A coil having $$n$$ turns and resistance $$R\Omega $$ is connected with a galvanometer of resistance $$4R\Omega .$$ This combination is moved in time $$t$$ seconds from a magnetic field $${W_1}$$ weber to $${W_2}$$ weber. The induced current in the circuit is | [{"identifier": "A", "content": "$${{\\left( {{W_2} - {W_1}} \\right)} \\over {Rnt}}$$ "}, {"identifier": "B", "content": "$$ - {{n\\left( {{W_2} - {W_1}} \\right)} \\over {5\\,\\,Rt}}$$ "}, {"identifier": "C", "content": "$$ - {{\\left( {{W_2} - {W_1}} \\right)} \\over {5\\,\\,Rnt}}$$ "}, {"identifier": "D", "content"... | ["B"] | null | $${{\Delta \phi } \over {\Delta t}} = {{\left( {{W_2} - {W_1}} \right)} \over t}$$
<br><br>$${R_{tot}} = \left( {R + 4R} \right)\Omega = 5R\Omega $$
<br><br>$$i = {{nd\phi } \over {{R_{tot}}dt}} = {{ - n\left( {{W_2} - {W_1}} \right)} \over {5Rt}}$$
<br><br>( as $${W_2}\,\,\& \,\,{W_1}$$ are magnetic flux ) | mcq | aieee-2004 | 10,215 |
ItdkA2UL44A1iA1b | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | The flux linked with a coil at any instant $$'t'$$ is given by
<br/>$$\phi = 10{t^2} - 50t + 250$$
<br/>The induced $$emf$$ at $$t=3s$$ is | [{"identifier": "A", "content": "$$-190$$ $$V$$ "}, {"identifier": "B", "content": "$$-10$$ $$V$$ "}, {"identifier": "C", "content": "$$10$$ $$V$$ "}, {"identifier": "D", "content": "$$190$$ $$V$$ "}] | ["B"] | null | $$\phi = 10{t^2} - 50t + 250$$
<br><br>$$e = - {{d\phi } \over {dt}} = - \left( {20t - 50} \right)$$
<br><br>$${e_{t = 3}} = - 10\,V$$ | mcq | aieee-2006 | 10,216 |
t535UqWrP4k8kuJP | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating; It is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to : | [{"identifier": "A", "content": "development of air current when the plate is placed "}, {"identifier": "B", "content": "induction of electrical charge on the plate "}, {"identifier": "C", "content": "shielding of magnetic lines of force as aluminium is a para-magnetic material."}, {"identifier": "D", "content": "elect... | ["D"] | null | Because of the Lenz's law of conservation of energy. | mcq | aieee-2012 | 10,217 |
Z5n0lsUsBLikpE7O6cb1T | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | A conducting metal circular-wire-loop of radius r is placed perpendicular to a
magnetic field which varies with time as
<br/>B = B<sub>0</sub>e$${^{{{ - t} \over r}}}$$ , where B<sub>0</sub> and $$\tau $$ are constants, at time t = 0. If the resistance of the loop is R then the heat generated in the loop after a long ... | [{"identifier": "A", "content": "$${{{\\pi ^2}{r^4}B_0^4} \\over {2\\tau R}}$$ "}, {"identifier": "B", "content": "$${{{\\pi ^2}{r^4}B_0^2} \\over {2\\tau R}}$$ "}, {"identifier": "C", "content": "$${{{\\pi ^2}{r^4}B_0^2R} \\over \\tau }$$ "}, {"identifier": "D", "content": "$${{{\\pi ^2}{r^4}B_0^2} \\over {\\tau R}}$$... | ["B"] | null | Given,
<br><br>B = B<sub>0</sub>e$$^{ - {t \over \tau }}$$
<br><br>Area of the circular loop, A = $$\pi $$ r<sup>2</sup>
<br><br>$$ \therefore $$ Flux $$\phi $$ = BA = $$\pi $$ r<sup>2</sup> B<sub>0</sub> e$$^{ - {t \over \tau }}$$
<br><br>Induced emf in the loop,
<br><br>$$\varepsilon $$ = $$-$$ $$... | mcq | jee-main-2016-online-10th-april-morning-slot | 10,219 |
Hxw4qb6KjT5MlmwSyyNmL | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | A uniform magnetic field B of 0.3 T is along the positive Z-direction. A rectangular loop (abcd) of sides 10 cm × 5 cm carries a current I of 12 A. Out of the following different orientations which one
corresponds to stable equilibrium ? | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265780/exam_images/c1zbqyemdzgyh3oixktw.webp\" style=\"max-width: 100%; height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2017 (Online) 9th April Morning Slot Physics - Electromagnetic Inducti... | ["C"] | null | . | mcq | jee-main-2017-online-9th-april-morning-slot | 10,220 |
BgWoF8xAN8g3YpLz | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | In a coil of resistance 100 $$\Omega $$, a current is induced by changing
the magnetic flux through it as shown in the figure. The
magnitude of change in flux through the coil is:<br/><br/>
<img src="data:image/png;base64,UklGRnwMAABXRUJQVlA4IHAMAADQyQCdASoAAxUCP4HA3GU2MS2nIjMo+sAwCWlu4XHhG/Pz9U+i7J/7hZrJvnzrKhfr/2XEH8... | [{"identifier": "A", "content": "275 Wb "}, {"identifier": "B", "content": "200 Wb"}, {"identifier": "C", "content": "225 Wb"}, {"identifier": "D", "content": "250 Wb"}] | ["D"] | null | According to Faraday's law of electromagnetic
induction,
<br><br>$$\varepsilon = {{d\phi } \over {dt}}$$
<br><br>Also, $$\varepsilon $$ = iR
<br><br>$$ \therefore $$ $${{d\phi } \over {dt}}$$ = iR
<br><br>$$ \Rightarrow $$ $$\int {d\phi } = R\int {idt} $$
<br><br>Magnitude of change in flux (d$$\phi $$) = R × area un... | mcq | jee-main-2017-offline | 10,221 |
ALgW7DH8UjJowW2rcQ2Rd | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | At the center of a fixed large circular coil of radius R, a much smaller circular coil of radius r is placed. The two coils are concentric and are in the same plane. The larger coil carries a current I. The smaller coil is set to rotate with a constant angular velocity $$\omega $$ about an axis along their common diame... | [{"identifier": "A", "content": "$${{{\\mu _o}{\\rm I}} \\over {2\\,R}}$$ $$\\omega $$ $$\\pi $$ r<sup>2</sup> sin$$\\omega $$ t"}, {"identifier": "B", "content": "$${{{\\mu _o}{\\rm I}} \\over {4\\,R}}$$ $$\\omega $$ $$\\pi $$ r<sup>2</sup> sin$$\\omega $$ t"}, {"identifier": "C", "content": "$${{{\\mu _o}{\\rm I}} ... | ["A"] | null | <p>We know that electric flux $$\phi = \overrightarrow B \,.\,\overrightarrow A $$</p>
<p>$$ \Rightarrow \phi = BA\cos \omega t$$</p>
<p>Now, $$B = {{{\mu _0}} \over 2}{I \over R}$$ is magnetic field due to circular coil of radius R and $$A = \pi {r^2}$$ is area of circular coil of radius r. Therefore,</p>
<p>$$\phi ... | mcq | jee-main-2018-online-15th-april-evening-slot | 10,222 |
k2z8MsUISoQwokP0YEwmv | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | A coil of cross-sectional area A having n turns is placed in a uniform magnetic field B. When it is rotated with an angular velocity $$\omega ,$$ the maxium e.m.f. induced in the coil will be: | [{"identifier": "A", "content": "3 nBA$$\\omega $$"}, {"identifier": "B", "content": "$${3 \\over 2}$$ nBA$$\\omega $$"}, {"identifier": "C", "content": "nBA$$\\omega $$"}, {"identifier": "D", "content": "$${1 \\over 2}$$ nBA$$\\omega $$"}] | ["C"] | null | Flux in the coil, $$\phi $$ = nBA sin($$\omega $$t)
<br><br>When n = no. of turns
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ A = Area of coil
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$$$\omega $$ = angular speed
<br><br>Induced emf,
<br><br>$$\left| e \right| = {{d\phi } \over {dt}}$$
<br><br>= nBA$$\omega $$ cos$$\omega... | mcq | jee-main-2018-online-16th-april-morning-slot | 10,223 |
Qndw2dUu3JFCJJmegFOuX | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | A conducting circular loop made of a thin wire, has area 3.5 $$ \times $$ 10<sup>$$-$$3</sup> m<sup>2</sup> and resistance 10 $$\Omega $$. It is placed perpendicular to a time dependent magnetic field B(t) = (0.4T)sin(50$$\pi $$t). The field is uniform in space. Then the net charge flowing through the loop during t = 0... | [{"identifier": "A", "content": "0.14 mC"}, {"identifier": "B", "content": "0.7 mC"}, {"identifier": "C", "content": "0.21 mC"}, {"identifier": "D", "content": "0.6 mC"}] | ["A"] | null | At t = 0 s
