When E and B are the amplitudes of the electric and magnetic fields in an electromagnetic wave in vacuum,the total average energy density in the wave is A) $\frac { 1 } { 4 } \varepsilon _ { 0 } E ^ { 2 }$ . B) $\frac { 1 } { 2 } \varepsilon _ { 0 } E ^ { 2 }$ . C)$\varepsilon$₀E². D) $\frac { B ^ { 2 } } { \mu _ { 0 } }$ . E) $\frac { E B } { \mu _ { 0 } }$ .

Question

Quizplus · Accepted Answer

The total average energy density in an electromagnetic wave in vacuum is given by:$$u=\frac{1}{2}\varepsilon_0E^2+\frac{1}{2\mu_0}B^2$$where:* u is the energy density (J/m^3)* $\varepsilon_0$ is the permittivity of free space (8.85 x 10^-12 F/m)* E is the amplitude of the electric field (V/m)* $\mu_0$ is the permeability of free space (4π x 10^-7 H/m)* B is the amplitude of the magnetic field (T)In this case, we are given that E and B are the amplitudes of the electric and magnetic fields, respectively. Therefore, the total average energy density in the wave is:$$u=\frac{1}{2}\varepsilon_0E^2+\frac{1}{2\mu_0}B^2=\frac{1}{2}\varepsilon_0E^2$$Therefore, the correct answer is B.

When E and B Are the Amplitudes of the Electric 14ε0E2\frac { 1 } { 4 } \varepsilon _ { 0 } E ^ { 2 }41​ε0​E2

When E and B Are the Amplitudes of the Electric $\frac { 1 } { 4 } \varepsilon _ { 0 } E ^ { 2 }$