Question:medium

In an electromagnetic wave, the ratio of energy densities of electric and magnetic fields is ______.
Fill in the blank with the correct answer from the options given below

Updated On: May 26, 2026
  • 1 : 1
  • 1 : c
  • c : 1
  • \(1 : c^2\)
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The Correct Option is A

Solution and Explanation

The energy density of an electromagnetic wave is derived from its electric and magnetic fields. The energy density of the electric field (\( u_E \)) and the magnetic field (\( u_B \)) are mathematically defined as:

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

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

where \( \epsilon_0 \) is the permittivity of free space, \( \mu_0 \) is the permeability of free space, \( E \) represents the electric field strength, and \( B \) signifies the magnetic field strength. In a vacuum, the electric and magnetic field strengths of an electromagnetic wave are related by \( E = cB \), with \( c \) being the speed of light.

By substituting \( E = cB \) into the equation for \( u_E \), we obtain:

\( u_E = \frac{1}{2} \epsilon_0 (cB)^2 = \frac{1}{2} \epsilon_0 c^2 B^2 \)

Given that \( c = \frac{1}{\sqrt{\epsilon_0 \mu_0}} \), we can substitute this relation into the \( u_E \) expression:

\( u_E = \frac{1}{2} \left(\frac{1}{\mu_0}\right) B^2 = \frac{1}{2} \frac{B^2}{\mu_0} = u_B \)

This derivation demonstrates that the energy density of the electric field is equal to that of the magnetic field, leading to an energy density ratio of \( \frac{u_E}{u_B} = 1:1 \).

Therefore, the definitive answer is 1:1.

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