<br><br>B(0) = 0.4 sin (0) = 0
<br><br>and at t = 10 ms
<br><br>B(10) = 0.4 sin (50$$\pi $$$$ \times $$10$$ \times $$10<sup>-3</sup>)
<br><br>= 0.4 sin $$\left( {{\pi \over 2}} \right)$$
<br><br>= 0.4
<br><br>As q = $${{\Delta \... | mcq | jee-main-2019-online-9th-january-morning-slot | 10,224 |
Iu3gwOOiPHUzNwyUk4EQh | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | A solid metal cube of edge length 2 cm is moving in a positive y-direction at a constant speed of 6 m/s. There is a uniform magnetic field of 0.1 T in the positive z-direction. The potential difference between the two faces of the cube perpendicular to the x-axis, is - | [{"identifier": "A", "content": "2mV"}, {"identifier": "B", "content": "12 mV"}, {"identifier": "C", "content": "6 mV"}, {"identifier": "D", "content": "1 mV"}] | ["B"] | null | <p>We can apply Faraday's law of electromagnetic induction to solve this problem. Faraday's law states that the induced electromotive force (emf) in any closed circuit is equal to the rate of change of the magnetic flux through the circuit.</p>
<p>The cube is moving through a magnetic field, so it's behavin... | mcq | jee-main-2019-online-10th-january-morning-slot | 10,225 |
PHlpNnPo3q6wxoYVWyUUy | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | A very long solenoid of radius R is carrying
current I(t) = kte<sup>–at</sup>(k > 0), as a function of time
(t $$ \ge $$ 0). counter clockwise current is taken to be
positive. A circular conducting coil of radius
2R is placed in the equatorial plane of the
solenoid and concentric with the solenoid. The
current induc... | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734267365/exam_images/q73qytm9hgypfu5fxnxm.webp\" style=\"max-width: 100%; height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2019 (Online) 9th April Evening Slot Physics - Electromagnetic Inducti... | ["A"] | null | $$\varepsilon = - {{d\phi } \over {dt}} = {{ - d} \over {dt}}\left( {{\mu _0}nl} \right)\left( {\pi {R^2}} \right)$$<br><br>
$$ = - \left( {{\mu _0}n\pi {R^2}} \right){{dl} \over {dt}}$$<br><br>
$$ = - {\mu _0}n\pi {R^2}{{d\left( {kt{e^{ - \omega t}}} \right)} \over {dt}}$$<br><br>
$$ = {\rm{ }}-C{\rm{ }}\left( {t\... | mcq | jee-main-2019-online-9th-april-evening-slot | 10,226 |
x9a3dtv5JzPJz409Y3jgy2xukfos38cw | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | An infinitely long, straight wire carrying current
I, one side opened rectangular loop and a
conductor C with a sliding connector are
located in the same plane, as shown, in the
figure. The connector has length $$l$$ and
resistance R. It slides to the right with a
velocity v. The resistance of the conductor and
the sel... | [{"identifier": "A", "content": "$${{{\\mu _0}} \\over {4\\pi }}{{Ivl} \\over {Rr}}$$"}, {"identifier": "B", "content": "$${{{\\mu _0}} \\over {\\pi }}{{Ivl} \\over {Rr}}$$"}, {"identifier": "C", "content": "$${{{\\mu _0}} \\over {2\\pi }}{{Ivl} \\over {Rr}}$$"}, {"identifier": "D", "content": "$${{2{\\mu _0}} \\over \... | ["C"] | null | B = $$\left( {{{{\mu _0}I} \over {2\pi r}}} \right)$$
<br><br>induced current <i>i</i> = $${e \over R} = {{Bvl} \over R}$$
<br><br>= $$\left( {{{{\mu _0}I} \over {2\pi r}}} \right){{vl} \over R}$$ | mcq | jee-main-2020-online-5th-september-evening-slot | 10,228 |
LNJbHig5OQrwaXNddxjgy2xukfhqrvfk | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | Two concentric circular coils, C<sub>1</sub> and C<sub>2</sub> are
<br/>placed in the XY plane. C<sub>1</sub> has 500 turns, and
<br/>a radius of 1 cm. C<sub>2</sub> has 200 turns and radius
<br/>of 20 cm. C<sub>2</sub> carries a time dependent current
<br/>I(t) = (5t<sup>2</sup> – 2t + 3) A where t is in s. The emf
<b... | [] | null | 5 | $${B_2} = {{{\mu _0}{I_2}{N_2}} \over {2{R_2}}}$$<br><br>Total flux $$\phi = {N_1}{B_2}\pi {R_1}^2 = {N_1}{N_2}{{{\mu _0}I} \over {2{R_2}}}\pi {R_1}^2$$<br><br>$$ \therefore $$ $$\phi = {{500 \times 200 \times 4\pi \times {{10}^{ - 7}} \times (5{t^2} - 2t - 3)\pi {{({{10}^{ - 2}})}^2}} \over {2 \times 20 \times {{10... | integer | jee-main-2020-online-5th-september-morning-slot | 10,229 |
1ytEXlOBxQUEHU0EXHjgy2xukf3vcegf | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | A uniform magnetic field B exists in a direction perpendicular to the plane of a square loop made of
a metal wire. The wire has a diameter of 4 mm and a total length of 30 cm. The magnetic field
changes with time at a steady rate $${{dB} \over {dt}}$$ = 0.032 Ts<sup>–1</sup>. The induced current in the loop is close to... | [{"identifier": "A", "content": "0.53 A"}, {"identifier": "B", "content": "0.43 A"}, {"identifier": "C", "content": "0.34 A"}, {"identifier": "D", "content": "0.61 A"}] | ["D"] | null | We know, $$\phi = BA$$<br><br>Also, $$E = {{d\phi } \over {dt}} = {{AdB} \over {dt}}$$<br><br>$$E = {l^2}{{dB} \over {dt}}$$<br><br>$$i = {E \over R} $$
<br><br>= $${{{l^2}{{dB} \over {dt}}} \over {{{\rho l} \over A}}}$$
<br><br>$$= {{{l^2}} \over {pl}}{{dB} \over {dt}}A$$
<br><br>= $${{{{l^2}\pi {R^2}} \over {\rho l}... | mcq | jee-main-2020-online-3rd-september-evening-slot | 10,230 |
SdtVDvgJi6waffHOI8jgy2xukf22whdt | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | An elliptical loop having resistance R, of semi major axis a, and semi minor axis b is placed in
magnetic field as shown in the figure. If the loop is rotated about the x-axis with angular frequency
$$\omega $$, the average power loss in the loop due to Joule heating is :
<img src="data:image/png;base64,UklGRnIOAABXRUJ... | [{"identifier": "A", "content": "$${{\\pi abB\\omega } \\over R}$$"}, {"identifier": "B", "content": "$${{{\\pi ^2}{a^2}{b^2}{B^2}{\\omega ^2}} \\over R}$$"}, {"identifier": "C", "content": "$${{{\\pi ^2}{a^2}{b^2}{B^2}{\\omega ^2}} \\over {2R}}$$"}, {"identifier": "D", "content": "Zero"}] | ["C"] | null | $$\phi $$ = BAcos$$\theta $$
<br><br>$$ \Rightarrow $$ $$\phi $$ = BAcos($$\omega t$$)
<br><br>$${{d\phi } \over {dt}} = - BA\omega \sin \left( {\omega t} \right)$$
<br><br>We know, $$\varepsilon $$ = $$ - {{d\phi } \over {dt}}$$ = $$BA\omega \sin \left( {\omega t} \right)$$
<br><br>Power (P) = $${{{\varepsilon ^2}} \... | mcq | jee-main-2020-online-3rd-september-morning-slot | 10,231 |
pPnzZj7VY2K8Ndq4lvjgy2xukev4xa4l | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | A circular coil of radius 10 cm is placed in a
uniform magnetic field of 3.0 $$ \times $$ 10<sup>–5</sup> T with its
plane perpendicular to the field initially. It is
rotated at constant angular speed about an
axis along the diameter of coil and
perpendicular to magnetic field so that it
undergoes half of rotation in 0... | [] | null | 15 | At any time
flux $$\phi $$ = BA cos $$\omega t$$
<br><br>|emf| = $$\left| {{{d\phi } \over {dt}}} \right|$$ = BA$$\omega$$ sin $$\omega t$$
<br><br>|emf|<sub>max</sub> = BA$$\omega$$ = BA$${{2\pi } \over T}$$
<br><br>= $${{3 \times {{10}^{ - 5}} \times \pi \times {{\left( {0.1} \right)}^2} \times 2\pi } \over {0.4}}$$... | integer | jee-main-2020-online-2nd-september-morning-slot | 10,232 |
mGKZBkNXIoIV2OoGRI7k9k2k5gy10ci | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | At time t = 0 magnetic field of 1000 Gauss is
passing perpendicularly through the area
defined by the closed loop shown in the figure.
