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Electromagnetic Waves — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsElectromagnetic Waves4 Marks
Question
In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0 \times 10^{10} Hz$ 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 $\left[ c =3 \times 10^8 m s ^{-1}\right]$.
✓

Answer

Given : $\quad v=2.0 \times 10^{10} Hz$
$E_0=48 V / m$
$\lambda=?$
$B _0=?$ and $c =3 \times 10^8 m / s$
(a) Wavelength, $\quad \lambda=\frac{c}{v}$
$=\frac{3 \times 10^8}{2 \times 10^{10}}=1.5 \times 10^{-2} m$
(b)$B_0=\frac{E_0}{c}$
On putting the values,
$B_0=\frac{48}{3 \times 10^8} T$
$B_0=16 \times 10^{-8} T=1.6 \times 10^{-7} T$
(c) Energy density in electric field $U_E=\frac{1}{2} \in_0 E^2$
Energy density in magnetic field $U_B=\frac{1}{2 \mu_0} B^2$
Energy density $\quad U _{ E }=\frac{1}{2} \in_0(c B)^2 \quad \because E =c B$
$U _{ E }=\frac{1}{2} \in_0 c^2 B^2=c^2\left(\frac{1}{2} \in_0 B^2\right)$
But $c=\frac{1}{\sqrt{\mu_0 \in_0}}$
$c^2=\frac{1}{\mu_0 \in_0}$
$\therefore \quad U _{ E }=\frac{1}{\mu_0 \in_0}\left(\frac{1}{2} \in_0 B^2\right)$
$=\frac{1}{2 \mu_0} B^2= U _{ B }$
$\therefore \quad U _{ E }= U _{ B }$
Hence the average energy density of the $\vec{E}$ field equals the averge energy density of the $\vec{B}$ field.

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