# The magnetic field of a plane electromagnetic wave is given by:

Question:

The magnetic field of a plane electromagnetic wave is given by:

$\vec{B}=B_{0} \hat{i}[\cos (k z-\omega t)]+B_{1} \hat{j} \cos (k z+\omega t)$

Where $\mathrm{B}_{0}=3 \times 10^{-5} \mathrm{~T}$ and $\mathrm{B}_{1}=2 \times 10^{-6} \mathrm{~T}$.

The rms value of the force experienced by a stationary charge $\mathrm{Q}=10^{-4} \mathrm{C}$ at $\mathrm{z}=0$ is closest to:

1. (1) $0.6 \mathrm{~N}$

2. (2) $0.1 \mathrm{~N}$

3. (3) $0.9 \mathrm{~N}$

4. (4) $3 \times 10^{-2} \mathrm{~N}$

Correct Option: 1,

Solution:

(1) $B_{0}=\sqrt{B_{0}^{2}+B_{1}^{2}}=\sqrt{30^{2}+2^{2}} \times 10^{-6}$

$\approx 30 \times 10^{-6} \mathrm{~T}$

$\therefore E_{0}=c B=3 \times 10^{8} \times 30 \times 10^{-6}$

$=9 \times 10^{3} \mathrm{~V} / \mathrm{m}$

$\mathrm{E}_{\text {mas }}=\frac{E_{0}}{\sqrt{2}}=\frac{9}{\sqrt{2}} \times 10^{3} \mathrm{~V} / \mathrm{m}$

Force on the charge,

$F=E_{r m s} Q=\frac{9}{\sqrt{2}} \times 10^{3} \times 10^{-4} \simeq 0.64 \mathrm{~N}$