If the magnetic field reduces linearly to
500 Gauss, in the next 5s, then induced EMF
in the loop is :
<img src="data:image/png;base64,UklGRtwMAABXRUJQVlA4INAMAAAwTQCdASpnAcUAPm02lkikIy... | [{"identifier": "A", "content": "48 \u03bcV"}, {"identifier": "B", "content": "28 \u03bcV"}, {"identifier": "C", "content": "56 \u03bcV"}, {"identifier": "D", "content": "36 \u03bcV"}] | ["C"] | null | Area of loop = ( 16 × 4 – 2 × Area of triangle) cm<sup>2</sup>
<br><br>= $$\left( {64 - 2 \times {1 \over 2} \times 2 \times 4} \right)$$ cm<sup>2</sup>
<br><br>= 56 × 10<sup>–4</sup> m<sup>2</sup>
<br><br>Using faraday law
Induced EMF
<br><br>|$$\varepsilon $$|$$ = |- A{{dB} \over {dt}}$$|
<br><br>= 56 × 10<sup>–4</s... | mcq | jee-main-2020-online-8th-january-morning-slot | 10,233 |
UY0mQiu9gQoZ1Uy5rm7k9k2k5djjnvg | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | A long solenoid of radius R carries a time (t) - dependent current <br/>I(t)=I<sub>0</sub>t(1 - t). A ring of radius 2R is placed coaxially near its middle. During the time interval 0 $$ \le $$ t $$ \le $$ 1, the induced current (I<sub>R</sub>) and the induced EMF(V<sub>R</sub>) in the ring change as : | [{"identifier": "A", "content": "Direction of I<sub>R</sub> remains unchanged and V<sub>R</sub> is zero at t = 0.25"}, {"identifier": "B", "content": "Direction of I<sub>R</sub> remains unchanged and V<sub>R</sub> is maximum at t = 0.5"}, {"identifier": "C", "content": "At t = 0.25 direction of I<sub>R</sub> reverses a... | ["D"] | null | I(t) = I<sub>0</sub>t(1 - t)
<br><br>We know, $$\phi $$ = BA
<br><br>$$ \Rightarrow $$ $$\phi $$ = $$\mu $$<sub>0</sub>nIA
<br><br>$$ \Rightarrow $$ $$\phi $$ = $$\mu $$<sub>0</sub>nAI<sub>0</sub>(t - t<sup>2</sup>)
<br><br>Also V<sub>R</sub> = $$ - {{d\phi } \over {dt}}$$
<br><br>= - $$\mu $$<sub>0</sub>nAI<sub>0</sub... | mcq | jee-main-2020-online-7th-january-morning-slot | 10,236 |
RjoDgmNZ1AdNATxKiz7k9k2k5dglzs4 | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | Consider a circular coil of wire carrying constant current I, forming a magnetic dipole. The magnetic flux through an infinite plane that contains the circular coil and excluding the circular coil area is given by $$\phi $$<sub>i</sub>. The magnetic flux through the area of the circular coil area is given by $$\phi $$<... | [{"identifier": "A", "content": "$$\\phi $$<sub>i</sub> = $$\\phi $$<sub>0</sub>"}, {"identifier": "B", "content": "$$\\phi $$<sub>i</sub> < $$\\phi $$<sub>0</sub>"}, {"identifier": "C", "content": "$$\\phi $$<sub>i</sub> $$>$$ $$\\phi $$<sub>0</sub>"}, {"identifier": "D", "content": "$$\\phi $$<sub>i</sub> = - $... | ["D"] | null | As, magnetic field lines forms a closed loop, hence each line from circular area will pass through outer area
in opposite direction hence $$\phi $$<sub>i</sub> = - $$\phi $$<sub>0</sub>. | mcq | jee-main-2020-online-7th-january-morning-slot | 10,237 |
wCji5PTHoWle4nQxlzjgy2xukeu17g8l | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | A small bar magnet is moved through a coil at constant speed from one end to the other. Which of the following series of observations will be seen on the galvanometer G attached across the coil?
<img src="data:image/png;base64,UklGRsAJAABXRUJQVlA4ILQJAADQNACdASpEAXAAPm00l0ekIyIhJxOKSIANiWlu4W8hG/Or8Z/zDtO/uPiL+JfNf1j8t... | [{"identifier": "A", "content": "<img src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734265638/exam_images/rg9nbnizawmitysfvqsa.webp\" style=\"max-width: 100%;height: auto;display: block;margin: 0 auto;\" loading=\"lazy\" alt=\"JEE Main 2020 (Online) 4th September Morning Slot Physics - Electromagnetic Induc... | ["C"] | null | When bar magnet is entering with constant speed, flux will change and an e.m.f. is induced, so galvanometer
will deflect in positive direction.
<br><br>When magnet is completely inside, flux will not change, so reading of galvanometer will be zero.
