Question

A charged particle $(Q={10}^{-4}C)$ is released from rest at $z=0$ in magnetic field given as $\overrightarrow{B}=B_0\cos (\omega t-kz)\hat{i}+B_1\cos (\omega t+kz)\hat{j}$ where $B_0=3\times {10}^{-5}T$ and $B_1=2\times {10}^{-6}T$. Then the rms value of force acting on particle is?

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

Answer 1

The correct option is B $0.6$

The electric field in the region is:
$\overrightarrow{E}=-c{B}_0\cos (\omega t-kz)\hat{j}-c{B}_1\cos (\omega t+kz)\hat{i}$
so for charge released from rest at $z=0$, the rms value of force is:
$F_{rms}=g\sqrt{{\left(\frac{c{B}_0}{\sqrt{2}}\right)}^2+{\left(\frac{c{B}_1}{\sqrt{2}}\right)}^2}$
$={10}^{-4}\times \frac{3\times {10}^8}{\sqrt{2}}\sqrt{(30\times {10}^{-6}{)}^2+(2\times {10}^{-6}{)}^2}$
$={10}^{-4}\times \frac{3\times {10}^8}{\sqrt{2}}\sqrt{904}\times {10}^{-6}=0.63$