A plane-polarized electromagnetic wave in vacuum has magnetic field $B_x(y,t)=B_m\cos\left[2\pi\left(\frac{y}{\lambda}-\frac{t}{T}\right)\right]$, $B_y=0$, $B_z=0$. (a) What direction does the wave travel? (b) What is the polarization direction? (c) Find $E_x(y,t)$, $E_y(y,t)$, and $E_z(y,t)$ in terms of $B_m$, $\lambda$, $T$, and $c$.

Reading the wave from its phase

A traveling wave has a phase like $ky-\omega t$. If you keep the phase constant and see which way you must move in space as time increases, you get the propagation direction.

(a) Direction of propagation

The given phase is $$2\pi\left(\frac{y}{\lambda}-\frac{t}{T}\right)=\frac{2\pi}{\lambda}y-\frac{2\pi}{T}t=ky-\omega t.$$ A surface of constant phase satisfies $y-(\lambda/T)t=\text{constant}$, which moves toward increasing $y$ as $t$ increases. So the wave travels in the $+y$ direction.

(b) What “polarization direction” means here

For a plane-polarized EM wave, the polarization direction refers to the direction of the electric field $\mathbf{E}$. In vacuum, $\mathbf{E}\perp\mathbf{B}$ and both are perpendicular to the propagation direction.

(c) Finding $\mathbf{E}$ from $\mathbf{B}$

Here $\mathbf{B}$ points along $+x$ (since only $B_x$ is nonzero) and the wave travels along $+y$. The only axis perpendicular to both $x$ and $y$ is $z$, so $\mathbf{E}$ must be along $\pm z$.

To get the sign, use the right-hand rule for the Poynting direction: $\mathbf{S}\propto \mathbf{E}\times\mathbf{B}$ must point along $+y$. Since $\hat{z}\times\hat{x}=\hat{y}$, $\mathbf{E}$ must be along $+z$.

In vacuum the magnitudes satisfy $E=cB$ and the fields are in phase, so $$E_z(y, t)=c\, B_x(y, t)=cB_m\cos\left[2\pi\left(\frac{y}{\lambda}-\frac{t}{T}\right)\right],$$ with $E_x=0$ and $E_y=0.