Question:
The magnetic field of a plane electromagnetic wave is
\(\vec{B} = 3\times 10^{-8}\sin\left[200\pi (y+ct)\right]\hat{i}\) T
where \(c = 3\times 10^{8}\ \mathrm{ms^{-1}}\) is the speed of light. The corresponding electric field is :
The magnetic field of a plane electromagnetic wave is
\(\vec{B} = 3\times 10^{-8}\sin\left[200\pi (y+ct)\right]\hat{i}\) T
where \(c = 3\times 10^{8}\ \mathrm{ms^{-1}}\) is the speed of light. The corresponding electric field is :
\(\vec{B} = 3\times 10^{-8}\sin\left[200\pi (y+ct)\right]\hat{i}\) T
where \(c = 3\times 10^{8}\ \mathrm{ms^{-1}}\) is the speed of light. The corresponding electric field is :
Show Hint
Amplitude ratio \(E_0 = cB_0\); the direction of \(\vec{E}\times\vec{B}\) must be the direction of propagation, here \(-\hat{j}\).
Updated On: Nov 5, 2026
- \(\vec{E} = -10^{-6}\sin\left[200\pi (y+ct)\right]\hat{k}\) V/m
- \(\vec{E} = -9\sin\left[200\pi (y+ct)\right]\hat{k}\) V/m
- \(\vec{E} = 9\sin\left[200\pi (y+ct)\right]\hat{k}\) V/m
- \(\vec{E} = 3\times 10^{-8}\sin\left[200\pi (y+ct)\right]\hat{k}\)
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The Correct Option is B
Solution and Explanation
The argument \(y+ct\) shows the wave travels along \(-\hat{j}\).
Amplitude: \(E_0 = cB_0 = 3\times10^{8}\times 3\times10^{-8} = 9\ \mathrm{V/m}\).
Direction: \(\vec{E}\times\vec{B}\) must point along \(-\hat{j}\). Since \(\vec{B}\) is along \(+\hat{i}\) and \(\hat{k}\times\hat{i} = +\hat{j}\), the electric field must be along \(-\hat{k}\).
Hence \(\vec{E} = -9\sin[200\pi(y+ct)]\hat{k}\) V/m.
Answer: (2) \(\vec{E} = -9\sin[200\pi(y+ct)]\hat{k}\) V/m
Amplitude: \(E_0 = cB_0 = 3\times10^{8}\times 3\times10^{-8} = 9\ \mathrm{V/m}\).
Direction: \(\vec{E}\times\vec{B}\) must point along \(-\hat{j}\). Since \(\vec{B}\) is along \(+\hat{i}\) and \(\hat{k}\times\hat{i} = +\hat{j}\), the electric field must be along \(-\hat{k}\).
Hence \(\vec{E} = -9\sin[200\pi(y+ct)]\hat{k}\) V/m.
Answer: (2) \(\vec{E} = -9\sin[200\pi(y+ct)]\hat{k}\) V/m
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