Question

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

• A. 1 : 1
• B. 1 : c
• C. c : 1
• D. \(1 : c^2\)

Answer 1

The Correct Option is A

In an electromagnetic wave, both electric and magnetic fields contribute to the energy density of the wave. The energy density associated with the electric field \( u_E \) and the magnetic field \( u_B \) can be expressed as: 

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

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

Here, \( \epsilon_0 \) is the permittivity of free space, \( \mu_0 \) is the permeability of free space, \( E \) is the electric field strength, and \( B \) is the magnetic field strength. In free space, the relationship between the magnitudes of the electric field and magnetic field in an electromagnetic wave is given by the equation \( E = cB \), where \( c \) is the speed of light in a vacuum.

Substituting \( E = cB \) into the expression for \( u_E \), we have:

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

Now, the speed of light can also be expressed as \( c = \frac{1}{\sqrt{\epsilon_0 \mu_0}} \). Substituting this into the expression for \( u_E \):

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

This shows that the energy density of the electric field is equal to the energy density of the magnetic field, resulting in the ratio \( \frac{u_E}{u_B} = 1:1 \).

Thus, the correct answer is 1:1.