Question

What is the ratio of amplitude of electric field and magnetic field in EM wave?

Answer 1

Hint: Consider an EM wave propagating in z-direction. Write an equation for electric and magnetic fields in terms of direction of propagation and time. Then use Ampere-Maxwell law for EM waves to determine ratio of amplitude of electric field and magnetic field in EM wave.
Formula Used:
Ampere-Maxwell’s law for electromagnetic waves, $\nabla \times \,\vec{B}=\dfrac{1}{{{c}^{2}}}\dfrac{\partial E}{\partial t}$

Complete answer:
Let us consider a plane electromagnetic wave travelling along the z-direction. Therefore, the electric and magnetic fields are functions of z-coordinate at a given time t. Electric and magnetic fields in electromagnetic waves are always perpendicular to each other as well as to the direction of propagation of the EM wave.
The electric field $E$ is along the x-axis and varies sinusoidally with z, at a given time. The magnetic field $B$ is along the y-axis and it also varies sinusoidally with z. We can write $E$ and $B$ as follows:
$E={{E}_{0}}\sin (kz-\omega t)$
$B={{B}_{0}}\sin (kz-\omega t)$
Here k is the magnitude of the propagation vector, $\vec{k}$ and its direction gives direction of propagation of the wave. k is given by
$k=\dfrac{2\pi }{\lambda }$
$\omega$ is the angular frequency of the wave.
Now,
$\dfrac{\partial E}{\partial t}=\omega {{E}_{0}}\cos (kz-\omega t)$
$\nabla \times \vec{B}=k{{B}_{0}}\cos (kz-\omega t)$
Using Ampere-Maxwell’s law for electromagnetic waves $\nabla \times \,\vec{B}=\dfrac{1}{{{c}^{2}}}\dfrac{\partial E}{\partial t}$, we get
$k{{B}_{0}}\cos (kz-\omega t)={{\mu }_{0}}\vec{J}+\dfrac{1}{{{c}^{2}}}\omega {{E}_{0}}\cos (kz-\omega t)\hat{i}$
$k{{B}_{0}}=\dfrac{\omega {{E}_{0}}}{{{c}^{2}}}$
$\dfrac{{{c}^{2}}k}{\omega }=\dfrac{{{E}_{0}}}{{{B}_{0}}}$
The velocity of a wave is given by $v=\dfrac{\omega }{k}$ and we know that velocity of EM waves is equal to velocity of light. Therefore,
$\dfrac{\omega }{k}=c$
Substituting this value in previous equation we get,
$\dfrac{{{E}_{0}}}{{{B}_{0}}}=c$
Hence, we conclude that the ratio of electric field and magnetic field in the EM wave is equal to speed of light in free space.

Note:
In an EM wave, the magnetic energy density is equal to the electrical energy density. EM waves do not require any medium to propagate which makes it different from other waves. The magnitude of velocity of propagation of an EM wave is equal to that of speed of light.