Question

For the plane electromagnetic wave given by \(E = E_o sin (ωt-kx)\) and \(B = B_o sin ( ωt-kx)\), the ratio of average electric energy density to average magnetic energy density is

• A. 2
• B. 1/2
• C. 1
• D. 4

Answer 1

The Correct Option is C

In EM waves, average electric energy density is equal to average magnetic energy density.
1/ 4 𝜖_0𝐸_0^2 = 1/4𝜇_0 𝐵_0

Answer 2

Electromagnetic Wave Energy Densities 

Step 1: Energy Densities

The average electric energy density (\( u_E \)) and average magnetic energy density (\( u_B \)) are given by:

\( u_E = \frac{1}{4} \epsilon_0 E_0^2 \)

\( u_B = \frac{1}{4\mu_0} B_0^2 \)

where \( \epsilon_0 \) is the permittivity of free space and \( \mu_0 \) is the permeability of free space.

Step 2: Relationship between E and B

For an electromagnetic wave, the electric and magnetic field amplitudes are related by:

\( E_0 = cB_0 \)

where \( c \) is the speed of light. Also, \( c = \sqrt{\frac{1}{\mu_0 \epsilon_0}} \).

Step 3: Ratio of Energy Densities

Substituting \( E_0 = cB_0 \) into the expression for \( u_E \):

\( u_E = \frac{1}{4} \epsilon_0 (cB_0)^2 = \frac{1}{4} \epsilon_0 c^2 B_0^2 = \frac{1}{4} \epsilon_0 \frac{1}{\mu_0 \epsilon_0} B_0^2 = \frac{1}{4\mu_0} B_0^2 \)

Therefore, \( u_E = u_B \). The ratio of average electric energy density to average magnetic energy density is:

\( \frac{u_E}{u_B} = 1 \)

Conclusion:

The ratio of average electric energy density to average magnetic energy density is 1 (Option 3).