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

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