1

### JEE Main 2019 (Online) 9th January Morning Slot

A plane electromagnetic wave of frequency 50 MHz travels in free space along the positive x-direction. At a particular point in space and time, $\overrightarrow E = 63\widehat j\,V/m.$ The corresponding magnetic field $\overrightarrow {B,}$ at that point will be :
A
18.9 $\times$ 10$-$8 $\widehat k$T
B
2.1 $\times$ 10$-$8 $\widehat k$T
C
6.3 $\times$ 10$-$8 $\widehat k$T
D
18.9 $\times$ 108 $\widehat k$T

## Explanation

Given, $\overrightarrow E = 6.3\widehat j$

$\therefore$   $\left| {\overrightarrow E } \right| = \sqrt {{{\left( {6.3} \right)}^2}} = 6.3$

$\therefore$   $\left| {\overrightarrow B } \right| = {{\left| {\overrightarrow E } \right|} \over C}$

$= {{6.3} \over {3 \times {{10}^8}}}$

$= 2.1 \times {10^{ - 8}}\,$ T

Now we have to find the direction of $\overrightarrow B$.

We know, $\widehat E \times \widehat B = \widehat C$ and given $\overrightarrow E$ is in y-direction and wave moving in positive x-direction.

$\therefore$   $\widehat J \times \widehat B = \widehat i$

$\Rightarrow$   $\widehat B = \widehat K$

$\therefore$   $\overrightarrow B = \left| {\overrightarrow B } \right|\widehat B$

$= 2.1 \times {10^{ - 8}}\,\widehat K\,T$
2

### JEE Main 2019 (Online) 9th January Morning Slot

A bar magnet is demagnetized by inserting it inside a solenoid of length 0.2 m, 100 turns, and carrying a current of 5.2 m, 100 turns, and carrying a current of 5.2 A. The corecivity of the bar magnet is :
A
285 A/m
B
2600 A/m
C
520 A/m
D
1200 A/m

## Explanation

Coercivity, H = ${B \over {{\mu _0}}}$

Inside solenoid the magnetic field,

B = $\mu$0ni

$\therefore$   H = ${{{\mu _0}ni} \over {{\mu _0}}}$

= ni

= ${N \over \ell } \times i$

= ${{100} \over {0.2}} \times 5.2$

= 2600 A/m
3

### JEE Main 2019 (Online) 9th January Evening Slot

The energy associated with electric field is (UE) and with magnetic field is (UB) for an electromagnetic wave in free space. Then :
A
${U_E} = {{{U_B}} \over 2}$
B
${U_E} > {U_B}$
C
${U_E} < {U_B}$
D
${U_E} = {U_B}$

## Explanation

Energy density of magnetic field (UB) = ${{{B^2}} \over {2{\mu _0}}}$

Also,

Energy density of electric field

UE = ${1 \over 2}$ $\varepsilon$0E2

= ${1 \over 2}$ $\varepsilon$0 B2 C2     [as   ${E \over B}$ = C ]

= ${1 \over 2}$ $\varepsilon$0 B2 $\times$ $\left( {{1 \over {{\varepsilon _0}{\mu _0}}}} \right)$    [as   C2 = ${{1 \over {{\varepsilon _0}{\mu _0}}}}$]

= ${{{B^2}} \over {2{\mu _0}}}$

$\therefore$  UB = UE
4

### JEE Main 2019 (Online) 9th January Evening Slot

A particle having the same charge as of electron moves in a ciurcular path of radius 0.5 cm under the influence of a magnetic field 0f 0.5 T. If an electric field of 100 V/m makes it to move in a straight path, then the mass of the particle is (Given charge of electron = 1.6 $\times$ 10$-$19C)
A
9.1 $\times$ 10$-$31 kg
B
1.6 $\times$ 10$-$27 kg
C
1.6 $\times$ 10$-$19 kg
D
2.0 $\times$ 10$-$24 kg

## Explanation

Given,
radius of circular path(r) = 0.5 cm
Magnetic field (B) = 0.5 T
Electric field (E) = 100 V/m

Charge of particle (q) = 1.6$\times$10$-$19 C

As particle is moving in a circular path so,

${{m{v^2}} \over r} = qvB$

$\Rightarrow$  r = ${{mv} \over {qB}}$ . . . . . . . (1)

When electric field of 100 v/m is applied on the particle then particle is moving in the straight line. So, the net force on the particle is zero.

$\therefore$    Fnet = 0

$\Rightarrow$   Fe = Fm

$\Rightarrow$  qE = qvB

$\Rightarrow$   E = vB  . . . . .(2)

From equation (1) and (2) we get,

r = ${m \over {qB}}\left( {{E \over B}} \right)$

= ${{mE} \over {q{B^2}}}$

$\Rightarrow$   m = ${{q{B^2}r} \over E}$

= ${{1.6 \times {{10}^{ - 19}} \times {{\left( {0.5} \right)}^2} \times 0.5 \times {{10}^{ - 2}}} \over {100}}$

= 2 $\times$ 10$-$24 kg