Question:
If the magnetic field of a plane electromagnetic wave is given by (the speed of light = \(3\times10^{8}\) B = \(100\times10^{-6}\) sin
\(\left[2\pi\times2\times10^{15}\left(t-\frac{x}{c}\right)\right]\)
then the maximum electric field associated with it is -
If the magnetic field of a plane electromagnetic wave is given by (the speed of light = \(3\times10^{8}\) B = \(100\times10^{-6}\) sin
\(\left[2\pi\times2\times10^{15}\left(t-\frac{x}{c}\right)\right]\)
then the maximum electric field associated with it is -
\(\left[2\pi\times2\times10^{15}\left(t-\frac{x}{c}\right)\right]\)
then the maximum electric field associated with it is -
Show Hint
For an electromagnetic wave in vacuum \(E_0 = cB_0\).
Updated On: Nov 5, 2026
- \(4.5\times10^4\) N/C
- \(4\times10^4\) N/C
- \(6\times10^4\) N/C
- \(3\times10^4\) N/C
Show Solution
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The Correct Option is D
Solution and Explanation
Amplitude of the magnetic field: \(B_0 = 100\times10^{-6}\) T \(= 10^{-4}\) T.
In vacuum \(\frac{E_0}{B_0} = c\), so \(E_0 = cB_0 = 3\times10^{8}\times10^{-4} = 3\times10^{4}\) N/C.
Answer: (4) \(3\times10^4\) N/C
In vacuum \(\frac{E_0}{B_0} = c\), so \(E_0 = cB_0 = 3\times10^{8}\times10^{-4} = 3\times10^{4}\) N/C.
Answer: (4) \(3\times10^4\) N/C
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