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Question 8.10: In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 10 10 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 × 10 8 m s −1 .]Class 12 - Physics - Electromagnetic Waves Electromagnetic Waves Page-286

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(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.


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