JEE Mains · Physics · STD 12 - 8. Electromagnetic waves
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
- A \(0.9\,N\)
- B \(3\times 10^{-2}\,N\)
- C \(0.1\,N\)
- D \(0.6\,N\)
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…
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