91Ó°ÊÓ

Energy density is the amount of energy stored in a region per unit volume. For electromagnetic waves, energy density can be divided into two parts: electric field energy density and magnetic field energy density. These energy densities can be calculated using the formula: Electric field energy density, \(u_E = \frac{1}{2} \epsilon_0 E^2\) Magnetic field energy density, \(u_B = \frac{1}{2} \frac{B^2}{\mu_0}\) Where \(E\) is the electric field strength, \(B\) is the magnetic field strength, \(\epsilon_0\) is the vacuum permittivity, and \(\mu_0\) is the vacuum permeability.

In an electromagnetic wave, the electric and magnetic fields are perpendicular to each other and to the direction of propagation. The magnitudes of these fields in an electromagnetic wave (in free space) are related by the equation \(E = cB\), where \(c\) is the speed of light in free space.

From the energy density formulas in step 1, let's find the ratio of the electric field energy density (\(u_E\)) to the magnetic field energy density (\(u_B\)): \( \frac{u_E}{u_B} = \frac{\frac{1}{2} \epsilon_0 E^2}{\frac{1}{2} \frac{B^2}{\mu_0}} \) Using the relationship between electric and magnetic fields from step 2 (\(E = cB\)), we can substitute \(E = cB\) into the equation: \( \frac{u_E}{u_B} = \frac{\frac{1}{2} \epsilon_0 (cB)^2}{\frac{1}{2} \frac{B^2}{\mu_0}} \) After simplification, we find: \( \frac{u_E}{u_B} = \frac{\epsilon_0 c^2 B^2}{\frac{B^2}{\mu_0}} = \frac{\epsilon_0 c^2}{\frac{1}{\mu_0}} \) Since \(c^2 = \frac{1}{\epsilon_0 \mu_0}\), we have: \( \frac{u_E}{u_B} = \frac{\epsilon_0}{\frac{1}{\mu_0 \epsilon_0}} = 1 \) Thus, the electric field energy density and magnetic field energy density are equal in an electromagnetic wave.

Based on our analysis, we can conclude that the average energy density in an electromagnetic wave is equally associated with the electric and magnetic fields. Therefore, the correct answer is: (C) equally with electric and magnetic fields.