JEE Mains · Physics · STD 12 - Electromagnetic waves
- JEE Mains 2019
- Hard
- MCQ
The magnetic field of a plane electromagnetic wave is given by \(\vec B\, = {B_0}\hat i\,[\cos \,(kz - \omega t)]\, + \,{B_1}\hat j\,\cos \,(kz - \omega t)\) 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 closet to
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Answer & Solution
Correct Answer
(D) \(0.6\,N\)
Step-by-step Solution
Detailed explanation
Maximum electric field \(E=(B)(C)\)
\(\overrightarrow{\mathrm{E}}_{0}=\left(3 \times 10^{-5}\right) \mathrm{c}(-\hat{\mathrm{j}})\)
\(\overrightarrow{\mathrm{E}}_{1}=\left(2 \times 10^{-6}\right) \mathrm{c}(-\hat{\mathrm{i}})\)
Maximum force
\(\overrightarrow{\mathrm{F}}_{\mathrm{net}}=10^{-4} \times 3 \times 10^{8} \sqrt{\left(3 \times 10^{-5}\right)^{2}+\left(2 \times 10^{-6}\right)^{2}}=0.9\, \mathrm{N}\)
\(\mathrm{F}_{\mathrm{ms}}=\frac{\mathrm{F}_{0}}{\sqrt{2}}=0.6\, \mathrm{N}\) (approx)
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