QJEEM21Ph3685: JEE Main 2021 Physics - Class 12 JEEM21Ph3685 Solution: The magnetic field vector of an electromagnetic wave is given…

Question

Verified Solution • 2025–26 Session moderate

The magnetic field vector of an electromagnetic wave is given by

B = {B_0}{{\widehat i + \widehat j} \over {\sqrt 2 }}\cos (kz - \omega t)

; where

\widehat i,\widehat j

represents unit vector along x and y-axis respectively. At t = 0s, two electric charges q₁ of 4

\pi

coulomb and q₂ of 2

\pi

coulomb located at

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

and

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

, respectively, have the same velocity of 0.5 c

\widehat i

, (where c is the velocity of light). The ratio of the force acting on charge q₁ to q₂ is :-

Accepted Answer

\overrightarrow F = q\left( {\overrightarrow V \times \overrightarrow B } \right)

{\overrightarrow F _1} = 4\pi \left[ {0.5c\widehat i \times {B_0}\left( {{{\widehat i + \widehat j} \over 2}} \right)\cos \left( {K.{\pi \over K} - 0} \right)} \right]

{\overrightarrow F _2} = 2\pi \left[ {0.5c\widehat i \times {B_0}\left( {{{\widehat i + \widehat j} \over 2}} \right)\cos \left( {K.{{3\pi } \over K} - 0} \right)} \right]

\cos \pi = - 1

,

\cos 3\pi = - 1

\therefore

{{{F_1}} \over {{F_2}}} = 2

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