Question

An electromagnetic wave is traveling through vacuum in the positive x direction. Its electric field vector is given by E⃗ =E0sin(kx−ωt)j^,where j^ is the unit vector in the y direction.
What is the Poynting vector S⃗ (x,t), that is, the power per unit area associated with the electromagnetic wave described in the problem introduction?

Answer 1

Given that,

The electric field is given by,

$\vec{E}=E_{0}\sin(kx-\omega t)\hat{j}$

Suppose, B is the amplitude of magnetic field vector.

We need to find the complete expression for the magnetic field vector of the wave

Using formula of magnetic field

Direction of $(\vec{E}\times\vec{B})$ vector is the direction of propagation of the wave .

Direction of magnetic field = $\hat{j}$

$B=B_{0}\sin(kx-\omega t)\hat{k}$

We need to calculate the poynting vector

Using formula of poynting

$\vec{S}=\dfrac{E\times B}{\mu_{0}}$

Put the value into the formula

$\vec{S}=\dfrac{E_{0}\sin(kx-\omega t)\hat{j}\timesB_{0}\sin(kx-\omega t)\hat{k}}{\mu_{0}}$

$\vec{S}=\dfrac{E_{0}B_{0}}{\mu_{0}}(\sin^2(kx-\omega t))\hat{i}$

Hence, The poynting vector is $\dfrac{E_{0}B_{0}}{\mu_{0}}(\sin^2(kx-\omega t))\hat{i}$