Question

What is the magnetic field amplitude of an electromagnetic wave whose electric field amplitude is $10\mathrm{V}/\mathrm{m}$ $?$

Short answer

The magnetic field amplitude of an electromagnetic wave is$3.33\times {10}^{-8}\mathrm{T}$.

Step-by-step solution

Step 1: 

As it stand's,

$\mathrm{E}=10\mathrm{V}/\mathrm{m}$

$\mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s}$

Electric field amplitude,

$\mathrm{E}=\mathrm{c}·\mathrm{B}$

$\mathrm{B}$-Magnitude field of amplitude.

localid="1648801389365" $\mathrm{c}$-In a vacuum, the speed of light

Calculation of magnetic field amplitude

Electric field amplitude is,

$\mathrm{E}=\mathrm{c}·\mathrm{B}$

from the formula,

both to be divided by$\mathrm{c}$,

$\frac{\mathrm{E}}{\mathrm{c}}=\frac{\mathrm{c}·\mathrm{B}}{\mathrm{c}}$

$\mathrm{B}=\frac{\mathrm{E}}{\mathrm{c}}$

$\mathrm{B}=\frac{10\mathrm{V}/\mathrm{m}}{3\times {10}^8\mathrm{m}/\mathrm{s}}$

$\mathrm{B}=3.33\times {10}^{-8}\mathrm{T}$