Question:
The electric field of a plane electromagnetic wave propagating along the x direction in vacuum is \(\vec{E} = E_0\hat{j}\cos(\omega t - kx)\).
The magnetic field \(\vec{B}\), at the moment t = 0 is :
The electric field of a plane electromagnetic wave propagating along the x direction in vacuum is \(\vec{E} = E_0\hat{j}\cos(\omega t - kx)\).
The magnetic field \(\vec{B}\), at the moment t = 0 is :
The magnetic field \(\vec{B}\), at the moment t = 0 is :
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\(\vec{E}\), \(\vec{B}\) and the propagation direction are mutually perpendicular with \(\vec{E}\times\vec{B}\) along the direction of travel; \(B_0 = E_0/c = E_0\sqrt{\mu_0\epsilon_0}\).
Updated On: Nov 5, 2026
- \(\vec{B} = \frac{E_0}{\sqrt{\mu_0\epsilon_0}}\cos(kx)\,\hat{j}\)
- \(\vec{B} = \frac{E_0}{\sqrt{\mu_0\epsilon_0}}\cos(kx)\,\hat{k}\)
- \(\vec{B} = E_0\sqrt{\mu_0\epsilon_0}\cos(kx)\,\hat{k}\)
- \(\vec{B} = E_0\sqrt{\mu_0\epsilon_0}\cos(kx)\,\hat{j}\)
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The Correct Option is C
Solution and Explanation
The wave travels along +x with \(\vec{E}\) along \(\hat{j}\). \(\vec{E}\times\vec{B}\) must point along \(+\hat{i}\): \(\hat{j}\times\hat{k} = \hat{i}\), so \(\vec{B}\) is along \(\hat{k}\).
Amplitude: \(B_0 = \frac{E_0}{c} = E_0\sqrt{\mu_0\epsilon_0}\) since \(c = \frac{1}{\sqrt{\mu_0\epsilon_0}}\).
\(\vec{B} = E_0\sqrt{\mu_0\epsilon_0}\cos(\omega t - kx)\,\hat{k}\). At t = 0: \(\cos(-kx) = \cos kx\).
\(\vec{B} = E_0\sqrt{\mu_0\epsilon_0}\cos(kx)\,\hat{k}\).
Answer: (3) \(\vec{B} = E_0\sqrt{\mu_0\epsilon_0}\cos(kx)\,\hat{k}\)
Amplitude: \(B_0 = \frac{E_0}{c} = E_0\sqrt{\mu_0\epsilon_0}\) since \(c = \frac{1}{\sqrt{\mu_0\epsilon_0}}\).
\(\vec{B} = E_0\sqrt{\mu_0\epsilon_0}\cos(\omega t - kx)\,\hat{k}\). At t = 0: \(\cos(-kx) = \cos kx\).
\(\vec{B} = E_0\sqrt{\mu_0\epsilon_0}\cos(kx)\,\hat{k}\).
Answer: (3) \(\vec{B} = E_0\sqrt{\mu_0\epsilon_0}\cos(kx)\,\hat{k}\)
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