The magnetic field vector of an electromagnetic wave is given by B = B _ 0 ^ i + ^ j 2 cos (kz - ) where ^ i , ^ j represents unit vector along x and y -axis respectively. At t = 0s, two electric charges q _ 1 of 4π

Correct answer: D. 2: 1 At t = 0 B at (0, 0, (π)/(k)) = B _ 0 ( ^ i + ^ j 2 ) (π) B at (0, 0, (3π)/(k)) = B _ 0 ( ^ i + ^ j 2 ) (3π) Force on charged particle q _ 1 F _ 1 = q _ 1 → V 1 × → B 1 = 4π(0 . 5c ^ i ) × [- B _ 0 ( ^ i + ^ j 2 )] = 4π B _ 0 c 2 2 (- ^ k ) Force on charged particle q _ 2 F _ 2 = q _ 2 → V 2 × → B 2 = 2π(0 . 5c ^ i ) × [- B _ 0 ( ^ i + ^ j 2 )] = 2π B _ 0 c 2 2 (- ^ k ) ⇒ F _ 1 F _ 2 = 2

JEE Main 202131 AugustEvening Shift

Exam: JEE Main
Year: 2021
Date: 31 August
Shift: Evening Shift
Subject: Physics
Chapter: Electromagnetic Waves

JEE Main 2021 · 31 August · Evening Shift · Physics · Electromagnetic Waves

The magnetic field vector of an electromagnetic wave is given by B...

The magnetic field vector of an electromagnetic wave is given by

B = {B}_{0}\dfrac{\overset{^}{i} + \overset{^}{j}}{\sqrt{2}}cos

\left(kz - \omegat\right)

where

\overset{^}{i}, \overset{^}{j}

represents unit vector along

x and

y

-axis respectively. At

t = 0s,

two electric charges

{q}_{1}

4\pi

coulomb and

{q}_{2}

2\pi

coulomb located at

\left(0, 0, \dfrac{\pi}{k}\right)

and

\left(0, 0, \dfrac{3\pi}{k}\right),

respectively, have the same velocity of

0 . 5c\overset{^}{i},

(where

c

is the velocity of light ). The ratio of the force acting on charge

{q}_{1}

{q}_{2}

is :

Options

A $1: \sqrt{2}$
B $2\sqrt{2}: 1$
C $\sqrt{2}: 1$
D $2: 1$

Correct answer

D. $2: 1$

Step-by-step solution

t = 0

B

at \left(0, 0, \dfrac{\pi}{k}\right) = {B}_{0}\left(\dfrac{\overset{^}{i} + \overset{^}{j}}{\sqrt{2}}\right)\cos \left(\pi\right)

<br />

at \left(0, 0, \dfrac{3\pi}{k}\right) = {B}_{0}\left(\dfrac{\overset{^}{i} + \overset{^}{j}}{\sqrt{2}}\right)\cos \left(3\pi\right)

Force on charged particle

{q}_{1}

{F}_{1} = {q}_{1}\overset{\to}{V}1 \times \overset{\to}{B}1

= 4\pi\left(0 . 5c\overset{^}{i}\right) \times \left[- {B}_{0}\left(\dfrac{\overset{^}{i} + \overset{^}{j}}{\sqrt{2}}\right)\right] = \dfrac{4\pi{B}_{0}c}{2\sqrt{2}}\left(- \overset{^}{k}\right)

Force on charged particle

{q}_{2}

{F}_{2} = {q}_{2}\overset{\to}{V}2 \times \overset{\to}{B}2

= 2\pi\left(0 . 5c\overset{^}{i}\right) \times \left[- {B}_{0}\left(\dfrac{\overset{^}{i} + \overset{^}{j}}{\sqrt{2}}\right)\right] = \dfrac{2\pi{B}_{0}c}{2\sqrt{2}}\left(- \overset{^}{k}\right) ⇒ \dfrac{{F}_{1}}{{F}_{2}} = 2

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