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.

\[  \nabla \times \vec{E} = \left(\frac{\partial E_z}{\partial y}-\frac{\partial E_y}{\partial z}\right)\vec{x}+\left(\frac{\partial E_x}{\partial z}-\frac{\partial E_z}{\partial x}\right)\vec{y}   \]

\[  = E_{0y}k\cos(\omega t-kz+\phi)\vec{x} – E_{0x}k\cos(\omega t-kz+\phi)\vec{y}  \]

This is equal to \(-\frac{\partial \vec{B}}{\partial t}\). Thus, integrate both sides with respect to time to find the magnetic field.

\[  \vec{B} = -\int (\nabla \times \vec{E}) dt  \]

\[  = -E_{0y}\frac{k}{\omega}\sin(\omega t-kz+\phi)\vec{x} + E_{0x}\frac{k}{\omega}\sin(\omega t-kz+\phi)\vec{y}  \]

\[  = (-E_{0y}\vec{x}+E_{0x}\vec{y})\frac{1}{c}\sin(\omega t – kz + \phi)  \]

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.