<br><br>When bar magnet is making on exit, again flux will change and o... | mcq | jee-main-2020-online-4th-september-morning-slot | 10,238 |
1ktjpr6q7 | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | A coil is placed in a magnetic field $$\overrightarrow B $$ as shown below :<br/><br/><img src="data:image/png;base64,UklGRhALAABXRUJQVlA4IAQLAAAQRQCdASqLAagAPm02mUikIyKhI3WJMIANiWlu/HyZN+tQzf0P/o3Z//aP6l+yfnv+I/O/178pv6v7EVTjqZdS/47+Pf3H/fej3+V/kn4m+mvtZ/VfUL9Uf5D+U/uF/UfSx+jX9V72IAH6N/G/83/Jf7h/1f7N5238l/JfUD6YfpP6qH... | [{"identifier": "A", "content": "Outward and decreasing with time"}, {"identifier": "B", "content": "Parallel to the plane of coil and decreasing with time"}, {"identifier": "C", "content": "Outward and increasing with time"}, {"identifier": "D", "content": "Parallel to the plane of coil and increasing with time"}] | ["A"] | null | $$\overrightarrow B $$ must not be parallel to the plane of coil for non zero flux and according to lenz law if B is outward it should be decreasing for anticlockwise induced current. | mcq | jee-main-2021-online-31st-august-evening-shift | 10,241 |
1l55kaare | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | <p>A coil is placed in a time varying magnetic field. If the number of turns in the coil were to be halved and the radius of wire doubled, the electrical power dissipated due to the current induced in the coil would be :</p>
<p>(Assume the coil to be short circuited.)</p> | [{"identifier": "A", "content": "Halved"}, {"identifier": "B", "content": "Quadrupled"}, {"identifier": "C", "content": "The same"}, {"identifier": "D", "content": "Doubled"}] | ["D"] | null | <p>The electrical power dissipated due to the current induced in a coil placed in a time-varying magnetic field can be determined by considering Faraday's Law of Induction and the resistance of the coil.</p>
<h3>Faraday's Law of Induction</h3>
<p>The induced EMF ($\mathcal{E}$) in a coil with $N$ turns experiencing a t... | mcq | jee-main-2022-online-28th-june-evening-shift | 10,242 |
1l56wngns | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | <p>A metallic rod of length 20 cm is placed in North-South direction and is moved at a constant speed of 20 m/s towards East. The horizontal component of the Earth's magnetic field at that place is 4 $$\times$$ 10<sup>$$-$$3</sup> T and the angle of dip is 45$$^\circ$$. The emf induced in the rod is ___________ mV.</p> | [] | null | 16 | <p>$$E = Blv$$</p>
<p>$$ = 4 \times {10^{ - 3}} \times {{20} \over {100}} \times 20$$ Volts</p>
<p>$$ = 16$$ mV</p> | integer | jee-main-2022-online-27th-june-evening-shift | 10,243 |
1l58buogs | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | <p>The magnetic flux through a coil perpendicular to its plane is varying according to the relation $$\phi = (5{t^3} + 4{t^2} + 2t - 5)$$ Weber. If the resistance of the coil is 5 ohm, then the induced current through the coil at t = 2 s will be,</p> | [{"identifier": "A", "content": "15.6 A"}, {"identifier": "B", "content": "16.6 A"}, {"identifier": "C", "content": "17.6 A"}, {"identifier": "D", "content": "18.6 A"}] | ["A"] | null | $\phi=5 \mathrm{t}^3+4 \mathrm{t}^2+2 \mathrm{t}-5$
<br/><br/>$|\mathrm{e}|=\frac{\mathrm{d} \phi}{\mathrm{dt}}=15 \mathrm{t}^2+8 \mathrm{t}+2$
<br/><br/>At $\mathrm{t}=2,|\mathrm{e}|=15 \times 2^2+8 \times 2+2$
<br/><br/>$\Rightarrow \mathrm{e}=78 \mathrm{~V} $
<br/><br/>$\Rightarrow \mathrm{I}=\frac{\mathrm{e}}{\math... | mcq | jee-main-2022-online-26th-june-morning-shift | 10,244 |
1l6f49mkk | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | <p>The electric current in a circular coil of 2 turns produces a magnetic induction B<sub>1</sub> at its centre. The coil is unwound and in rewound into a circular coil of 5 tuns and the same current produces a magnetic induction B<sub>2</sub> at its centre. The ratio of $${{{B_2}} \over {{B_1}}}$$ is</p> | [{"identifier": "A", "content": "$${5 \\over 2}$$"}, {"identifier": "B", "content": "$${25 \\over 4}$$"}, {"identifier": "C", "content": "$${5 \\over 4}$$"}, {"identifier": "D", "content": "$${25 \\over 2}$$"}] | ["B"] | null | <p>$$B = {{n{\mu _0}I} \over {2R}}$$</p>
<p>$${B_1} = {{2{\mu _0}I} \over {2{R_1}}}$$</p>
<p>$${B_2} = {{5{\mu _0}I} \over {2{R_2}}}$$</p>
<p>$${R_2} = {{2{R_1}} \over 5}$$</p>
<p>$$ \Rightarrow {{{B_2}} \over {{B_1}}} = {5 \over 2} \times {{{R_1}} \over {{R_2}}} = {{25} \over 4}$$</p> | mcq | jee-main-2022-online-25th-july-evening-shift | 10,245 |
1l6f5xtfr | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | <p>Magnetic flux (in weber) in a closed circuit of resistance 20 $$\Omega$$ varies with time t(s) at $$\phi$$ = 8t<sup>2</sup> $$-$$ 9t + 5. The magnitude of the induced current at t = 0.25 s will be ____________ mA.</p> | [] | null | 250 | <p>$$R = 20\,\Omega $$</p>
<p>$$\phi = 8{t^2} - 9t + 5$$</p>
<p>$$\varepsilon = \left| { - {{d\phi } \over {dt}}} \right| = |16t - 9| = |16(0.25) - 9| = 5$$</p>
<p>$$i = {\varepsilon \over R} = {5 \over {20}} = 0.25\,A = {{0.25} \over {{{10}^3}}} \times {10^3}\,A = 250\,mA$$</p> | integer | jee-main-2022-online-25th-july-evening-shift | 10,246 |
1l6i3vj9u | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | <p>In a coil of resistance $$8 \,\Omega$$, the magnetic flux due to an external magnetic field varies with time as $$\phi=\frac{2}{3}\left(9-t^{2}\right)$$. The value of total heat produced in the coil, till the flux becomes zero, will be _____________ $$J$$.</p> | [] | null | 2 | <p>$$R = 8\,\Omega $$</p>
<p>$$\phi = {2 \over 3}(9 - {t^2})$$</p>
<p>At $$t = 3$$, $$\phi = 0$$</p>
<p>$$\varepsilon = \left| { - {{d\phi } \over {dt}}} \right| = {4 \over 3}t$$</p>
<p>$$H = \int_0^3 {{{{V^2}} \over R}dt = \int_0^3 {{1 \over 8} \times {{16} \over 9}{t^2}dt} } $$</p>
<p>$$ = {2 \over 9} \times \left... | integer | jee-main-2022-online-26th-july-evening-shift | 10,247 |
1l6ko9x70 | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | <p>A conducting circular loop is placed in $$X-Y$$ plane in presence of magnetic field $$\overrightarrow{\mathrm{B}}=\left(3 \mathrm{t}^{3} \,\hat{j}+3 \mathrm{t}^{2}\, \hat{k}\right)$$ in SI unit. If the radius of the loop is $$1 \mathrm{~m}$$, the induced emf in the loop, at time, $$\mathrm{t}=2 \mathrm{~s}$$ is $$\m... | [] | null | 12 | <p>$${B_ \bot } = 3{t^2}$$</p>
<p>$${{d{B_ \bot }} \over {dt}} = 6t - 12$$ at $$t = 2$$</p>
<p>$${{d{\phi _1}} \over {dt}} = 12 \times \pi {(1)^2} = 12\pi $$</p> | integer | jee-main-2022-online-27th-july-evening-shift | 10,248 |
1lds9pijm | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | <p>A square loop of area 25 cm$$^2$$ has a resistance of 10 $$\Omega$$. The loop is placed in uniform magnetic field of magnitude 40.0 T. The plane of loop is perpendicular to the magnetic field. The work done in pulling the loop out of the magnetic field slowly and uniformly in 1.0 sec, will be</p> | [{"identifier": "A", "content": "$$\\mathrm{1.0\\times10^{-3}~J}$$"}, {"identifier": "B", "content": "$$\\mathrm{5\\times10^{-3}~J}$$"}, {"identifier": "C", "content": "$$\\mathrm{2.5\\times10^{-3}~J}$$"}, {"identifier": "D", "content": "$$\\mathrm{1.0\\times10^{-4}~J}$$"}] | ["A"] | null | <p>From energy conservation</p>
<p>Work done to pull the loop out = Energy is lost in the resistance</p>
<p>Emf in the loop $$ = {{d\phi } \over {dt}} = {{B \times A} \over t} = {{40 \times 25 \times {{10}^{ - 4}}} \over {1s}} = 0.1\,V$$</p>
<p>Energy lost $$ = {{em{f^2}} \over R} = {{{{(0.1)}^2}} \over {10}} = {10^{ -... | mcq | jee-main-2023-online-29th-january-evening-shift | 10,250 |
1lgp13fau | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | <p>An insulated copper wire of 100 turns is wrapped around a wooden cylindrical core of the cross-sectional area $$24 \mathrm{~cm}^{2}$$. The two ends of the wire are connected to a resistor. The total resistance in the circuit is $$12 ~\Omega$$. If an externally applied uniform magnetic field in the core along its axi... | [] | null | 60 | The induced emf in the circuit is given by Faraday's law of electromagnetic induction, which is $\mathcal{E}=-d\phi/dt$, where $\phi$ is the magnetic flux through the circuit.
