Electromagnetic waves: Maxwell's equations

Question:
The electric field of an electromagnetic wave is given as
\[
\vec{E}=(E_{0x}\vec{x} + E_{0y}\vec{y})\sin(\omega t - kz + \phi)
\]
Knowing that this is traveling along the z-axis. Find the magnetic field from this.

Answer:
From Maxwell's equations, we have
\[
\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}
\]
This is known as Faraday's law. Calculate the left hand side.
\begin{eqnarray}
\nabla \times \vec{E} &=& \left(\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}\right)\vec{x}+\left(\frac{\partial E_x}{\partial z}-\frac{\partial E_z}{\partial x}\right)\vec{y} \\
&=& E_{0y}k\cos(\omega t-kz+\phi)\vec{x} - E_{0x}k\cos(\omega t-kz+\phi)\vec{y}
\end{eqnarray}
This is equal to $-\frac{\partial \vec{B}}{\partial t}$. Thus, integrate both sides with respect to time to find the magnetic field.
\begin{eqnarray}
\vec{B} &=& -\int (\nabla \times \vec{E}) dt  \\
  &=& -E_{0y}\frac{k}{\omega}\sin(\omega t-kz+\phi)\vec{x} + E_{0x}\frac{k}{\omega}\sin(\omega t-kz+\phi)\vec{y}  \\
  &=& (-E_{0y}\vec{x}+E_{0x}\vec{y})\frac{1}{c}\sin(\omega t - kz + \phi)
\end{eqnarray}
Note that $\omega=2\pi f$ and $k=\frac{2\pi}{\lambda}$. Therefore, $\frac{k}{\omega}=\frac{1}{c}$ since $c=\lambda f $. The magnetic field propagates perpendicularly to the electric field with a shift of $\pi$ as you can see from the equations.

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