Question

The magnetic field of a plane electromagnetic wave is given by:
$\vec{B}=B_0\hat{i}\left[cos\left(kz-\omega t\right)\right]+B_1\hat{j}cos\left(kz+\omega t\right)$
Where $B_0=3\times 10^{-5}T$ and $B_1=2\times 10^{-6}T$
The rms value of the force (in newton) experienced by a stationary charge $Q=10^{-4}C$ at $z=0$ is:

Answer 1

Answer: 0.64

$B_0=\sqrt{B_0^2+B_1^2}=\sqrt{30^2+2^2}\times 10^{-6}$
$\approx 30\times 10^{-6}T$
$\therefore E_0=cB=3\times 10^8\times 30\times 10^{-6}$
$=9\times 10^{3}V/m$
$E_{rms}=\frac{E_0}{\sqrt{2}}=\frac{9}{\sqrt{2}}\times 10^{3}V/m$
Force on the charge,
$F=E_{rms}Q=\frac{9}{\sqrt{2}}\times 10^3\times 10^{-4}\simeq 0.64N$