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 :-
Solution
$$\overrightarrow F = q\left( {\overrightarrow V \times \overrightarrow B } \right)$$<br><br>$${\overrightarrow F _1} = 4\pi \left[ {0.5c\widehat i \times {B_0}\left( {{{\widehat i + \widehat j} \over 2}} \right)\cos \left( {K.{\pi \over K} - 0} \right)} \right]$$<br><br>$${\overrightarrow F _2} = 2\pi \left[ {0.5c\widehat i \times {B_0}\left( {{{\widehat i + \widehat j} \over 2}} \right)\cos \left( {K.{{3\pi } \over K} - 0} \right)} \right]$$<br><br>$\cos \pi = - 1$, $\cos 3\pi = - 1$<br><br>$\therefore$ ${{{F_1}} \over {{F_2}}} = 2$
About this question
Subject: Physics · Chapter: Electromagnetic Waves · Topic: Properties of EM Waves
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