Question:easy

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

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In a plane electromagnetic wave travelling in free space, the electric and magnetic fields equally share the energy: \[ u_E=u_B \]
Updated On: Jun 24, 2026
  • \(K_E=K_B\)
  • \(K_E=K_B=0\)
  • \(K_E\gt K_B\)
  • \(K_E\lt K_B\)
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The Correct Option is A

Solution and Explanation

Step 1: Write the electric and magnetic energy densities.
In an electromagnetic wave, at any point:
Electric energy density: $u_E = \frac{1}{2}\varepsilon_0 E^2$
Magnetic energy density: $u_B = \frac{B^2}{2\mu_0}$

Step 2: Use the EM wave relation $E = cB$.
For an electromagnetic wave in free space, the electric and magnetic fields are always related by:
\[ E = cB \quad \text{where } c = \frac{1}{\sqrt{\mu_0\varepsilon_0}} \]

Step 3: Substitute into the expression for electric energy density.
\[ u_E = \frac{1}{2}\varepsilon_0 E^2 = \frac{1}{2}\varepsilon_0 c^2 B^2 \] Since $c^2 = 1/(\mu_0\varepsilon_0)$:
\[ u_E = \frac{1}{2}\varepsilon_0 \times \frac{1}{\mu_0\varepsilon_0} \times B^2 = \frac{B^2}{2\mu_0} = u_B \]

Step 4: Interpret the result.
At every instant, the energy stored in the electric field equals the energy stored in the magnetic field. This is a fundamental property of EM waves, independent of their frequency.

Step 5: Extend to average energy densities.
Since the instantaneous values are always equal ($u_E = u_B$ at all times), their time-averaged values are also equal:
\[ K_E = \langle u_E \rangle = \langle u_B \rangle = K_B \]

Step 6: State the answer.
\[ \boxed{K_E = K_B} \]
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