1

### JEE Main 2019 (Online) 10th January Morning Slot

If the magnetic field of a plane electromagnetic wave is given by (the speed of light = 3 × 108 B = 100 × 10–6 sin $\left[ {2\pi \times 2 \times {{10}^{15}}\left( {t - {x \over c}} \right)} \right]$ then the maximum electric field associated with it is -
A
4.5 $\times$ 104 N/C
B
4 $\times$ 104 N/C
C
6 $\times$ 104 N/C
D
3 $\times$ 104 N/C

## Explanation

E0 = B0 $\times$ C

= 100 $\times$ 10$-$6 $\times$ 3 $\times$ 108

= 3 $\times$ 104 N/C

$\therefore$  correct answer is 3 $\times$ 104 N/C
2

### JEE Main 2019 (Online) 10th January Evening Slot

The self induced emf of a coil is 25 volts. When the current in it is changed at uniform rate from 10 A to 25 A in 1s, the change in the energy of the inductance is -
A
740 J
B
637.5 J
C
540 J
D
437.5 J

## Explanation

$L{{di} \over {dt}} = 25$

$L \times {{15} \over 1} = 25$

$L = {5 \over 3}H$

$\Delta E = {1 \over 2} \times {5 \over 3} \times ({25^2} - {10^2})$

$= {5 \over 6} \times 525 = 437.5$ J
3

### JEE Main 2019 (Online) 10th January Evening Slot

The electric field of a plane polarized electromagnetic wave in free space at time t = 0 is given by an expression $\overrightarrow E \left( {x,y} \right) = 10\widehat j\cos \left[ {\left( {6x + 8z} \right)} \right].$ The magnetic field $\overrightarrow B$(x,z, t) is given by $-$ (c is the velocity of light)
A
${1 \over c}\left( {6\hat k + 8\widehat i} \right)\cos \left[ {\left( {6x + 8z - 10ct} \right)} \right]$
B
${1 \over c}\left( {6\widehat k - 8\widehat i} \right)\cos \left[ {\left( {6x + 8z - 10ct} \right)} \right]$
C
${1 \over c}\left( {6\hat k + 8\widehat i} \right)\cos \left[ {\left( {6x - 8z + 10ct} \right)} \right]$
D
${1 \over c}\left( {6\hat k - 8\widehat i} \right)\cos \left[ {\left( {6x - 8z + 10ct} \right)} \right]$

## Explanation

$\overrightarrow E = 10\widehat j\cos \left[ {\left( {6\widehat i + 8\widehat k} \right).\left( {x\widehat i + z\widehat k} \right)} \right]$

$= 10\widehat j\,\cos \left[ {\overrightarrow K .\overrightarrow r } \right]$

$\therefore$   $\overrightarrow K = 6\widehat i + 8\widehat k;$ direction of waves travel.

i.e., direction of 'c'.

$\therefore$   Direction of $\widehat B$ will be along

$\widehat C \times \widehat E = {{ - 4\widehat i + 3\widehat k} \over 5}$

Mag. of $\overrightarrow B$ will be along

$\widehat C \times \widehat E = {{ - 4\widehat i + 3\widehat k} \over 5}$

Mag. of $\overrightarrow B = {E \over C} = {{10} \over C}$

$\therefore$   $\overrightarrow B = {{10} \over C}\left( {{{ - 4\widehat i + 3\widehat k} \over 5}} \right) = {{\left( { - 8\widehat i + 6\widehat k} \right)} \over C}$
4

### JEE Main 2019 (Online) 11th January Morning Slot

An electromagnetic wave of intensity 50 Wm–2 enters in a medium of refractive index 'n' without any loss. The ratio of the magnitudes of electric, and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively, given by:
A
$\left( {{1 \over {\sqrt n }},{1 \over {\sqrt n }}} \right)$
B
$\left( {\sqrt n ,\sqrt n } \right)$
C
$\left( {\sqrt n ,{1 \over {\sqrt n }}} \right)$
D
$\left( {{1 \over {\sqrt n }},\sqrt n } \right)$

## Explanation

C $= {1 \over {\sqrt {{\mu _0}{ \in _0}} }}$

V = $= {1 \over {\sqrt {k{ \in _0}{\mu _0}} }}$ [For transparent medium $\mu$r $\approx$ $\mu$0]

$\therefore$  ${C \over V}$ $=$ $\sqrt k =$ n

${1 \over 2} \in {}_0\,E_0^2$C $=$ intensity $=$ ${1 \over 2}$$\in$0 kE2v

$\therefore$   E$_0^2$C $=$ kE2v

$\Rightarrow$  ${{E_0^2} \over {{E^2}}} = {{kV} \over C} = {{{n^2}} \over n} \Rightarrow {{{E_0}} \over E} = \sqrt n$

similarly

${{B_0^2C} \over {2{\mu _0}}} = {{{B^2}v} \over {2{\mu _0}}} \Rightarrow {{{B_0}} \over B} = {1 \over {\sqrt n }}$