8.Electromagnetic waves
hard

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 :-

A

$2 \sqrt{2}: 1$

B

$1: \sqrt{2}$

C

$2: 1$

D

$\sqrt{2}: 1$

(JEE MAIN-2021)

Solution

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

$\overrightarrow{{F}}_{1}=4 \pi\left[0.5\,c \hat{{i}} \times {B}_{0}\left(\frac{\hat{{i}}+\hat{{j}}}{2}\right) \cos \left({K} \cdot \frac{\pi}{{K}}-0\right)\right]$

$\overrightarrow{{F}}_{2}=2 \pi\left[0.5 {c} \hat{{i}} \times {B}_{0}\left(\frac{\hat{{i}}+\hat{{j}}}{2}\right) \cos \left({K} \cdot \frac{3 \pi}{{K}}-0\right)\right]$

$\cos \pi=-1, \quad \cos 3 \pi=-1$

$\therefore \frac{{F}_{1}}{{F}_{2}}=2$

Standard 12
Physics

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