Question

The magnetic field of a plane electromagnetic wave is given by $\overset{ \rightarrow }{B}=B_{0}\hat{i}\left[c o s \left(k z - \omega t\right)\right]+B_{1}\hat{j}cos\left(k z + \omega t\right)$ , where $B_{0}=3\times 10^{- 5} \, T$ and $B_{1}=2\times 10^{- 6} \, T$ . The $RMS$ value of the force experienced by a stationary charge $Q=10^{- 4} \, C$ at $z=0$ is closest to

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

Answer 1

Answer: (D)

$\overset{ \rightarrow }{B}=B_{0}\hat{i}\left[c o s \left(k z - \omega t\right)\right]+B_{1}\hat{j}cos\left(k z + \omega t\right)$
$B_{0}=3\times 10^{- 5} \, T \, \&B_{1}=2\times 10^{- 6}T$
Electric filed associated with it is,
$\overset{ \rightarrow }{E}=-B_{0}c\hat{j}cos\left(k z - \omega t\right)-B_{1}c\hat{i}cos\left(k z + \omega t\right)$
Here, $c$ is the speed of light in vacuum.
At $z=0,$
Here $c$ is speed of light in vacuum.
$ \begin{array}{l} =\sqrt{\frac{\left(10^{-4} \times 3 \times 10^{-5} \times 3 \times 10^{8}\right)^{2}+\left(10^{-4} \times 2 \times 10^{-6} \times 3 \times 10^{8}\right)^{2}}{2}} \\ =\sqrt{\frac{0.81+0.0036}{2}}=0.6 N \end{array} $