Question

The amplitude of the magnetic field part of a harmonic electromagnetic wave in a vacuum is $B_{\circ }=510nT$. What is the amplitude of the electric field part of the wave?

• A. $140N{C}^{-1}$
• B. $153N{C}^{-1}$
• C. $163N{C}^{-1}$
• D. $133N{C}^{-1}$

Answer 1

The correct option is B

$153N{C}^{-1}$

Step1: Given data

The amplitude of the magnetic field = $B_o=510nT$

Speed of light=$c=3\times {10}^{8}m{s}^{-1}$

Step2: Formula used

We know that the magnitude of the electric field is given by the formula $E_{\circ }=B_{\circ }c$

Where $c=speedoflight,B_{\circ }=amplitudeofmagneticfield$.

Step3: Calculating the amplitude of the electric field

Substitute the known values in the formula $E_{\circ }=B_{\circ }c$,

$$\begin{gathered} E_{\circ }=510\times {10}^{-9}\times 3\times {10}^8 \\ {E}_{\circ }=153N{C}^{-1} \end{gathered}$$

So, the amplitude of the electric field part of the wave will be $153N{C}^{-1}$.

Hence, option B is the correct answer.