Animated Solution for Physics - Electromagnetic Waves: An electromagnetic wave is represented by the electric field $\mathbf{E}=E_0\hat{\mathbf{n}}\sin [\omega t+(6y-8z)]$. Taking unit vectors in $x$, $y$ and $z$-directions to be $\hat{\mathbf{i}},\hat{\mathbf{j}},\hat{\mathbf{k}}$, the direction of propagation $\hat{\mathbf{s}}$, is

The propagation of an electromagnetic wave is entirely encoded within its phase. When we look at the equation of a wave, the argument of the sine or cosine function tells us not just how fast it oscillates, but exactly where it is going.

Let's start by recalling the standard equation of a plane electromagnetic wave. The electric field vector is given by:

Here, $E_0$ is the amplitude, $\hat{\mathbf{n}}$ is the unit vector in the direction of polarization, $\omega$ is the angular frequency, and most importantly, $\mathbf{k}$ is the propagation vector (or wave vector). The direction of $\mathbf{k}$ is the direction in which the wave travels.

Now, let's look at the equation given in our problem:

Notice the plus sign inside the argument? To properly compare this with our standard form, we need the term after $\omega t$ to be subtracted. We can achieve this by factoring out a minus sign from the spatial part of the phase:

This small rearrangement is crucial. It aligns our given equation perfectly with the standard form, allowing us to extract the wave vector without any sign errors.

By comparing our rearranged equation with the standard form, we can directly equate the spatial terms:

We know that the general position vector $\mathbf{r}$ in 3D space is:

And the wave vector $\mathbf{k}$ can be written in terms of its components:

So, their dot product is simply:

Comparing this with our extracted term $-6y+8z$, we can match the coefficients: - $k_x=0$ - $k_y=-6$ - $k_z=8$

Therefore, our propagation vector $\mathbf{k}$ is:

The problem asks for the direction of propagation, which is represented by the unit vector $\hat{\mathbf{s}}$. To find a unit vector, we simply divide the vector by its magnitude.

First, let's calculate the magnitude of $\mathbf{k}$:

Now, we divide the vector $\mathbf{k}$ by its magnitude to get $\hat{\mathbf{s}}$:

Finally, we simplify the fraction by dividing the numerator and denominator by 2:

And there we have it! The wave is propagating in the direction of $\frac{-3\hat{\mathbf{j}}+4\hat{\mathbf{k}}}{5}$. This matches option (c).