Question

In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of $2.0\times {{10}^{10}}Hz$ and Amplitude $48V{{m}^{-1}}$ .
(a) What is the wavelength of the wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average density of the E field equals the average density of the B field.
$\left[ c=3\times {{10}^{8}}m{{s}^{-1}} \right]$

Answer 1

We know for an electromagnetic wave moving in space, the wavelength of the wave is given as speed of wave upon its frequency. In the second part, for the amplitude of the oscillating magnetic field, we know that the ratio of Amplitude of electric field to the Amplitude of magnetic field is given by the speed of wave. Here, the speed of the wave is equal to the speed of light.

Complete step-by-step answer:
(a) In the first part of the question, in order to find the wavelength of the electromagnetic wave, we use the formula written below:
$\Rightarrow \lambda =\dfrac{c}{f}$
Where,
$c$is the speed of wave whose value is equal to $3\times {{10}^{8}}m{{s}^{-1}}$ .
And, $f$ is the frequency of the electromagnetic wave whose value is given as: $2.0\times {{10}^{10}}Hz$.
Putting these values in the above equation, we get:
$\begin{aligned} & \Rightarrow \lambda =\dfrac{3\times {{10}^{8}}}{2\times {{10}^{10}}}m \\\ & \Rightarrow \lambda =1.5\times {{10}^{-2}}m \\\ \end{aligned}$
Hence, the wavelength of electromagnetic wave comes out to be $1.5\times {{10}^{-2}}m$.
(b) The Amplitude of the magnetic field can be calculated using the formula:
$\Rightarrow \dfrac{{{E}_{0}}}{{{B}_{0}}}=c$
Where,
${{E}_{0}}$ is the amplitude of electric field and its value is given to be $48V{{m}^{-1}}$.
${{B}_{0}}$ is the amplitude of magnetic field.
Putting these values in the above equation, we get:
$\begin{aligned} & \Rightarrow {{B}_{0}}=\dfrac{{{E}_{0}}}{c} \\\ & \Rightarrow {{B}_{0}}=\dfrac{48}{3\times {{10}^{8}}}T \\\ & \Rightarrow {{B}_{0}}=1.6\times {{10}^{-7}}T \\\ \end{aligned}$
Hence, the Amplitude of oscillating magnetic field is found to be $1.6\times {{10}^{-7}}T$ .
(c)
In the last part of the question, we have to proof that the average energy density of electric field is equal to the average energy field density of magnetic field.
So, we have:
The formula for average density of electric field is: $\dfrac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}$
And, the formula for average density of magnetic field is: $\dfrac{{{B}^{2}}}{2{{\mu }_{0}}}$
Where,
${{\varepsilon }_{0}}$ is the permittivity of free space, and
${{\mu }_{0}}$ is the permeability of free space
Now, we have the relation between E and B as:
$\Rightarrow E=cB$
And using the formula,
$\Rightarrow c=\dfrac{1}{\sqrt{{{\varepsilon }_{0}}{{\mu }_{0}}}}$
We can simplify the relation between E and B as:
$\Rightarrow E=\dfrac{B}{\sqrt{{{\varepsilon }_{0}}{{\mu }_{0}}}}$
On squaring the terms and multiplying them by $\dfrac{1}{2}$ , we get:
$\begin{aligned} & \Rightarrow \dfrac{{{E}^{2}}}{2}=\dfrac{{{B}^{2}}}{2{{\varepsilon }_{0}}{{\mu }_{0}}} \\\ & \Rightarrow \dfrac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}=\dfrac{{{B}^{2}}}{2{{\mu }_{0}}} \\\ \end{aligned}$
Hence, it is proved that the average density of electric field and magnetic field is equal.

Note: We used the basic properties of an electromagnetic wave and used them to solve the above problem. As it was greatly a numerical based problem, one should be very careful during calculations. Also, the unit used for amplitude of oscillating magnetic fields is Tesla.