Question:
Magnetic field in a plane electromagnetic wave is given by \(\vec B=B_0\sin(kx+\omega t)\,\hat j\ \mathrm{T}\). Expression for corresponding electric field will be : Where \(c\) is speed of light.
Magnetic field in a plane electromagnetic wave is given by \(\vec B=B_0\sin(kx+\omega t)\,\hat j\ \mathrm{T}\). Expression for corresponding electric field will be : Where \(c\) is speed of light.
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The wave travels along \(-x\); \(\vec E\times\vec B\) must point along \(-\hat i\), and \(E_0=cB_0\).
Updated On: Nov 11, 2026
- \(\vec E=B_0c\sin(kx+\omega t)\,\hat k\ \mathrm{V/m}\)
- \(\vec E=\frac{B_0}{c}\sin(kx+\omega t)\,\hat k\ \mathrm{V/m}\)
- \(\vec E=-B_0c\sin(kx+\omega t)\,\hat k\ \mathrm{V/m}\)
- \(\vec E=B_0c\sin(kx-\omega t)\,\hat k\ \mathrm{V/m}\)
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The Correct Option is A
Solution and Explanation
The phase \(kx+\omega t\) means the wave propagates along \(-\hat i\).
The direction of propagation is along \(\vec E\times\vec B\). With \(\vec B\) along \(\hat j\): \(\hat k\times\hat j=-\hat i\), so \(\vec E\) is along \(+\hat k\).
Amplitude \(E_0=cB_0\), and E and B are in phase.
\(\vec E=B_0c\sin(kx+\omega t)\,\hat k\ \mathrm{V/m}\).
Answer: (1) \(\vec E=B_0c\sin(kx+\omega t)\,\hat k\ \mathrm{V/m}\)
The direction of propagation is along \(\vec E\times\vec B\). With \(\vec B\) along \(\hat j\): \(\hat k\times\hat j=-\hat i\), so \(\vec E\) is along \(+\hat k\).
Amplitude \(E_0=cB_0\), and E and B are in phase.
\(\vec E=B_0c\sin(kx+\omega t)\,\hat k\ \mathrm{V/m}\).
Answer: (1) \(\vec E=B_0c\sin(kx+\omega t)\,\hat k\ \mathrm{V/m}\)
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