Question

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0 × 10^10 \ Hz$ and amplitude $48\ V m^{−1}$.

1. What is the wavelength of the wave?
2. What is the amplitude of the oscillating magnetic field?
3. Show that the average energy density of the E field equals the average energy density of the B field. $[c = 3 × 10^8 m s^{−1}]$

Answer 1

Frequency of the electromagnetic wave, $ν = 2.0 × 10^{10} Hz$
Electric field amplitude, $E_0 = 48\ V m^{−1 }$
Speed of light, $c = 3 × 10^8 \ m/s$

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$(a)$ Wavelength of a wave is given as:

$λ = \frac {c}{v}$

$λ =\frac { 3\times 10^8}{2\times 10^{10}}$

$λ = 0.015\ m$

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$(b)$ Magnetic field strength is given as:

$B_o =\frac { E_o}{c}$

$B_o = \frac{48}{3\times 10^8}$

$B_0 = 1.6\times 10^{-7} T$

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$(c)$Energy density of the electric field is given as:

$U_E = \frac {1}{2} ε_oE^2$

And, energy density of the magnetic field is given as:

$U_B = \frac {1}{2 µ_0} B^2$

Where,
$ε_0$ = Permittivity of free space
$µ_0$ = Permeability of free space
We have the relation connecting E and B as:
E = cB .… (1)
Where,

$c = \frac {1}{\sqrt {ε_0µ_0}}$ .....(2)

Putting equation (2) in equation (1), we get

$E = \frac {1}{\sqrt {ε_0µ_0}} B$

Squaring both sides, we get

$E^2 = \frac {1}{ {ε_0µ_0}} B^2$

$ε_0E^2 = \frac {B^2}{µ_0}$

$\frac 12 ε_0E^2 = \frac 12 \frac {B^2}{µ_0}$

$⇒ U_E = U_B$