Question

A plane electromagnetic wave of frequency \(50 \, \text{MHz}\) travels in free space. If the average energy densities in the electric and magnetic fields are \(K_E\) and \(K_B\), then:

• A. \( K_E = K_B \)
• B. \( K_E = K_B = 0 \)
• C. \( K_E>K_B \)
• D. \( K_E<K_B \)

Hint:

In free space, electromagnetic waves have equal energy densities in electric and magnetic fields: \( K_E = K_B \).

Answer 1

The Correct Option is A

In an electromagnetic wave traveling in vacuum or free space:
- Electric and magnetic fields are perpendicular and in phase.
- Energy densities in both fields are equal:
\[ K_E = \frac{1}{2} \varepsilon_0 E^2, \quad K_B = \frac{1}{2\mu_0} B^2 \] And, \[ K_E = K_B \]