<br/><br/>
The magnetic flux through the circuit is proportional to the magnetic field through the core, so we can write $\phi=NBA$, where $N$ i... | integer | jee-main-2023-online-13th-april-evening-shift | 10,252 |
1lguyla4y | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | <p>The magnetic field B crossing normally a square metallic plate of area $$4 \mathrm{~m}^{2}$$ is changing with time as shown in figure. The magnitude of induced emf in the plate during $$\mathrm{t}=2 s$$ to $$\mathrm{t}=4 s$$, is __________ $$\mathrm{mV}$$.</p>
<p><img src="data:image/png;base64,UklGRvYRAABXRUJQVlA4I... | [] | null | 8 | $$
\begin{aligned}
& \mathrm{m}=\tan \theta=\frac{10}{5}=2 \\\\
& \mathrm{~B}=\mathrm{mt} \\\\
& \mathrm{B}=2 \mathrm{t}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \varepsilon=\left|\frac{\mathrm{d} \phi}{\mathrm{dt}}\right|=\frac{\mathrm{d}(\mathrm{BA})}{\mathrm{dt}}=\frac{\mathrm{AdB}}{\mathrm{dt}} \\\\
& \varep... | integer | jee-main-2023-online-11th-april-morning-shift | 10,253 |
1lgxwolyh | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | <p>Given below are two statements:</p>
<p>Statement I : If the number of turns in the coil of a moving coil galvanometer is doubled then the current sensitivity becomes double.</p>
<p>Statement II : Increasing current sensitivity of a moving coil galvanometer by only increasing the number of turns in the coil will also... | [{"identifier": "A", "content": "Statement I is true but Statement II is false"}, {"identifier": "B", "content": "Statement I is false but Statement II is true"}, {"identifier": "C", "content": "Both Statement I and Statement II are false"}, {"identifier": "D", "content": "Both Statement I and Statement II are true"}] | ["A"] | null | <p>Statement I: If the number of turns in the coil of a moving coil galvanometer is doubled then the current sensitivity becomes double.</p>
<p>This statement is true. The formula for current sensitivity ($I_s$) of a moving coil galvanometer is given by:</p>
<p>$I_s = \frac{NAB}{k}$</p>
<p>where:</p>
<ul>
<li>$N$ is th... | mcq | jee-main-2023-online-10th-april-morning-shift | 10,254 |
1lh248pml | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | <p>The induced emf can be produced in a coil by</p>
<p>A. moving the coil with uniform speed inside uniform magnetic field</p>
<p>B. moving the coil with non uniform speed inside uniform magnetic field</p>
<p>C. rotating the coil inside the uniform magnetic field</p>
<p>D. changing the area of the coil inside the unifo... | [{"identifier": "A", "content": "A and C only"}, {"identifier": "B", "content": "C and D only"}, {"identifier": "C", "content": "B and D only"}, {"identifier": "D", "content": "B and C only"}] | ["B"] | null | If the coil is simply moved at uniform or non-uniform speed in a uniform magnetic field without changing the orientation of the coil or the area of the coil enclosed by the magnetic field, the magnetic flux through the coil does not change, and no emf is induced according to Faraday's Law of electromagnetic induction. | mcq | jee-main-2023-online-6th-april-morning-shift | 10,255 |
jaoe38c1lsd8o0gk | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | <p>The magnetic flux $$\phi$$ (in weber) linked with a closed circuit of resistance $$8 \Omega$$ varies with time (in seconds) as $$\phi=5 t^2-36 t+1$$. The induced current in the circuit at $$t=2 \mathrm{~s}$$ is __________ A.</p> | [] | null | 2 | <p>$$\begin{aligned}
& \varepsilon=-\left(\frac{\mathrm{d} \phi}{\mathrm{dt}}\right)=10 \mathrm{t}-36 \\
& \text { at } \mathrm{t}=2, \varepsilon=16 \mathrm{~V} \\
& \mathrm{i}=\frac{\varepsilon}{\mathrm{R}}=\frac{16}{8}=2 \mathrm{~A}
\end{aligned}$$</p> | integer | jee-main-2024-online-31st-january-evening-shift | 10,257 |
jaoe38c1lse6gmkn | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | <p>A coil is places perpendicular to a magnetic field of $$5000 \mathrm{~T}$$. When the field is changed to $$3000 \mathrm{~T}$$ in $$2 \mathrm{~s}$$, an induced emf of $$22 \mathrm{~V}$$ is produced in the coil. If the diameter of the coil is $$0.02 \mathrm{~m}$$, then the number of turns in the coil is:</p> | [{"identifier": "A", "content": "35"}, {"identifier": "B", "content": "70"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "140"}] | ["B"] | null | <p>$$\begin{aligned}
\varepsilon & =\mathrm{N}\left(\frac{\Delta \phi}{\mathrm{t}}\right) \\
\Delta \phi & =(\Delta \mathrm{B}) \mathrm{A} \\
\mathrm{B}_{\mathrm{i}} & =5000 \mathrm{~T}, \\
\mathrm{~B}_{\mathrm{f}} & =3000 \mathrm{~T} \\
\mathrm{~d} & =0.02 \mathrm{~m} \\
\mathrm{r} & =0.01 \mathrm{~m} \\
\Delta \phi &... | mcq | jee-main-2024-online-31st-january-morning-shift | 10,258 |
jaoe38c1lsf1p9jn | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | <p>Match List I with List II</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;bo... | [{"identifier": "A", "content": "A-IV, B-III, C-I, D-II\n"}, {"identifier": "B", "content": "A-I, B-II, C-III, D-IV\n"}, {"identifier": "C", "content": "A-IV, B-I, C-III, D-II\n"}, {"identifier": "D", "content": "A-II, B-III, C-I, D-IV"}] | ["A"] | null | <p>Ampere-Maxwell law</p>
<p>$$\rightarrow \oint \overrightarrow{\mathrm{B}} \cdot \overrightarrow{\mathrm{dl}}=\mu_0 \mathrm{i}_{\mathrm{c}}+\mu_0 \varepsilon_0 \frac{\mathrm{d} \phi_{\mathrm{E}}}{\mathrm{dt}}$$</p>
<p>Faraday law $$\rightarrow \oint \vec{E} \cdot \overrightarrow{d l}=\frac{d \phi_{\mathrm{B}}}{d t}$$... | mcq | jee-main-2024-online-29th-january-morning-shift | 10,259 |
jaoe38c1lsf27ly3 | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | <p>A square loop of side $$10 \mathrm{~cm}$$ and resistance $$0.7 \Omega$$ is placed vertically in east-west plane. A uniform magnetic field of $$0.20 T$$ is set up across the plane in north east direction. The magnetic field is decreased to zero in $$1 \mathrm{~s}$$ at a steady rate. Then, magnitude of induced emf is ... | [] | null | 2 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt2vwplj/57f7acea-fc42-4254-b91c-29d33a047dfc/3dea4a70-d49e-11ee-8574-81dc091f6c89/file-1lt2vwplk.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt2vwplj/57f7acea-fc42-4254-b91c-29d33a047dfc/3dea4a70-d49e-11ee-8574-81dc091f6c89... | integer | jee-main-2024-online-29th-january-morning-shift | 10,260 |
