Question

In a plane electromagnetic wave, $U_E$ and $U_B$ are average energy densities of electric field and magnetic field respectively, then the correct option in the following is

• A. $U_E = \frac{U_B}{2}$
• B. $U_E = 2U_B$
• C. $U_E = U_B$
• D. $U_E \ne U_B$

Hint:

Energy density of electric field: $u_E = \frac{1}{2}\epsilon_0 E^2$.
Energy density of magnetic field: $u_B = \frac{1}{2\mu_0}B^2$.
In an EM wave: $E = cB$ and $c = 1/\sqrt{\epsilon_0 \mu_0}$.
This leads to $u_E = u_B$ at any instant, so their average values $U_E$ and $U_B$ are also equal.
The total average energy density of the EM wave is $U = U_E + U_B = 2U_E = 2U_B$.

Answer 1

The Correct Option is C

In a plane electromagnetic wave propagating in vacuum or a non-dispersive medium, the energy is equally shared between the electric field and the magnetic field. The instantaneous energy density of the electric field ($u_E$) is given by $u_E = \frac{1}{2}\epsilon_0 E^2$, where $E$ is the instantaneous electric field strength and $\epsilon_0$ is the permittivity of free space. The instantaneous energy density of the magnetic field ($u_B$) is given by $u_B = \frac{1}{2\mu_0}B^2$, where $B$ is the instantaneous magnetic field strength and $\mu_0$ is the permeability of free space. For an electromagnetic wave, the magnitudes of the electric and magnetic fields are related by $E = cB$, where $c$ is the speed of light in vacuum. Also, $c = \frac{1}{\sqrt{\epsilon_0 \mu_0}}$. Substitute $E=cB$ into the expression for $u_E$: $u_E = \frac{1}{2}\epsilon_0 (cB)^2 = \frac{1}{2}\epsilon_0 c^2 B^2$. Since $c^2 = \frac{1}{\epsilon_0 \mu_0}$, $u_E = \frac{1}{2}\epsilon_0 \left(\frac{1}{\epsilon_0 \mu_0}\right) B^2 = \frac{1}{2\mu_0}B^2$. This is exactly the expression for $u_B$. So, $u_E = u_B$ for the instantaneous energy densities. If $U_E$ and $U_B$ are the average energy densities, they are obtained by averaging the instantaneous values over one period or a long time. Since $u_E(t) = u_B(t)$ at every instant, their averages must also be equal. $U_E = \langle u_E \rangle = \langle \frac{1}{2}\epsilon_0 E^2 \rangle$. $U_B = \langle u_B \rangle = \langle \frac{1}{2\mu_0}B^2 \rangle$. Since $u_E = u_B$ at all times, it follows that $U_E = U_B$. \[ \boxed{U_E = U_B} \]