Question

Suppose that the electromagnetic wave in vacuum is
$\overrightarrow{E}=\{\left(3.1\frac{N}{C}\right)\cos \left[(1.8\frac{rad}{m}\right)y+(5.4\times {10}^6\frac{rad}{s})t]\hat{i}$
What is the amplitude of the magnetic field part of the wave?

• A. $1.03\times {10}^{-8}$T
• B. $2.03\times {10}^{-5}$ T
• C. $1.03\times {10}^{-5}$ T
• D. $2.03\times {10}^{-3}$ T

Answer 1

The correct option is A $1.03\times {10}^{-8}$T

Comparing the given equation with general form of wave
$E=E_0\cos \left(ky-\omega t\right)$
we get
$E_0=3.1$ units, which is the amplitude of electric field
So amplitude of electric field is given by
$B_0=\frac{E_0}{c}$
where
$c$ is the speed of light
$B_0=\frac{3.1}{3\times {10}^8}=1.03\times {10}^{-8}\text{T}$