1lsg5x0jc | physics | electromagnetic-induction | magnetic-flux,-faraday's-and-lenz's-laws | <p>Match List I with List II</p>
<p><style type="text/css">
.tg {border-collapse:collapse;border-spacing:0;}
.tg td{border-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px;
overflow:hidden;padding:10px 5px;word-break:normal;}
.tg th{border-color:black;border-style:solid;bo... | [{"identifier": "A", "content": "A-III, B-IV, C-I, D-II\n"}, {"identifier": "B", "content": "A-IV, B-II, C-III, D-I\n"}, {"identifier": "C", "content": "A-II, B-III, C-IV, D-I\n"}, {"identifier": "D", "content": "A-I, B-III, C-IV, D-II"}] | ["C"] | null | <p>Let's identify each law listed in List I and match it with the corresponding mathematical expression listed in List II.</p>
<strong>Gauss's law of magnetostatics</strong> states that the total magnetic flux through a closed surface is zero, as magnetic monopoles do not exist. This is given by the formula:
<p>$$ \o... | mcq | jee-main-2024-online-30th-january-evening-shift | 10,261 |
0WO4ORKXEf4tgLR0 | physics | electromagnetic-induction | motional-emf-and-eddy-current | A conducting square loop of side $$L$$ and resistance $$R$$ moves in its plane with a uniform velocity $$v$$ perpendicular to one of its sides. A magnetic induction $$B$$ constant in time and space, pointing perpendicular and into the plane at the loop exists everywhere with half the loop outside the field, as shown in... | [{"identifier": "A", "content": "zero "}, {"identifier": "B", "content": "$$RvB$$ "}, {"identifier": "C", "content": "$$vBL/R$$ "}, {"identifier": "D", "content": "$$vBL$$ "}] | ["D"] | null | The induced $$emf$$ is
<br><br>$$e = {{ - d\phi } \over {dt}} = - {{d\left( {\overrightarrow B .\overrightarrow A } \right)} \over {dt}}$$
<br><br>$$ = {{ - d\left( {BA\cos {0^ \circ }} \right)} \over {dt}}$$
<br><br><img class="question-image" src="https://imagex.cdn.examgoal.net/W8jgZfIH9c5t2KhtD/C8Sm8wVsJkdaC7K3Ff... | mcq | aieee-2002 | 10,262 |
BNakOfrO39szUS3t | physics | electromagnetic-induction | motional-emf-and-eddy-current | A metal conductor of length $$1$$ $$m$$ rotates vertically about one of its ends at angular velocity $$5$$ radians per second. If the horizontal component of earth's magnetic field is $$0.2 \times {10^{ - 4}}T,$$ then the $$e.m.f.$$ developed between the two ends of the conductor is | [{"identifier": "A", "content": "$$5mV$$ "}, {"identifier": "B", "content": "$$50\\mu V$$ "}, {"identifier": "C", "content": "$$5\\mu V$$"}, {"identifier": "D", "content": "$$50mV$$ "}] | ["B"] | null | $$\ell = 1m,\,\,\omega = 5\,rad/s,\,\,B = 0.2 \times {10^{ - 4}}T$$
<br><br>$$\varepsilon = {{B\omega {\ell ^2}} \over 2} = {{0.2 \times {{10}^{ - 4}} \times 5 \times 1} \over 2} = 50\mu V$$ | mcq | aieee-2004 | 10,263 |
xbGCYTaI1XIAKd8K | physics | electromagnetic-induction | motional-emf-and-eddy-current | One conducting $$U$$ tube can slide inside another as shown in figure, maintaining electrical contacts between the tubes. The magnetic field $$B$$ is perpendicular to the plane of the figure. If each tube moves towards the other at a constant speed $$v,$$ then the $$emf$$ induced in the circuit in terms of $$B,l$$ and ... | [{"identifier": "A", "content": "$$-Blv$$ "}, {"identifier": "B", "content": "$$blv$$"}, {"identifier": "C", "content": "$$2$$ $$Blv$$"}, {"identifier": "D", "content": "zero "}] | ["C"] | null | Relative velocity $$ = v + v = 2v$$
<br><br>$$\therefore$$ $$emf.$$ $$ = B.l\left( {2v} \right)$$ | mcq | aieee-2005 | 10,264 |
8ZYOzABkujn3sdfm | physics | electromagnetic-induction | motional-emf-and-eddy-current | A boat is moving due east in a region where the earth's magnetic fields is $$5.0 \times {10^{ - 5}}$$ $$N{A^{ - 1}}\,{m^{ - 1}}$$ due north and horizontal. The best carries a vertical aerial $$2$$ $$m$$ long. If the speed of the boat is $$1.50\,m{s^{ - 1}},$$ the magnitude of the induced $$emf$$ in the wire of aerial i... | [{"identifier": "A", "content": "$$0.75$$ $$mV$$ "}, {"identifier": "B", "content": "$$0.50$$ $$mV$$ "}, {"identifier": "C", "content": "$$0.15$$ $$mV$$ "}, {"identifier": "D", "content": "$$1$$ $$mV$$ "}] | ["C"] | null | Induced $$emf$$ $$ = v{B_H}l = 1.5 \times 5 \times {10^{ - 5}} \times 2$$
<br><br>$$ = 15 \times {10^{ - 5}} = 0.15\,mV$$ | mcq | aieee-2011 | 10,266 |
CnUgW4A7b5n2YqdytnIvN | physics | electromagnetic-induction | motional-emf-and-eddy-current | A 10 m long horizontal wire extends from North East to South West. It is falling with a speed of 5.0 ms<sup>–1</sup>, at right angles to the horizontal component of the earth's magnetic field of 0.3 $$ \times $$ 10<sup>–4</sup><sup></sup> Wb/m<sup>2</sup>. The value of the induced emf in wire is : | [{"identifier": "A", "content": "0.3 $$ \\times $$ 10<sup>\u20133</sup> V"}, {"identifier": "B", "content": "2.5 $$ \\times $$ 10<sup>\u20133</sup> V"}, {"identifier": "C", "content": "1.5 $$ \\times $$ 10<sup>\u20133</sup> V"}, {"identifier": "D", "content": "1.1 $$ \\times $$ 10<sup>\u20133</sup> V"}] | ["D"] | null | Induied emf = Bv$$\ell $$ sin 45<sup>o</sup>
<br><br>= 0.3 $$ \times $$ 10<sup>$$-$$4</sup> $$ \times $$ 5 $$ \times $$ 10 $$ \times $$ sin 45<sup>o</sup>
<br><br>= 1.1 $$ \times $$ 10<sup>$$-$$3</sup> V | mcq | jee-main-2019-online-12th-january-evening-slot | 10,270 |
UXz0RHVxtrsGZBjzX23rsa0w2w9jx3ncwig | physics | electromagnetic-induction | motional-emf-and-eddy-current | The figure shows a square loop L of side 5 cm which is connected to a network of resistances. The whole setup is moving towards right with a constant speed of 1 cm s<sup>-1</sup>. At some instant, a part of L is in a uniform
magnetic field of 1 T, perpendicular to the plane of the loop. If the resistance of L is 1.7 $$... | [{"identifier": "A", "content": "115 $$\\mu $$A"}, {"identifier": "B", "content": "170 $$\\mu $$A"}, {"identifier": "C", "content": "60 $$\\mu $$A"}, {"identifier": "D", "content": "150 $$\\mu $$A"}] | ["B"] | null | Since it is a balanced wheatstone bridge, its equivalent resistance = $${4 \over 3}\Omega $$<br><br>
$$\varepsilon = B\ell v = 5 \times {10^{ - 4}}V$$<br><br>
So total resistance<br>
$$R = {4 \over 3} + 1.7 \approx 3\Omega $$<br><br>
