Physics Ncert Solutions Class 12th Electromagnetic Waves Class 12th Physics

**8.10 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.]

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Payal Gupta | Contributor-Level 10

7 months ago

8.10 Frequency of the electromagnetic wave, $ν$ = 2.0 $* 10^{10}$ Hz

Electric field amplitude, $E_{0}$ = 48 V/m

Speed of light, c = 3 $* 10^{8}$ m/s

Wavelength of a wave is given by,

$λ = \frac{c}{ν}$ = $\frac{3 * 10^{8}}{2.0 * 10^{10}}$ = 0.015 m

Magnetic field strength is given by

$B_{0} =$ $\frac{E_{0}}{c}$ = $\frac{48}{3 * 10^{8}}$ = 1.6 $10^{- 12}$ T

Energy density of the electric field is given as,

$U_{E}$ = $\frac{1}{2} ?_{0} E^{2}$ and energy density of magnetic

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42.9 Ans

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s? = -? + k? ∴ s? = (-? +k? )/√2

A point source surrounded by vacuum emits an electromagnetic wave of frequency of 900 kHz. If the power emitted by the source is 15 W, the amplitude of the electric field of the wave (in 10^-3V/m) at a distance 15 km from the source is :
( (1 / 4πε₀) = 9×10⁹ Nm²/C² and speed of light in vacuum = 3 × 10⁸ ms^-1)

I = P/ (4πr²) . (i)
and I = (1/2)ε? E? ²c . (ii)
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P/ (4πr²) = (1/2)ε? E? ²c
E? = √ (2P/ (4πε? r²c) ; E? = √ (9×10? ×2×15)/ (15×15×10? ×3×10? ) = 2×10? ³ V/m

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Physics Ncert Solutions Class 12th 2023

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