Animated Solution for Physics - Electromagnetic Waves: The magnetic field of an electromagnetic wave is given by $\mathbf{B}=1.6\times 10^{-6}\cos (2\times 10^{7}z+6\times 10^{15}t)(2\hat{i}+\hat{j}){\text{ Wbm}}^{-2}$ The associated electric field will be

This problem is a beautiful exercise in the anatomy of an electromagnetic wave. It tests your ability to extract physical meaning from a mathematical equation and apply the fundamental constraints of electromagnetism. Let's break it down step-by-step.

The first thing we should always look at is the argument of the cosine function, known as the phase of the wave. Here, the phase is given by $(2\times 10^{7}z+6\times 10^{15}t)$.

In general, a wave traveling in the $+z$ direction has a phase of the form $(kz-\omega t)$, where the spatial and temporal terms have opposite signs. Conversely, if the terms have the same sign, like $(kz+\omega t)$, the wave is propagating in the negative direction.

Therefore, our wave is traveling along the $-z$ axis. We can write the unit vector for the direction of propagation as $\hat{n}=-\hat{k}$.

Next, we need the amplitude of the electric field, $E_0$. The fundamental relationship between the electric and magnetic field amplitudes in a vacuum is $E_0=c{B}_0$, where $c$ is the speed of light.

Looking at the magnetic field vector $\mathbf{B}=1.6\times 10^{-6}(2\hat{i}+\hat{j})$, we must be careful to find its true magnitude. The length of the vector $(2\hat{i}+\hat{j})$ is $\sqrt{2^2+1^2}=\sqrt{5}$.

So, the peak magnetic field is $B_0=1.6\times 10^{-6}\sqrt{5}\text{ T}$.

Substituting this into our master equation:

Notice how the options are formatted: they leave the $\sqrt{5}$ inside the direction vector. This is a common trick to make the options look cleaner.

Now comes the most critical part: determining the direction of the electric field vector $\mathbf{E}$. We have two strict geometric constraints:

1. Transversality: The electric field must be perpendicular to the magnetic field. Mathematically, $\mathbf{E}\cdot \mathbf{B}=0$. 2. Poynting Vector: The cross product of $\mathbf{E}$ and $\mathbf{B}$ must point in the direction of wave propagation. Mathematically, $\hat{E}\times \hat{B}=\hat{n}=-\hat{k}$.

Since $\mathbf{B}$ is proportional to $(2\hat{i}+\hat{j})$, any vector perpendicular to it in the $xy$-plane must swap the components and flip one sign. This leaves us with two candidates for the direction of $\mathbf{E}$: either $(\hat{i}-2\hat{j})$ or $(-\hat{i}+2\hat{j})$.

Let's test the second candidate, $(-\hat{i}+2\hat{j})$, using the cross product constraint:

Since $\hat{i}\times \hat{j}=\hat{k}$ and $\hat{j}\times \hat{i}=-\hat{k}$, this evaluates to:

This result points exactly in the $-\hat{k}$ direction, which perfectly matches our propagation direction!

Combining the amplitude, the phase, and the correct direction vector, we arrive at the final expression for the electric field:

This matches option (d). A brilliant problem that rewards careful attention to vector mathematics!