Question

The amplitude of the magnetic field part of a harmonic electromagnetic wave in vacuum is $B_0=510nT$. What is the amplitude of the electric field part of the wave?

Answer 1

Formula used: $E_0=c{B}_0$
Given, Amplitude of magnetic field of an electromagnetic wave in vacuum, $B_0=510nT=510\times {10}^{-9}T$
Amplitude of electric field of the electromagnetic wave, $E_0=c{B}_0$
Here, 𝑐=speed of light $=3\times {10}^{8}m/s$
$\Rightarrow E_0=3\times {10}^8\times 510\times {10}^{-9}$
$\Rightarrow E_0=153N/C$
Therefore, the amplitude of the electric field part of the wave is $153N/C$.
Final answer: $153N/C$.