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

\(E^2 = \frac {1}{ {ε_0µ_0}} B^2\)

\(ε_0E^2 = \frac {B^2}{µ_0}\)

\(\frac 12 ε_0E^2 = \frac 12 \frac {B^2}{µ_0}\)

\(⇒ U_E = U_B\)