Question:
A plane electromagnetic wave of frequency 25 GHz is propagating in vacuum along the z-direction. At a particular point in space and time, the magnetic field is given by \(\vec{B} = 5 \times 10^{-8}\,\hat{j}\,\mathrm{T}\). The corresponding electric field \(\vec{E}\) is (speed of light \(c = 3 \times 10^8\) ms-1)
A plane electromagnetic wave of frequency 25 GHz is propagating in vacuum along the z-direction. At a particular point in space and time, the magnetic field is given by \(\vec{B} = 5 \times 10^{-8}\,\hat{j}\,\mathrm{T}\). The corresponding electric field \(\vec{E}\) is (speed of light \(c = 3 \times 10^8\) ms-1)
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\(|E| = cB\) and \(\vec{E} \times \vec{B}\) must point along the propagation direction \(+\hat{k}\): \(\hat{i} \times \hat{j} = \hat{k}\), so \(\vec{E}\) is along \(+\hat{i}\).
Updated On: Nov 5, 2026
- \(15\,\hat{i}\) V / m
- \(-15\,\hat{i}\) V / m
- \(1.66 \times 10^{-16}\,\hat{i}\) V / m
- \(-1.66 \times 10^{-16}\,\hat{i}\) V / m
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The Correct Option is A
Solution and Explanation
Magnitude: \(E = cB = 3 \times 10^8 \times 5 \times 10^{-8} = 15\) V/m.
Direction: in an EM wave, \(\vec{E} \times \vec{B}\) is along the direction of propagation (\(+\hat{k}\)). With \(\vec{B}\) along \(+\hat{j}\), we need \(\hat{E} \times \hat{j} = \hat{k}\). Since \(\hat{i} \times \hat{j} = \hat{k}\), \(\hat{E} = +\hat{i}\).
Hence \(\vec{E} = 15\,\hat{i}\) V/m (the frequency of 25 GHz is not needed).
Answer: (1) \(15\,\hat{i}\) V / m
Direction: in an EM wave, \(\vec{E} \times \vec{B}\) is along the direction of propagation (\(+\hat{k}\)). With \(\vec{B}\) along \(+\hat{j}\), we need \(\hat{E} \times \hat{j} = \hat{k}\). Since \(\hat{i} \times \hat{j} = \hat{k}\), \(\hat{E} = +\hat{i}\).
Hence \(\vec{E} = 15\,\hat{i}\) V/m (the frequency of 25 GHz is not needed).
Answer: (1) \(15\,\hat{i}\) V / m
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