Question 1 of 8intermediate🔧 ApplyNumerical5 marks

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0×10102.0 \times 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 = 3×1083 \times 108 m s–1].

(a)
Answer

The wavelength of the wave is 1.5×101.5 \times 10⁻² m or 1.5 cm.

(b)
Answer

The amplitude of the oscillating magnetic field is 1.6×101.6 \times 10⁻⁷ T or 160 nT.

(c)
Answer

The average energy density of the E field equals the average energy density of the B field, as shown in the derivation below.

Explanation

This question tests understanding of electromagnetic wave properties including wavelength calculation, relationship between electric and magnetic field amplitudes, and energy density distribution in EM waves. The solution uses fundamental relationships: λ = c/ν for wavelength, B₀ = E₀/c for magnetic field amplitude, and energy density formulas for electric and magnetic fields.

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