In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 V m−1.
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the E field equals the average energy density of the B field. [c = 3 × 108 m s−1.]
Answers
a) wavelength of electromagnetic wave is \lambda=\frac{c}{\nu}, here, c is speed of light .e.g.,c = 3 × 10^8 m/s v is the frequency .e.g., v = 2 × 10^10 Hz , so, \lambda=3 × 10^8/2 × 10^10 = 1.5 × 10^-2 m = 1.5 cm . (b) amplitude of magnetic field is given by c=\frac{E_0}{B_0}\\B_0=\frac{E_0}{c}=\frac{48Vm^{1}}{3\times10^8m/s}\\B_0=1.6\times10^{-8}T (c) Energy density as electric field, U_E=\frac{1}{2}\epsilon_0 E^2 here, \frac{E}{B}=c so,U_E=\frac{1}{2}\epsilon_0c^2B^2 also we know, speed of electromagnetic wave, c=\frac{1}{\sqrt{\mu_0\epsilon_0}} so,U_E=\frac{1}{2}\frac{\epsilon_0}{\mu_0\epsilon_0}B^2=\frac{B^2}{2\mu_0}=U_B here, U_B is energy density as magnetic field.
The frequency of electromagnetic waves is determined by electric field.
In fact, this is similar to electric speed and it carried out by maximum speed of light.
Therefore, the wavelength of wave is given as 0.015m.
It is permitted to get squaring on both sides such as Ue=Ub.