Question:
For an electromagnetic wave travelling in free space, the relation between average energy densities due to electric (\(U_e\)) and magnetic (\(U_m\)) fields is :
For an electromagnetic wave travelling in free space, the relation between average energy densities due to electric (\(U_e\)) and magnetic (\(U_m\)) fields is :
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Use \(E = cB\) and \(c^2 = 1/(\mu_0\epsilon_0)\) to compare \(\frac{1}{2}\epsilon_0E^2\) with \(\frac{B^2}{2\mu_0}\).
Updated On: Nov 4, 2026
- \(U_e = U_m\)
- \(U_e \ne U_m\)
- \(U_e \lt U_m\)
- \(U_e \gt U_m\)
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The Correct Option is A
Solution and Explanation
Energy density of the electric field: \(u_e = \frac{1}{2}\epsilon_0 E^2\). Energy density of the magnetic field: \(u_m = \frac{B^2}{2\mu_0}\).
For a plane electromagnetic wave in free space \(E = cB\) with \(c = \frac{1}{\sqrt{\mu_0\epsilon_0}}\).
\(u_e = \frac{1}{2}\epsilon_0 c^2 B^2 = \frac{1}{2}\epsilon_0\cdot\frac{1}{\mu_0\epsilon_0}B^2 = \frac{B^2}{2\mu_0} = u_m\).
The same holds for the averages over a cycle, so \(U_e = U_m\).
Answer: (1) \(U_e = U_m\)
For a plane electromagnetic wave in free space \(E = cB\) with \(c = \frac{1}{\sqrt{\mu_0\epsilon_0}}\).
\(u_e = \frac{1}{2}\epsilon_0 c^2 B^2 = \frac{1}{2}\epsilon_0\cdot\frac{1}{\mu_0\epsilon_0}B^2 = \frac{B^2}{2\mu_0} = u_m\).
The same holds for the averages over a cycle, so \(U_e = U_m\).
Answer: (1) \(U_e = U_m\)
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