Question

In a plane electromagnetic wave, the electric field oscillates sinusoidaly at a frequency of 2.0 × 10¹⁰Hz. if the peak value of electric field is 60 Vm^–1 the average energy density (in Jm^–3) of the magnetic field of the wave will be (given μo = 4π × 10^–7 Tm/A) 

(A) 2π × 10^–7 

(B) (1 / 2π) × 10^–7 

(C) 4π × 10^–7 

(D) (1 / 4π) × 10^–7

Answer 1

The correct option is (D) (1 / 4π) × 10^–7.

Explanation:

f = 2 × 10¹⁰Hz

E_o = 60V/m

(E_o / B_o) = c  hence B_o = (E_o / c) = [60 / (3 × 10⁸)] = 20 × 10^–8T

average energy density of magnetic field = μm = (B_o² / 4μ_o)

= [(20 × 10^–8)² / (4 × 4π × 10^–7)]

∴ μm = [(4 × 10^–14) / (4 × 4π × 10^–7)] = (1 / 4π) × 10^–7J/m³