In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of `2.0xx10^10 H_z` and amplitude `48V_m^-1`
(a) What is the wavelength o f the wave?
(b) What is the amplitude of the oscillating magnetic field.
(c) Show that the average energy density of the field `E` equals the average energy density of the field `B.[c=3xx10^8 ms^-1]`.

Options

A. `(1)/(16)xx10^(-8) Wb//m^(2)`
B. `16xx10^(-8)wb//m^(2)`
C. `12xx10^(-7) wb//m^(2)`
D. `(1)/(12)xx10^(-7) Wb//m^(2)`

Answer

B

Explanation

Oscillating magnetic field
B = `("E")/("C") = (48)/(3xx10^(8)) = 16 xx 10^(-8)) Wb/m^(2)`