Question

A) In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0\times {10}^{10}\text{Hz}$ and amplitude $48\text{V/m}$. What is the wavelength of the wave? $[c=3\times {10}^{8}m/s.]$

B) In a plane electromagnetic wave, the electric field oscillatessinusoidally at a frequency of $2.0\times {10}^{10}$ and amplitude $48V/m$. What is the amplitude of the oscillating magnetic field? $[c=3\times {10}^{8}m/s.]$

C) In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0\times {10}^{10}\text{Hz}$ and amplitude $48V/m$. Show that the average energy density of the E field equals the average energy density of the B field. $[c=3\times {10}^{8}m/s]$.

Answer 1

A) Formula used: $\lambda =\frac{c}{v}$
Given, Frequency of the electromagnetic wave, $v=2\times {10}^{10}\text{Hz}$
Wavelength of the wave, $\lambda =\frac{c}{v}$
Here,c = speed of light $=3\times {10}^{8}m/s$
$\lambda =\frac{3\times {10}^8}{2\times {10}^{10}}$
$\lambda =1.5\times {10}^{-2}m$
Final answer: $1.5\times {10}^{-2}m$

B) Formula used: $B_0=\frac{E_0}{c}$
Given, Amplitude of electric field,
$E_0=48V/m$
Magnitude of amplitude of magnetic field strength, $B_0=\frac{E_0}{c}$
Here, c = speed of light $=3\times {10}^{8}m/s$
$\Rightarrow B_0=\frac{48}{3\times {10}^8}=1.6\times {10}^{-7}T$
Final answer : $1.6\times {10}^{-7}T$

C) Energy density of the electric field, $U_E=\frac{1}{2}{ϵ}_0{E}^2$
And, energy density of the magnetic field, $U_B=\frac{1}{2{\mu }_0}{B}^2$
Where,
${ϵ}_0$ = Permitivity of free space
${\mu }_0$ = Permeability of free space
Also, we have the relation between E and B
$E=cB....\left(1\right)$
Here, c = speed of light
$c=\frac{1}{\sqrt{{ϵ}_0{\mu }_0}}....\left(2\right)$
Putting the value of 𝑐 from equation (2) in equation (1), we get,
$E=\frac{B}{\sqrt{{ϵ}_0{\mu }_0}}$
Squaring both sides, we get,
$E^2=\frac{B^2}{{ϵ}_0{\mu }_0}$
${ϵ}_0{E}^2=\frac{B^2}{{\mu }_0}$
$\frac{1}{2}{ϵ}_0{E}^2=\frac{1B^2}{2{\mu }_0}$
$U_E=U_B$
Hence, the average energy density of the E field equals the average energy density of the B field.
Final answer: $U_E=U_B$