$$ \therefore i = {\varepsilon \over R} \approx 166\,\mu A \approx 170\,\mu A$$ | mcq | jee-main-2019-online-12th-april-morning-slot | 10,271 |
TOC7q3W8KVUjoKaYLu1kluke710 | physics | electromagnetic-induction | motional-emf-and-eddy-current | An aeroplane, with its wings spread 10 m, is flying at a speed of 180 km/h in a horizontal direction. The total intensity of earth's field at that part is 2.5 $$\times$$ 10<sup>$$-$$4</sup> Wb/m<sup>2</sup> and the angle of dip is 60$$^\circ$$. The emf induced between the tips of the plane wings will be __________. | [{"identifier": "A", "content": "88.37 mV"}, {"identifier": "B", "content": "62.50 mV"}, {"identifier": "C", "content": "54.125 mV"}, {"identifier": "D", "content": "108.25 mV"}] | ["D"] | null | $$\varepsilon $$<sub>ind</sub> = (B<sub>v</sub>) LV and B<sub>v</sub> = B<sub>Total</sub> sin60<sup>o</sup><br><br>$$ \therefore $$ $$\varepsilon $$<sub>ind</sub> = (2.5 $$\times$$ 10<sup>$$-$$4</sup>)(sin 60<sup>o</sup>) $$\times$$ 10 $$\times$$ 180 $$\times$$ $${5 \over {18}}$$<br><br>= 108.25 mV | mcq | jee-main-2021-online-26th-february-evening-slot | 10,272 |
ntwj86FvNpV3x7grxX1kmipin64 | physics | electromagnetic-induction | motional-emf-and-eddy-current | The magnetic field in a region is given by $$\overrightarrow B = {B_o}\left( {{x \over a}} \right)\widehat k$$. A square loop of side d is placed with its edges along the x and y axes. The loop is moved with a constant velocity $$\overrightarrow v $$ = v<sub>0</sub>$$\widehat i$$. The emf induced in the loop is :<br/>... | [{"identifier": "A", "content": "$${{{B_o}{v_o}{d^2}} \\over {2a}}$$"}, {"identifier": "B", "content": "$${{{B_o}v_o^2d} \\over {2a}}$$"}, {"identifier": "C", "content": "$${{{B_o}{v_o}d} \\over {2a}}$$"}, {"identifier": "D", "content": "$${{{B_o}{v_o}{d^2}} \\over a}$$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265958/exam_images/bw9pqxobfnnzzeds0uc3.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Evening Shift Physics - Electromagnetic Induction Question 68 English Explanation"><br>... | mcq | jee-main-2021-online-16th-march-evening-shift | 10,274 |
bFDGWBeDs2KAPlgQGx1krpmlxyu | physics | electromagnetic-induction | motional-emf-and-eddy-current | The arm PQ of a rectangular conductor is moving from x = 0 to x = 2b outwards and then inwards from x = 2b to x = 0 as shown in the figure. A uniform magnetic field perpendicular to the plane is acting from x = 0 to x = b. Identify the graph showing the variation of different quantities with distance.<br/><br/><picture... | [{"identifier": "A", "content": "A-Flux, B-Power dissipated, C-EMF"}, {"identifier": "B", "content": "A-Flux, B-EMF, C-Power dissipated"}, {"identifier": "C", "content": "A-Power dissipated, B-Flux, C-EMF"}, {"identifier": "D", "content": "A-EMF, B-Power dissipated, C-Flux"}] | ["B"] | null | As the rectangular conductor moves in field area, so flux is increasing up to x = b, then flux is generated on return journey from x = b to x = 0. The flux is shown by plot A of the graph.<br/><br/>As, emf, $$e = - {{d\phi } \over {dt}}$$, which is shown by curve B and power dissipated, P = VI which is shown by curve ... | mcq | jee-main-2021-online-20th-july-morning-shift | 10,275 |
1krun4u4x | physics | electromagnetic-induction | motional-emf-and-eddy-current | A circular conducting coil of radius 1 m is being heated by the change of magnetic field $$\overrightarrow B $$ passing perpendicular to the plane in which the coil is laid. The resistance of the coil is 2 $$\mu$$$$\Omega$$. The magnetic field is slowly switched off such that its magnitude changes in time as <br/><br/>... | [] | null | 80 | $$\phi = \overrightarrow B .\overrightarrow S $$<br><br>$$\phi = {4 \over \pi } \times {10^{ - 3}}\left( {1 - {t \over {100}}} \right).\pi {R^2}$$<br><br>$$\phi = 4 \times {10^{ - 3}} \times {(1)^2}\left( {1 - {t \over {100}}} \right)$$<br><br>$$\varepsilon = {{ - d\phi } \over {dt}}$$<br><br>$$\varepsilon = {{ - ... | integer | jee-main-2021-online-25th-july-morning-shift | 10,276 |
1ktbvdl4e | physics | electromagnetic-induction | motional-emf-and-eddy-current | A circular coil of radius 8.0 cm and 20 turns is rotated about its vertical diameter with an angular speed of 50 rad s<sup>$$-$$1</sup> in a uniform horizontal magnetic field of 3.0 $$\times$$ 10<sup>$$-$$2</sup> T. The maximum emf induced the coil will be ................. $$\times$$ 10<sup>$$-$$2</sup> volt (rounded ... | [] | null | 60 | Maximum emf $$\varepsilon = N\,\omega AB$$<br><br>N = 20, $$\omega$$ = 50, B = 3 $$\times$$ 10<sup>$$-$$2</sup> T<br><br>$$\varepsilon $$ = 20 $$\times$$ 50 $$\times$$ $$\pi$$ $$\times$$ (0.08)<sup>2</sup> $$\times$$ 3 $$\times$$ 10<sup>$$-$$2</sup> = 60.28 $$\times$$ 10<sup>$$-$$2</sup><br><br>Rounded off to nearest ... | integer | jee-main-2021-online-26th-august-evening-shift | 10,277 |
1ktmlqvlm | physics | electromagnetic-induction | motional-emf-and-eddy-current | A square loop of side 20 cm and resistance 1$$\Omega$$ is moved towards right with a constant speed v<sub>0</sub>. The right arm of the loop is in a uniform magnetic field of 5T. The field is perpendicular to the plane of the loop and is going into it. The loop is connected to a network of resistors each of value 4$$\O... | [{"identifier": "A", "content": "1 m/s"}, {"identifier": "B", "content": "1 cm/s"}, {"identifier": "C", "content": "10<sup>2</sup> m/s"}, {"identifier": "D", "content": "10<sup>$$-$$2</sup> cm/s"}] | ["B"] | null | According to given circuit diagram, equivalent resistance between point P and Q.<br><br>$${R_{PQ}} = (4 + 4)||(4 + 4)$$<br><br>$$ = {{8 \times 8} \over {8 + 8}} = 4\,\Omega $$<br><br>The equivalent circuit can be drawn as,<br><br> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kx9wir0q/8ef16fdb-1a5f-4... | mcq | jee-main-2021-online-1st-september-evening-shift | 10,280 |
1l58hyygn | physics | electromagnetic-induction | motional-emf-and-eddy-current | <p>A metallic conductor of length 1 m rotates in a vertical plane parallel to east-west direction about one of its end with angular velocity 5 rad s<sup>$$-$$1</sup>. If the horizontal component of earth's magnetic field is 0.2 $$\times$$ 10<sup>$$-$$4</sup> T, then emf induced between the two ends of the conductor is ... | [{"identifier": "A", "content": "5 $$\\mu$$V"}, {"identifier": "B", "content": "50 $$\\mu$$V"}, {"identifier": "C", "content": "5 mV"}, {"identifier": "D", "content": "50 mv"}] | ["B"] | null | <p>$$Emf = {1 \over 2}B\omega {l^2}$$</p>
