Question

The magnetic field in a plane electromagnetic wave is given by
B = (200 µT) sin [(4.0 × 1015s−1)(t−x/c)].
Find the maximum electric field and the average energy density corresponding to the electric field.

Answer 1

The maximum electric field and the average energy density corresponding to the electric field are  $6 \times 10^{4} N C^{-1}$ and  $0.016 \frac{J}{m^{3}}$

Explanation:

Step 1:

Given data in the question  :

Total Magnetic Field Value, $B_0$ = 200 μT

An electromagnetic wave has a speed of c.

Step 2:

So, total electric field value

$\begin{aligned}&E_{0}=c B_{0}\\&E_{0}=c \times B_{0}=200 \times 10^{-6} \times 3 \times 10^{8}\end{aligned}$

$E_{0}=6 \times 10^{4} N C^{-1}$

Step 3:

Average Magnetic Field Energy Density,

$U_{a v}=\frac{1}{2 \mu_{0}} B_{0}^{2}=\frac{\left(200 \times 10^{-6}\right)^{2}}{\left(2 \times 4 \pi \times 10^{-7}\right)}$

$U_{a v}=\frac{4 \times 10^{-8}}{8 \pi \times 10^{-7}}=\frac{1}{20 \pi}$

$U_{a v}=0.0159 \approx 0.016 \frac{J}{m^{3}}$

For an electromagnetic wave, both the electrical and magnetic fields exchange energy equally.

Therefore the electric field energy density will be equal to the magnetic field energy density.