Animated Solution for Physics - Electromagnetic Waves: In an electromagnetic wave, the electric field vector and magnetic field vector are given as $\mathbf{E}=E_0\hat{\mathbf{i}}$ and $\mathbf{B}=B_0\hat{\mathbf{k}}$, respectively. The direction of propagation of electromagnetic wave is along

Imagine you are standing in a completely empty void, and suddenly, a beam of light shoots past you. What exactly is that beam made of? According to James Clerk Maxwell's brilliant synthesis of electromagnetism, light is an electromagnetic wave—a self-sustaining dance of oscillating electric and magnetic fields.

These fields don't just oscillate randomly; they follow a very strict, highly choreographed geometric relationship. The electric field $\mathbf{E}$ and the magnetic field $\mathbf{B}$ are always perfectly perpendicular to each other. But more importantly, they are both perpendicular to the direction in which the wave is traveling. This makes electromagnetic waves transverse waves.

The Compass of Energy

To find out exactly which way the wave is moving, physicists use a mathematical tool called the Poynting vector, denoted by $\mathbf{S}$. The Poynting vector represents the directional energy flux (the rate of energy transfer per unit area) of an electromagnetic field.

The formula for the Poynting vector in a vacuum is:

Notice the cross product $\mathbf{E}\times \mathbf{B}$. Because ${\mu }_0$ (the permeability of free space) is just a positive scalar constant, the direction of the wave's propagation is entirely dictated by the direction of the cross product $\mathbf{E}\times \mathbf{B}$. This is the master key to solving our problem.

Let's look at the specific vectors given in our problem. We are told that the electric field vector is pointing along the positive x-axis:

Here, $E_0$ is the amplitude (a positive number), and $\hat{\mathbf{i}}$ is the unit vector in the x-direction.

Similarly, the magnetic field vector is pointing along the positive z-axis:

Where $B_0$ is the amplitude, and $\hat{\mathbf{k}}$ is the unit vector in the z-direction.

To find the direction of propagation, we need to evaluate the cross product of these two vectors:

Since $E_0$ and $B_0$ are just scalar magnitudes, we can pull them out to the front:

The entire problem now boils down to evaluating the cross product of the unit vectors: $\hat{\mathbf{i}}\times \hat{\mathbf{k}}$.

How do we do this? We use the Right-Hand Rule.

1. Point the fingers of your right hand in the direction of the first vector, $\hat{\mathbf{i}}$ (the positive x-axis). 2. Curl your fingers towards the direction of the second vector, $\hat{\mathbf{k}}$ (the positive z-axis). 3. Look at your thumb. It is pointing straight down, along the negative y-axis!

Therefore, mathematically:

If you don't want to contort your hand during an exam, there is a foolproof mental trick you can use: the Cyclic Permutation Circle.

Draw a circle and place $\hat{\mathbf{i}}$, $\hat{\mathbf{j}}$, and $\hat{\mathbf{k}}$ evenly spaced around it in a clockwise direction.

- If you multiply two vectors in the clockwise direction, the result is positive. For example, $\hat{\mathbf{i}}\times \hat{\mathbf{j}}=\hat{\mathbf{k}}$, and $\hat{\mathbf{j}}\times \hat{\mathbf{k}}=\hat{\mathbf{i}}$. - If you multiply two vectors in the counter-clockwise direction, the result is negative.

In our case, we need $\hat{\mathbf{i}}\times \hat{\mathbf{k}}$. Moving from $\hat{\mathbf{i}}$ to $\hat{\mathbf{k}}$ on our imaginary circle goes against the clockwise flow (it's counter-clockwise). Therefore, the result must be negative, giving us $-\hat{\mathbf{j}}$.

Substituting this back into our equation, we get:

Since $E_0{B}_0$ is a positive magnitude, the wave is propagating purely in the $-\hat{\mathbf{j}}$ direction. The electromagnetic wave is traveling along the negative y-axis.

This elegant geometric relationship is a cornerstone of classical electrodynamics, ensuring that light always travels exactly as Maxwell's equations demand.