The electric fields of two plane electromagnetic plane waves in vacuum are given by → E _ 1 = E _ 0 ^ j ( - kx) and → E _ 2 = E _ 0 ^ k ( - ky) At t = 0, a particle of charge q is at origin with a velocity → v = 08c ^ j

Correct answer: D. E _ 0 q(0.8 ^ i + ^ j + 0.2 ^ k ) Magnetic field vectors associated with this electromagnetic wave are given by → B _ 1 = E _ 0 c ^ k (kx - ) & → B _ 2 = E _ 0 c ^ i (ky - ) → F = q → E + q( → V × → B ) = q( → E _ 1 + → E _ 2 ) + q( → V × ( → B _ 1 + → B _ 2 )) By putting the value of → E _ 1 , → E _ 2 , → B _ 1 and → B _ 2 The net Lorentz force on the charged particles is → F = q E _ 0 [0.8 (kx - ) ^ i + (ky - ) ^ j + 0.2 (ky - ) ^ k ] At t = 0 and at x = y = 0 → F = q E _ 0 [0.8 ^ i + ^ j + 0.2 ^ k ]

JEE Main 20209 JanuaryMorning Shift

Exam: JEE Main
Year: 2020
Date: 9 January
Shift: Morning Shift
Subject: Physics
Chapter: Electromagnetic Waves

JEE Main 2020 · 9 January · Morning Shift · Physics · Electromagnetic Waves

The electric fields of two plane electromagnetic plane waves in...

The electric fields of two plane electromagnetic plane waves in vacuum are given by

\overset{\to}{{E}_{1}} = {E}_{0}\overset{^}{j}\cos \left(\omegat - kx\right)

and

\overset{\to}{{E}_{2}} = {E}_{0}\overset{^}{k}\cos \left(\omegat - ky\right)

t = 0,

a particle of charge

q

is at origin with a velocity

\overset{\to}{v} = 08c\overset{^}{j}

(

c

is the speed of light in vaccum). The instantaneous force experienced by the particle is:

Options

A ${\text{E}}_{0}\text{q}\left(0.8\overset{^}{i} - \overset{^}{j} + 0.4\overset{^}{k}\right)$
B ${\text{E}}_{0}\text{q}\left(0.4\overset{^}{i} - 3\overset{^}{j} + 0.8\overset{^}{k}\right)$
C ${\text{E}}_{0}\text{q}\left(- 0.8\overset{^}{i} + \overset{^}{j} + \overset{^}{k}\right)$
D ${\text{E}}_{0}\text{q}\left(0.8\overset{^}{i} + \overset{^}{j} + 0.2\overset{^}{k}\right)$

Correct answer

D. ${\text{E}}_{0}\text{q}\left(0.8\overset{^}{i} + \overset{^}{j} + 0.2\overset{^}{k}\right)$

Step-by-step solution

Magnetic field vectors associated with this electromagnetic wave are given by

{\overset{\to}{B}}_{1} = \dfrac{{E}_{0}}{c}\overset{^}{k}\cos \left(kx - \omegat\right)

{\overset{\to}{B}}_{2} = \dfrac{{E}_{0}}{c}\overset{^}{i}\cos \left(ky - \omegat\right)

\overset{\to}{F} = q\overset{\to}{E} + q\left(\overset{\to}{V} \times \overset{\to}{B}\right)

= q\left({\overset{\to}{E}}_{1} + {\overset{\to}{E}}_{2}\right) + q\left(\overset{\to}{V} \times \left({\overset{\to}{B}}_{1} + {\overset{\to}{B}}_{2}\right)\right)

By putting the value of

{\overset{\to}{E}}_{1}, {\overset{\to}{E}}_{2}, {\overset{\to}{B}}_{1} \text{ and } {\overset{\to}{B}}_{2}

The net Lorentz force on the charged particles is

\overset{\to}{F} = q{E}_{0}\left[0.8\cos \left(kx - \omegat\right)\overset{^}{i} + \cos \left(ky - \omegat\right)\overset{^}{j} + 0.2\cos \left(ky - \omegat\right)\overset{^}{k}\right]

t = 0

and at

x = y = 0

\overset{\to}{F} = q{E}_{0}\left[0.8\overset{^}{i} + \overset{^}{j} + 0.2\overset{^}{k}\right]

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