1
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 9th April Morning Slot

A series LR circuit is connected to a voltage source with
V(t) = V0 sin$$\Omega $$t. After very large time, current I(t) behaves as
(t0 >> $${L \over R}$$) :
A
B
C
D

Explanation

Current in LR circuit,

$${\rm I} = {{{V_0}} \over {\sqrt {{R^2} + {w^2}{L^2}} }}\sin \left( {\omega t - {\pi \over 2}} \right)$$

it will be a sinusoidal wave.
2
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 10th April Morning Slot

A conducting metal circular-wire-loop of radius r is placed perpendicular to a magnetic field which varies with time as
B = B0e$${^{{{ - t} \over r}}}$$ , where B0 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 time (t $$ \to $$ $$\infty $$) is :
A
$${{{\pi ^2}{r^4}B_0^4} \over {2\tau R}}$$
B
$${{{\pi ^2}{r^4}B_0^2} \over {2\tau R}}$$
C
$${{{\pi ^2}{r^4}B_0^2R} \over \tau }$$
D
$${{{\pi ^2}{r^4}B_0^2} \over {\tau R}}$$

Explanation

Given,

B = B0e$$^{ - {t \over \tau }}$$

Area of the circular loop, A = $$\pi $$ r2

$$ \therefore $$   Flux $$\phi $$ = BA = $$\pi $$ r2 B0 e$$^{ - {t \over \tau }}$$

Induced emf in the loop,

$$\varepsilon $$ = $$-$$ $${{d\phi } \over {dt}}$$ = $$\pi $$ r2B0$${1 \over \tau }$$e$$^{ - {t \over \tau }}$$

Heat generated

= $$\int\limits_0^ \propto {{i^2}R\,dt} $$

= $$\int\limits_0^ \propto {{{{\varepsilon ^2}} \over R}} \,dt$$

= $${1 \over R}{{{\pi ^2}{r^4}B_0^2} \over {{\tau ^2}}}\int\limits_0^ \propto {{e^{ - {{2t} \over \tau }}}} \,dt$$

= $${{{\pi ^2}{r^4}B_0^2} \over {{\tau ^2}R}} \times {1 \over {\left( { - {2 \over \tau }} \right)}}\left[ {{e^{ - {{2t} \over \tau }}}} \right]_0^ \propto $$

= $${{ - {\pi ^2}{r^4}B_0^2} \over {2{\tau ^2}R}} \times \tau \left( {0 - 1} \right)$$

= $${{{\pi ^2}{r^4}B_0^2} \over {2\tau R}}$$
3
MCQ (Single Correct Answer)

JEE Main 2016 (Online) 10th April Morning Slot

Consider an electromagnetic wave propagating in vacuum. Choose the correct statement :
A
For an electromagnetic wave propagating in +x direction the electric field is $$\vec E = {1 \over {\sqrt 2 }}{E_{yz}}{\mkern 1mu} \left( {x,t} \right)\left( {\hat y - \hat z} \right)$$

and the magnetic field is $$\vec B = {1 \over {\sqrt 2 }}{B_{yz}}{\mkern 1mu} \left( {x,t} \right)\left( {\hat y + \hat z} \right)$$
B
For an electromagnetic wave propagating in +x direction the electric field is $$\vec E = {1 \over {\sqrt 2 }}{E_{yz{\mkern 1mu} }}\left( {y,z,t} \right)\left( {\hat y + \hat z} \right)$$

and the magnetic field is $$\vec B = {1 \over {\sqrt 2 }}{B_{yz{\mkern 1mu} }}\left( {y,z,t} \right)\left( {\hat y + \hat z} \right)$$
C
For an electromagnetic wave propagating in + y direction the electric field is $$\overrightarrow E = {1 \over {\sqrt 2 }}{E_{yz{\mkern 1mu} }}\left( {x,t} \right)\widehat y$$
and the magnetic field is $$\vec B = {1 \over {\sqrt 2 }}{B_{yz{\mkern 1mu} }}\left( {x,t} \right)\widehat z$$
D
For an electromagnetic wave propagating in + y direction the electric field is $$\overrightarrow E = {1 \over {\sqrt 2 }}{E_{yz{\mkern 1mu} }}\left( {x,t} \right)\widehat z$$
and the magnetic field is $$\overrightarrow B = {1 \over {\sqrt 2 }}{B_{z{\mkern 1mu} }}\left( {x,t} \right)\widehat y$$

Explanation

As wave is propagating in   + x   direction, then   $$\overrightarrow E $$  and   $$\overrightarrow B $$  should be function of   $$\left( {x,t} \right)$$  and must be in   y $$-$$ z   plane.
4
MCQ (Single Correct Answer)

JEE Main 2016 (Offline)

An arc lamp requires a direct current of 10 A at 80 V to function. If it is connected to a 220 V (rms), 50 Hz AC supply, the series inductor needed for it to work is close to :
A
0.044 H
B
0.065 H
C
80 H
D
0.08 H

Questions Asked from Alternating Current and Electromagnetic Induction

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