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A beam of light travelling along $X$-axis is described by the electric field $E_Y=600 \frac{\mathrm{V}}{\mathrm{m}}$. $\sin \omega\left(t-\frac{x}{c}\right)$, the maximum magnetic force on a charge $q=2 e$, moving along $Y$-axis with the speed of $3 \times 10^8 \mathrm{~m} / \mathrm{s}$ is
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The correct answer is:
$1.92 \times 10^{-17} \mathrm{~N}$
Given, electric field, $E_y=600 \frac{\mathrm{V}}{\mathrm{m}} \cdot \sin \omega\left(t-\frac{x}{\mathrm{c}}\right)$...(i)
We know that,
$E=E_0 \sin \omega\left(t-\frac{x}{c}\right)$.....(ii)
On comparing Eqs. (i) and (ii), we get $E_0=600 \mathrm{~V} / \mathrm{m}$ and $C=3 \times 10^8 \mathrm{~m} / \mathrm{s}$
Maximum magnetic field,
$\begin{aligned} & B_0=\frac{E_0}{C} \\ & B_0=\frac{600}{3 \times 10^8}=2 \times 10^{-6} \mathrm{~T}\end{aligned}$
$\begin{aligned} \therefore \quad F_m & =q v B_0=2 \mathrm{ev} B_0 \\ & =2 \times 1.6 \times 10^{-19} \times 3 \times 10^8 \times 2 \times 10^{-6} \\ & =19.2 \times 10^{-17} \mathrm{~N}\end{aligned}$
We know that,
$E=E_0 \sin \omega\left(t-\frac{x}{c}\right)$.....(ii)
On comparing Eqs. (i) and (ii), we get $E_0=600 \mathrm{~V} / \mathrm{m}$ and $C=3 \times 10^8 \mathrm{~m} / \mathrm{s}$
Maximum magnetic field,
$\begin{aligned} & B_0=\frac{E_0}{C} \\ & B_0=\frac{600}{3 \times 10^8}=2 \times 10^{-6} \mathrm{~T}\end{aligned}$
$\begin{aligned} \therefore \quad F_m & =q v B_0=2 \mathrm{ev} B_0 \\ & =2 \times 1.6 \times 10^{-19} \times 3 \times 10^8 \times 2 \times 10^{-6} \\ & =19.2 \times 10^{-17} \mathrm{~N}\end{aligned}$
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