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NEET
Exam
NEET
Subject
Physics
Chapter
Electromagnetic Waves

NEET · Physics · Electromagnetic Waves

Given below are two statements: one is labelled as Assertion (A)...

Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).

Assertion (A): In an electromagnetic wave propagating in vacuum, the average energy density of the electric field is equal to the average energy density of the magnetic field.
Reason (R): In vacuum, the amplitudes of the electric and magnetic fields are related by E0=cB0E_0 = cB_0.

In the light of the above statements, choose the most appropriate answer from the options given below:

Options

  1. ABoth (A) and (R) are correct and (R) is the correct explanation of (A)
  2. BBoth (A) and (R) are correct but (R) is not the correct explanation of (A)
  3. C(A) is correct but (R) is not correct
  4. D(A) is not correct but (R) is correct

Correct answer

A. Both (A) and (R) are correct and (R) is the correct explanation of (A)

Step-by-step solution

The average energy density of the electric field is uE=14ϵ0E02u_E = \dfrac{1}{4}\epsilon_0 E_0^2.

The average energy density of the magnetic field is uB=B024μ0u_B = \dfrac{B_0^2}{4\mu_0}.

In vacuum, the speed of light is c=1μ0ϵ0c = \dfrac{1}{\sqrt{\mu_0\epsilon_0}}, and the field amplitudes are related by E0=cB0E_0 = cB_0.

Substituting E0E_0 into the expression for uEu_E:
uE=14ϵ0(cB0)2=14ϵ0(1μ0ϵ0)B02=B024μ0=uBu_E = \dfrac{1}{4}\epsilon_0 (cB_0)^2 = \dfrac{1}{4}\epsilon_0 \left(\dfrac{1}{\mu_0\epsilon_0}\right) B_0^2 = \dfrac{B_0^2}{4\mu_0} = u_B.

Thus, the average energy densities are equal, making the Assertion (A) correct. The Reason (R) is also correct and provides the exact mathematical relationship required to prove the Assertion. Therefore, (R) is the correct explanation of (A).

Answer: Both (A) and (R) are correct and (R) is the correct explanation of (A)
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