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In a plane electromagnetic wave, the ele...

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]`.

Text Solution

Verified by Experts

The correct Answer is:
A, B

(a) `lambda=c/f`
(b) `c=(E_0)/(B_0)`
or
`B_0=(E_0)/c`
(c) `u_E=1/2epsilon_0E^2=1/2epsilon_0((E_0)/sqrt2)^2=1/4epsilon_0E_0^2`
similarly, `u_b=(B_0^2)/(4pi_0)`
After substituting the values we get,
`u_E=u_B`.
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