Question:

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: Recall the energy densities in an electromagnetic wave.
For an electromagnetic wave, the electric energy density is \[ u_E=\frac{1}{2}\varepsilon_0 E^2 \] and the magnetic energy density is \[ u_B=\frac{1}{2\mu_0}B^2 \]

Step 2: Use the relation between electric and magnetic fields.
In free space, \[ E=cB \] where \(c\) is the speed of light. Also, \[ c^2=\frac{1}{\mu_0\varepsilon_0} \] Substituting \(E=cB\) into the electric energy density expression, \[ u_E=\frac{1}{2}\varepsilon_0(cB)^2 \] \[ u_E=\frac{1}{2}\varepsilon_0 c^2 B^2 \] Using \[ c^2=\frac{1}{\mu_0\varepsilon_0}, \] we get \[ u_E=\frac{1}{2\mu_0}B^2 \] Thus, \[ u_E=u_B \]

Step 3: Compare the average energy densities.
Since the instantaneous electric and magnetic energy densities are equal, their average values are also equal.
Hence, \[ K_E=K_B \]

Step 4: Final conclusion.
Therefore, the correct relation is \[ \boxed{K_E=K_B} \]
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