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]

Solution

Here, $Freque n cy of the plane EM wave, v = 2 \times 10^{10} Hz Amplitude of electric field, E_{0} = 48 {Vm}^{- 1}$

Therefore,

Wavelength of the wave,

$λ = \frac{c}{v} = \frac{3 \times 10^{8}}{2 \times 10^{10}} m = 1.5 \times 10^{- 2} m$

Amplitude of oscillating magnetic field,

$B_{0} = \frac{E_{0}}{c} = \frac{48}{3 \times 10^{8}} T = 1.6 \times 10^{- 7} T$

Energy density in electric field,

$μ_{E} = \frac{1}{2} ε_{0} E^{2}$

Energy density in magnetic field,
$μ_{B} = \frac{1}{2 μ_{0}} B^{2}$

Using the relation, we have

$E = cB, u_{E} = \frac{1}{2} ε_{0} (cB)^{2} = c^{2} (\frac{1}{2} ε_{0} B^{2})$

But, $c = \frac{1}{\sqrt{μ_{0} ε_{0}}}$

$∴$ $μ_{E} = \frac{1}{μ_{0} ε_{0}} (\frac{1}{2} ε_{0} B^{2}) = \frac{1}{2 μ_{0}} B^{2} = μ_{B}$
Hence, the average energy density of electric field equals the average energy density of magnetic field.

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