Animated Solution for Physics - Electromagnetic Waves: The electric field of a plane electromagnetic wave is given by $\mathbf{E}=E_0(\hat{\mathbf{x}}+\hat{\mathbf{y}})\sin (kz-\omega t)$ Its magnetic field will be given by

Imagine you are standing in a three-dimensional space, watching an electromagnetic wave ripple past you. The first thing we need to figure out is where this wave is going. We look at the phase of the wave, which is given by the term $(kz-\omega t)$.

This specific mathematical form is a dead giveaway. Because it is a minus sign between the spatial part $kz$ and the temporal part $\omega t$, it tells us that the wave is propagating along the positive $z$-axis. If it were $(kz+\omega t)$, it would be moving in the negative $z$-direction. So, our propagation vector is simply $\hat{\mathbf{c}}=\hat{\mathbf{z}}$.

Next, we examine the electric field vector itself. The equation tells us that $\mathbf{E}=E_0(\hat{\mathbf{x}}+\hat{\mathbf{y}})\sin (kz-\omega t)$. The direction of this electric field is along the vector $(\hat{\mathbf{x}}+\hat{\mathbf{y}})$.

If you visualize this, the electric field is oscillating in the $xy$-plane, pointing exactly diagonally between the positive $x$ and positive $y$ axes. To be mathematically precise, the unit vector for the electric field is $\hat{\mathbf{E}}=\frac{\hat{\mathbf{x}}+\hat{\mathbf{y}}}{\sqrt{2}}$.

Now, we bring in the fundamental property of electromagnetic waves in a vacuum. The electric field $\mathbf{E}$, the magnetic field $\mathbf{B}$, and the direction of propagation $\hat{\mathbf{c}}$ are all mutually perpendicular to each other.

They don't just sit at right angles randomly; they follow a strict right-hand rule. Mathematically, this is expressed by the cross product:

This equation is our master key. We already know $\hat{\mathbf{E}}$ and we know $\hat{\mathbf{c}}$. We just need to find the correct $\hat{\mathbf{B}}$ that satisfies this relationship. Let's substitute what we know:

Instead of guessing, let's systematically test the direction given in option (b), which suggests the magnetic field points along $(-\hat{\mathbf{x}}+\hat{\mathbf{y}})$. The unit vector for this direction would be $\hat{\mathbf{B}}=\frac{-\hat{\mathbf{x}}+\hat{\mathbf{y}}}{\sqrt{2}}$.

Let's plug this into our cross product and see what happens:

When we expand this cross product, we distribute the terms just like standard algebra, but keeping the cross product rules in mind:

We know that the cross product of any vector with itself is zero, so $\hat{\mathbf{x}}\times \hat{\mathbf{x}}=0$ and $\hat{\mathbf{y}}\times \hat{\mathbf{y}}=0$. We also know from the right-hand rule that $\hat{\mathbf{x}}\times \hat{\mathbf{y}}=\hat{\mathbf{z}}$ and $\hat{\mathbf{y}}\times \hat{\mathbf{x}}=-\hat{\mathbf{z}}$. Substituting these in:

Boom! The result is exactly $\hat{\mathbf{z}}$, which perfectly matches our propagation direction. This confirms that the magnetic field must point along $(-\hat{\mathbf{x}}+\hat{\mathbf{y}})$.

We have the direction, but what about the magnitude? In an electromagnetic wave, the amplitude of the magnetic field $B_0$ is related to the amplitude of the electric field $E_0$ by the speed of light $c$:

Since the electric field expression has a coefficient of $E_0$, the magnetic field expression will simply have a coefficient of $\frac{E_0}{c}$. Furthermore, the magnetic field is always in phase with the electric field in a vacuum, so it will share the exact same $\sin (kz-\omega t)$ term.

Combining the magnitude, the verified direction, and the phase, we arrive at our final, elegant equation for the magnetic field:

This perfectly matches option (b). The dance of the fields is complete!