1

### 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.
2

### 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
3

### JEE Main 2017 (Offline)

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: A
275 Wb
B
200 Wb
C
225 Wb
D
250 Wb

## Explanation

According to Faraday's law of electromagnetic induction,

$\varepsilon = {{d\phi } \over {dt}}$

Also, $\varepsilon$ = iR

$\therefore$ ${{d\phi } \over {dt}}$ = iR

$\Rightarrow$ $\int {d\phi } = R\int {idt}$

Magnitude of change in flux (d$\phi$) = R × area under current vs time graph

$\Rightarrow$ d$\phi$ = $100 \times {1 \over 2} \times {1 \over 2} \times 10$ = 250 Wb
4

### JEE Main 2017 (Online) 8th April Morning Slot

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 = Io cos ($\omega$t). The emf induced in the smaller inner loop is nearly :
A
${{\pi {\mu _o}{I_o}} \over 2}.{{{a^2}} \over b}\,\omega \sin \left( {\omega t} \right)$
B
${{\pi {\mu _o}{I_o}} \over 2}.{{{a^2}} \over b}\,\omega \cos \left( {\omega t} \right)$
C
$\pi {\mu _o}{I_o}\,{{{a^2}} \over b}\omega \sin \left( {\omega t} \right)$
D
${{\pi {\mu _o}{I_o}\,{b^2}} \over a}\omega \cos \left( {\omega t} \right)$

## Explanation

Mutual inductance,

M = ${{{\mu _0}\pi {N_1}{N_2}\,{a^2}} \over {2b}}$

here ${{N_1}}$ = N2 = 1

$\therefore\,\,\,$ M = ${{{\mu _0}\pi {a^2}} \over {2b}}$

Current I = I0 cos ($\omega$t)

e = $-$ M ${{dI} \over {dt}}$

= $-$ ${{{\mu _0}\pi {a^2}} \over {2b}}$ ${d \over {dt}}$ (I0 cos $\omega$t)

= + ${{{\mu _0}\pi {a^2}} \over {2b}}$ I0 $\omega$ sin $\omega$t

= ${{\pi {\mu _0}{I_0}} \over 2}$ . ${{{a^2}} \over b}$ $\omega$ sin $\omega$ t