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
- 2
- 1/2
- 1
- 4
The Correct Option is C
Approach Solution - 1
1/ 4 𝜖_0𝐸_0^2 = 1/4𝜇_0 𝐵_0
Approach Solution -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).
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