Question

The oscillating magnetic field in a plane electromagnetic wave is given by ${B_z} = (8 \times {10^{ - 6}})\sin [2 \times {10^{11}}t + 300\pi x]T$. Calculate:
(A) Wavelength and frequency of the wave.
(B) Write down the expression for the oscillating electric field.

Answer 1

Hint: Electromagnetic waves are produced due to an accelerated charge. Electromagnetic waves consist of oscillating electric field component and magnetic field component and they are mutually perpendicular to each other and perpendicular to the direction of propagation of the wave.

Formula Used:
General expression for magnetic field is
${B_z} = {B_0}\sin (\omega t + kx)$
Where,
$k = \dfrac{{2\pi }}{\lambda } \\
\Rightarrow \dfrac{{{E_0}}}{{{B_0}}} = c \\
\Rightarrow {E_y} = {E_0}\sin (\omega t + kx)V{m^{ - 1}}$

Complete step by step answer:
We have given the expression for magnetic field as ${B_z} = {B_0}\sin (\omega t + kx)$ $ \to (i)$ which means magnetic field is in $z$ direction and wave is travelling in $x$ direction then electric field will be in
$y$ Direction.
Compare equation $(i)$ with the given magnetic field expression ${B_z} = (8 \times {10^{ - 6}})\sin [2 \times {10^{11}}t + 300\pi x]T$
We get,
${B_0} = 8 \times {10^{ - 6}}T \\
\Rightarrow \omega = 2 \times {10^{11}}radian{\sec ^{ - 1}} \\
\Rightarrow k = 300\pi {m^{ - 1}} \\ $
Now putting value of $k$ in $\lambda = \dfrac{{2\pi }}{k}$ us get,
$\lambda = \dfrac{2}{{300}} \\
\therefore \lambda = 0.0067\,m$
$\omega $ Is the frequency of the electromagnetic wave, so frequency of the wave is $\omega = 2 \times {10^{11}}\,rad{s ^{ - 1}}$.

Hence, wavelength and frequency of wave are $0.0067\,m$ and $\omega = 2 \times {10^{11}}rad{s ^{ - 1}}$.

(B) In an electromagnetic wave, the ratio of electric amplitude and magnetic amplitude is always constant which equals the velocity of light in free space.So, we have
$\dfrac{{{E_0}}}{{{B_0}}} = c \\
\Rightarrow {E_0} = 3 \times {10^8} \times 8 \times {10^{ - 6}} \\
\Rightarrow {E_0} = 2400V{m^{ - 1}} \\ $
Hence, expression for electric field can be written as
${E_y} = {E_0}\sin (\omega t + kx)V{m^{ - 1}} \\
\therefore {E_y} = 2400\sin (\omega t + kx)V{m^{ - 1}} \\ $
Hence, the expression for the oscillating electric field is given by ${E_y} = 2400\sin (\omega t + kx)V{m^{ - 1}}$.

Note:Electromagnetic waves are transverse in nature because they propagate with varying electric and magnetic fields such that two fields are mutually perpendicular to each other and perpendicular to the direction of propagation. These waves don’t require any medium to travel for their propagation.