<p>$$ = {1 \over 2} \times 0.2 \times {10^{ - 4}} \times 5 \times {1^2}$$ V</p>
<p>= 0.5 $$\times$$ 10<sup>$$-$$4</sup> V</p>
<p>= 50 $$\mu$$V</p> | mcq | jee-main-2022-online-26th-june-evening-shift | 10,281 |
1l5bcms89 | physics | electromagnetic-induction | motional-emf-and-eddy-current | <p>A circular coil of 1000 turns each with area 1m<sup>2</sup> is rotated about its vertical diameter at the rate of one revolution per second in a uniform horizontal magnetic field of 0.07T. The maximum voltage generation will be ___________ V.</p> | [] | null | 440 | <p>$${V_{\max }} = NAB\omega $$</p>
<p>$$ = 1000 \times 1 \times 0.07 \times (2\pi \times 1)$$</p>
<p>$$ \simeq 440$$ volts</p> | integer | jee-main-2022-online-24th-june-evening-shift | 10,282 |
1ldtxnbho | physics | electromagnetic-induction | motional-emf-and-eddy-current | <p>A wire of length 1m moving with velocity 8 m/s at right angles to a magnetic field of 2T. The magnitude of induced emf, between the ends of wire will be __________.</p> | [{"identifier": "A", "content": "20 V"}, {"identifier": "B", "content": "8 V"}, {"identifier": "C", "content": "16 V"}, {"identifier": "D", "content": "12 V"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1ledl0tks/6f0abe55-b10a-42f6-b8a8-2562be02de8e/0c43cfc0-b189-11ed-94f5-239612b3abb5/file-1ledl0tkt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1ledl0tks/6f0abe55-b10a-42f6-b8a8-2562be02de8e/0c43cfc0-b189-11ed-94f5-239612b3abb5/fi... | mcq | jee-main-2023-online-25th-january-evening-shift | 10,285 |
1ldwrc9ak | physics | electromagnetic-induction | motional-emf-and-eddy-current | <p>A metallic rod of length 'L' is rotated with an angular speed of '$$\omega$$' normal to a uniform magnetic field 'B' about an axis passing through one end of rod as shown in figure. The induced emf will be :</p>
<p><img src="data:image/png;base64,UklGRm4xAABXRUJQVlA4IGIxAADwEQKdASoAA2kCP4G20WK2LiqmJJPcCsAwCWlu/GPYaH... | [{"identifier": "A", "content": "$$\\mathrm{\\frac{1}{2}B^2L^2\\omega}$$"}, {"identifier": "B", "content": "$$\\mathrm{\\frac{1}{2}BL^2\\omega}$$"}, {"identifier": "C", "content": "$$\\mathrm{\\frac{1}{4}BL^2\\omega}$$"}, {"identifier": "D", "content": "$$\\mathrm{\\frac{1}{4}B^2L\\omega}$$"}] | ["B"] | null | Velocity of centre of $\operatorname{rod} v=\frac{\omega L}{2}$
<br><br>
So, $\mathrm{emf}=B \cdot v L$ $$
=\frac{B \omega L^{2}}{2}
$$
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le4bpg6w/e90105a5-c32b-4a16-88ea-feaa85527ea7/90048290-ac71-11ed-a551-772121b6ed7f/file-1le4bpg6x.png?format=png" da... | mcq | jee-main-2023-online-24th-january-evening-shift | 10,286 |
1lgsxldcn | physics | electromagnetic-induction | motional-emf-and-eddy-current | <p>A metallic cube of side $$15 \mathrm{~cm}$$ moving along $$y$$-axis at a uniform velocity of $$2 \mathrm{~ms}^{-1}$$. In a region of uniform magnetic field of magnitude $$0.5 \mathrm{~T}$$ directed along $$z$$-axis. In equilibrium the potential difference between the faces of higher and lower potential developed bec... | [] | null | 150 | $$
\begin{aligned}
& q E=e V B \\\\
& E=V B
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
\Delta V & =(E . d) \\\\
& =(V B) \times 0.15 \\\\
& =2 \times \frac{1}{2} \times 0.15 \mathrm{~V} \\\\
& =0.15 \mathrm{~V}=150 \mathrm{mV}
\end{aligned}
$$ | integer | jee-main-2023-online-11th-april-evening-shift | 10,290 |
1lgyq6f6x | physics | electromagnetic-induction | motional-emf-and-eddy-current | <p>An emf of $$0.08 \mathrm{~V}$$ is induced in a metal rod of length $$10 \mathrm{~cm}$$ held normal to a uniform magnetic field of $$0.4 \mathrm{~T}$$, when moves with a velocity of:</p> | [{"identifier": "A", "content": "$$20 \\mathrm{~ms}^{-1}$$"}, {"identifier": "B", "content": "$$2 \\mathrm{~ms}^{-1}$$"}, {"identifier": "C", "content": "$$3.2 \\mathrm{~ms}^{-1}$$"}, {"identifier": "D", "content": "$$0.5 \\mathrm{~ms}^{-1}$$"}] | ["B"] | null | <p>The emf induced in a rod moving through a magnetic field is given by Faraday's law of electromagnetic induction, specifically, in the form of motional emf, which states that:</p>
<p>$ \text{emf} = B \cdot L \cdot v $</p>
<p>where:</p>
<ul>
<li>(B) is the magnetic field strength,</li>
<li>(L) is the length of the... | mcq | jee-main-2023-online-8th-april-evening-shift | 10,293 |
lsank91r | physics | electromagnetic-induction | motional-emf-and-eddy-current | A coil of 200 turns and area $0.20 \mathrm{~m}^2$ is rotated at half a revolution per second and is placed in uniform magnetic field of $0.01 \mathrm{~T}$ perpendicular to axis of rotation of the coil. The maximum voltage generated in the coil is $\frac{2 \pi}{\beta}$ volt. The value of $\beta$ is _______. | [] | null | 5 | $\begin{aligned} & \phi=\mathrm{NAB} \cos (\omega \mathrm{t}) \\\\ & \varepsilon=-\frac{\mathrm{d} \phi}{\mathrm{dt}}=\mathrm{NAB} \omega \sin (\omega \mathrm{t}) \\\\ & \varepsilon_{\max }=\mathrm{NAB} \omega \\\\ & =200 \times 0.2 \times 0.01 \times \pi \\\\ & =\frac{4 \pi}{10}=\frac{2 \pi}{5} \mathrm{volt}\end{align... | integer | jee-main-2024-online-1st-february-evening-shift | 10,295 |
lsblf9zb | physics | electromagnetic-induction | motional-emf-and-eddy-current | A rectangular loop of sides $12 \mathrm{~cm}$ and $5 \mathrm{~cm}$, with its sides parallel to the $x$-axis and $y$-axis respectively, moves with a velocity of $5 \mathrm{~cm} / \mathrm{s}$ in the positive $x$ axis direction, in a space containing a variable magnetic field in the positive $z$ direction. The field has a... | [] | null | 216 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsrmpn5k/8bb0421b-5dee-44c6-a49e-5ddfb61af2af/a80e9870-ce6d-11ee-991d-5322ace9b43e/file-6y3zli1lsrmpn5l.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsrmpn5k/8bb0421b-5dee-44c6-a49e-5ddfb61af2af/a80e9870-ce6d-11ee-99... | integer | jee-main-2024-online-1st-february-morning-shift | 10,296 |
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