JEE Mains · Physics · STD 12 - Electromagnetic waves
- JEE Mains 2021
- Hard
- MCQ
The magnetic field vector of an electromagnetic wave is given by \({B}={B}_{o} \frac{\hat{{i}}+\hat{{j}}}{\sqrt{2}} \cos ({kz}-\omega {t})\); where \(\hat{i}, \hat{j}\) represents unit vector along \({x}\) and \({y}\)-axis respectively. At \(t=0\, {s}\), two electric charges \(q_{1}\) of \(4\, \pi\) coulomb and \({q}_{2}\) of \(2 \,\pi\) coulomb located at \(\left(0,0, \frac{\pi}{{k}}\right)\) and \(\left(0,0, \frac{3 \pi}{{k}}\right)\), respectively, have the same velocity of \(0.5 \,{c} \hat{{i}}\), (where \({c}\) is the velocity of light). The ratio of the force acting on charge \({q}_{1}\) to \({q}_{2}\) is :-
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Answer & Solution
Correct Answer
(C) \(2: 1\)
Step-by-step Solution
Detailed explanation
\(\overrightarrow{{F}}={q}(\overrightarrow{{V}} \times \overrightarrow{{B}})\)
\(\overrightarrow{{F}}_{1}=4 \pi\left[0.5\,c \hat{{i}} \times {B}_{0}\left(\frac{\hat{{i}}+\hat{{j}}}{2}\right) \cos \left({K} \cdot \frac{\pi}{{K}}-0\right)\right]\)
\(\overrightarrow{{F}}_{2}=2 \pi\left[0.5 {c} \hat{{i}} \times {B}_{0}\left(\frac{\hat{{i}}+\hat{{j}}}{2}\right) \cos \left({K} \cdot \frac{3 \pi}{{K}}-0\right)\right]\)
\(\cos \pi=-1, \quad \cos 3 \pi=-1\)
\(\therefore \frac{{F}_{1}}{{F}_{2}}=2\)
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