Question:
For a plane electromagnetic wave, the magnetic field at a point x and time t is
\(\vec{B}(x,t)=\left[1.2\times10^{-7}\sin\left(0.5\times10^{3}x+1.5\times10^{11}t\right)\hat{k}\right]\) T
The instantaneous electric field \(\vec{E}\) corresponding to \(\vec{B}\) is :
(speed of light c = \(3\times10^{8}\) ms\(^{-1}\))
For a plane electromagnetic wave, the magnetic field at a point x and time t is
\(\vec{B}(x,t)=\left[1.2\times10^{-7}\sin\left(0.5\times10^{3}x+1.5\times10^{11}t\right)\hat{k}\right]\) T
The instantaneous electric field \(\vec{E}\) corresponding to \(\vec{B}\) is :
(speed of light c = \(3\times10^{8}\) ms\(^{-1}\))
\(\vec{B}(x,t)=\left[1.2\times10^{-7}\sin\left(0.5\times10^{3}x+1.5\times10^{11}t\right)\hat{k}\right]\) T
The instantaneous electric field \(\vec{E}\) corresponding to \(\vec{B}\) is :
(speed of light c = \(3\times10^{8}\) ms\(^{-1}\))
Show Hint
The phase \(kx+\omega t\) means the wave travels along \(-\hat{i}\); \(\vec{E}\times\vec{B}\) must point along the propagation direction, and \(E_0=cB_0\).
Updated On: Nov 5, 2026
- \(\vec{E}(x,t)=\left[36\sin\left(1\times10^{3}x+1.5\times10^{11}t\right)\hat{i}\right]\frac{V}{m}\)
- \(\vec{E}(x,t)=\left[36\sin\left(0.5\times10^{3}x+1.5\times10^{11}t\right)\hat{k}\right]\frac{V}{m}\)
- \(\vec{E}(x,t)=\left[36\sin\left(1\times10^{3}x+0.5\times10^{11}t\right)\hat{j}\right]\frac{V}{m}\)
- \(\vec{E}(x,t)=\left[-36\sin\left(0.5\times10^{3}x+1.5\times10^{11}t\right)\hat{j}\right]\frac{V}{m}\)
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The Correct Option is D
Solution and Explanation
Phase is \(0.5\times10^{3}x+1.5\times10^{11}t\), so \(k=0.5\times10^{3}\) rad/m and \(\omega=1.5\times10^{11}\) rad/s. The wave travels in the \(-x\) direction (\(kx\) and \(\omega t\) have the same sign). Speed check: \(\omega/k=3\times10^{8}\) m/s = c.
Amplitude: \(E_0=cB_0=3\times10^{8}\times1.2\times10^{-7}=36\) V/m.
Direction: \(\vec{E}\times\vec{B}\) is along the propagation direction \(-\hat{i}\). \(\vec{B}\) is along \(\hat{k}\). Since \(\hat{j}\times\hat{k}=\hat{i}\), we need \(\vec{E}\) along \(-\hat{j}\) (because \((-\hat{j})\times\hat{k}=-\hat{i}\)).
E and B are in phase, so \(\vec{E}=-36\sin(0.5\times10^{3}x+1.5\times10^{11}t)\hat{j}\) V/m.
Answer: (4) \(\vec{E}(x,t)=[-36\sin(0.5\times10^{3}x+1.5\times10^{11}t)\hat{j}]\) V/m
Amplitude: \(E_0=cB_0=3\times10^{8}\times1.2\times10^{-7}=36\) V/m.
Direction: \(\vec{E}\times\vec{B}\) is along the propagation direction \(-\hat{i}\). \(\vec{B}\) is along \(\hat{k}\). Since \(\hat{j}\times\hat{k}=\hat{i}\), we need \(\vec{E}\) along \(-\hat{j}\) (because \((-\hat{j})\times\hat{k}=-\hat{i}\)).
E and B are in phase, so \(\vec{E}=-36\sin(0.5\times10^{3}x+1.5\times10^{11}t)\hat{j}\) V/m.
Answer: (4) \(\vec{E}(x,t)=[-36\sin(0.5\times10^{3}x+1.5\times10^{11}t)\hat{j}]\) V/m
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