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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    Answered by

    Payal Gupta | Contributor-Level 10

    3 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 = c ν = 3 × 10 8 2.0 × 10 10 = 0.015 m

    Magnetic field strength is given by

    B 0 = E 0 c = 48 3 × 10 8 = 1.6 10-12 T

    Energy density of the electric field is given as,

    U E 1 2 ? 0 E 2 and energy density of magnetic field is given by,

    U B = 1 2 B 2 μ 0

    w h e r e

    E 0  = Permittivity of free space

    μ 0 =  Permeability of free space

    We have the relation connecting E and B as:

    E = cB, where c = 1 ? 0 μ 0

    Hence E = B ? 0 μ 0 i ?

    E 2  = B 2 ? 0 μ 0

    ? 0 E 2 = B 2 μ 0

    2 U E = 2 U B  or U E = U B

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I = P/ (4πr²) . (i)
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