Question

In a plane electromagnetic wave, \(U_E\) and \(U_B\) are average energy densities of electric field and magnetic field respectively, then

• A. \(U_E=\dfrac{U_B}{2}\)
• B. \(U_E=2U_B\)
• C. \(U_E=U_B\)
• D. \(U_E\neq U_B\)

Hint:

For an electromagnetic wave, \[ E=cB. \] As a result, \[ U_E=\frac{1}{2}\varepsilon_0E^2 = \frac{B^2}{2\mu_0} = U_B. \] Thus, the electric and magnetic fields contribute equally to the total energy density.

Answer 1

The Correct Option is C

Step 1: Recall energy density expressions for EM waves.
Electric energy density: \( u_E = \frac{1}{2}\varepsilon_0 E^2 \). Magnetic energy density: \( u_B = \frac{B^2}{2\mu_0} \).

Step 2: Use EM wave relation \( E = cB \) and \( c = 1/\sqrt{\mu_0\varepsilon_0} \).
\( u_B = \frac{B^2}{2\mu_0} = \frac{E^2/c^2}{2\mu_0} = \frac{E^2\varepsilon_0\mu_0}{2\mu_0} = \frac{1}{2}\varepsilon_0 E^2 = u_E \). So average \( U_E = U_B \).

\[ \boxed{U_E = U_B} \]