Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

The ratio of contributions made by the electric field and magnetic field components to the intensity of an electromagnetic wave is: ( $c$ = speed of electromagnetic waves)

1 : 1

1 : c

1 : c²

c : 1

Step-by-Step Solution

Energy Density: The intensity of an electromagnetic wave is the energy transported per unit area per unit time. This energy is stored in the electric and magnetic fields of the wave.
Equipartition of Energy: In a plane electromagnetic wave traveling in a vacuum, the average electric energy density ( $⟨u_E⟩$ ) and the average magnetic energy density ( $⟨u_B⟩$ ) are equal. $⟨u_E⟩ = \frac{1}{4}\epsilon_0 E_0^2$ $⟨u_B⟩ = \frac{1}{4\mu_0} B_0^2$
Proof: Using the relation $E_0 = cB_0$ and $c = 1/\sqrt{\mu_0\epsilon_0}$ , we can substitute into the expression for $⟨u_E⟩$ : $⟨u_E⟩ = \frac{1}{4}\epsilon_0 (cB_0)^2 = \frac{1}{4}\epsilon_0 (\frac{1}{\mu_0\epsilon_0}) B_0^2 = \frac{1}{4\mu_0} B_0^2 = ⟨u_B⟩$ .
Conclusion: Since the energy densities are equal, the electric and magnetic fields contribute equally to the total energy and thus to the intensity of the wave. The ratio is 1:1.

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started