Question

The magnetic field of a plane electromagnetic wave is given by:
$\overrightarrow{B}=B_0\hat{i}\left[\cos \right(kz-\omega t\left)\right]+B_1\hat{j}\cos (kz+\omega t),$ Where $B_0=3\times {10}^{-5}\text{T}\text{and}{B}_1=2\times {10}^{-6}\text{T}$
The rms value of the force experienced by a stationary charge $Q={10}^{-4}\text{C}\text{at}z=0$ is closest to

• A. $0.1\text{N}$
• B. $0.6\text{N}$
• C. $3\times {10}^{-2}\text{N}$
• D. $0.9\text{N}$

Answer 1

The correct option is B $0.6\text{N}$

Given:
$B_0=3\times {10}^{-5}\text{T},B_1=2\times {10}^{-6}\text{T}$
$Q={10}^{-4}\text{C}\text{at}z=0$

Net magnetic field
$B_0=\sqrt{B_0^2+B_1^2}=\sqrt{{30}^2+2^2}\times {10}^{-6}\approx 30\times {10}^{-6}\text{T}$

So, Electric field
$E_0=cB=3\times {10}^8\times 30\times {10}^{-6}=9\times {10}^3\text{V/m}$

Now, rms value of the electric field
$E_{rms}=\frac{E_0}{\sqrt{2}}=\frac{9}{\sqrt{2}}\times {10}^3\text{V/m}$

Force on the charge,
$F=E_{rms}Q$
$F=\frac{9}{\sqrt{2}}\times {10}^3\times {10}^{-4}=0.64\text{N}$
$F\simeq 0.6\text{N}$
Hence option $\left(A\right)$ is correct.