Question

During the propagation of electromagnetic waves in a medium,

• A. electric energy density is double of the magnetic energy density
• B. electric energy density is half of the magnetic energy density
• C. electric energy density is equal to the magnetic energy density
• D. Both electric and magnetic energy densities are zero

Hint:

Energy is equally divided between the electric and magnetic fields. The Electric energy density is equivalent to the magnetic energy density.

Answer 1

The Correct Option is C

Energy is equally divided between the electric and magnetic fields. The Electric energy density is equivalent to the magnetic energy density.

Here, u_E= ½ ϵ₀E² 

u_B=B²/2μ0

Also, E=cB and c=1/√μ₀ϵ₀

Therefore, we get u_E = ½ ϵ₀E² = ½ ϵ₀(cB)²

=½ ϵ₀\(1 \over \mu_o \epsilon_o\)B²

u_E= u_B

So, the correct answer is C) During the propagation of electromagnetic waves in a medium, electric energy density is equal to the magnetic energy density.

Answer 2

The formulas used in order to solve this question are given by

1. Electric energy density, \(U_{E}=\frac{1}{2}εE^{2}\)
2. \(U_{B}=\frac{1}{2μ}B^{2}\)
3. \(c=\frac{1}{\sqrt{με}}\)

Here, U_E and U_B respectively denote the electric and the magnetic field densities corresponding to an EM wave, c is the speed of the electromagnetic field in a medium with magnetic permeability is \(μ\) and the electric permittivity is \(ε\).

Complete step-by-step solution:

Electric energy density - \(U_E=\frac{1}{2}εE^2\)  (1)

Magnetic field - \(U_B=\frac{1}{2μ}B^2\)  (2)

Dividing (1) by (2) we get

\(\frac{U_E}{U_B}=\frac{\frac{1}{2}εE^2}{\frac{1}{2μ}B^2}\)

\(⇒\frac{U_E}{U_B}=με(\frac{E}{B})^2\)  (3)

We know that the speed of an electromagnetic field is related to the amplitudes of the electric and the magnetic field by \(\frac{E}{B}=c\)

Putting it in (3) we get

⇒\(\frac{U_E}{U_B}=μεc^2\)  (4)

Speed of the electromagnetic wave in a medium is given by \(c=\frac{1}{\sqrtμε}\)

On squaring both sides, we get

\(c=\frac{1}{\sqrtμε}\) (5)

Putting (5) in (4) we finally get

\(\frac{U_E}{U_B}=με(\frac{1}{με})\)

\(⇒\frac{U_E}{U_B}=1\)

This can also be written as

\(U_E=U_B\)

Thus, the electric energy density of an electromagnetic wave is equal to its magnetic energy density.

Hence, the correct answer is option C.