Animated Solution for Physics - Electromagnetic Waves: The magnetic field of a plane electromagnetic wave is given by $\mathbf{B}=B_0[\cos (kz-\omega t)]\hat{\mathbf{i}}+B_1\cos (kz+\omega t)\hat{\mathbf{j}}$ where, $B_0=3\times 10^{-5}$ T and $B_1=2\times 10^{-6}$ T. The rms value of the force experienced by a stationary charge $Q=10^{-4}$ C at $z=0$ is closest to

Imagine a tiny, stationary charge $Q=10^{-4}$ C resting peacefully at the origin of our 3D coordinate system ($z=0$). Suddenly, a complex electromagnetic wave washes over it. Our mission is to find the root-mean-square (RMS) force experienced by this charge.

The very first conceptual hurdle is understanding the nature of the Lorentz force, given by $\mathbf{F}=Q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$. Because our charge is perfectly stationary, its velocity $\mathbf{v}$ is exactly zero. This is a crucial realization: a stationary charge feels absolutely zero magnetic force. It is completely blind to the magnetic field $\mathbf{B}$ and will only be pushed and pulled by the electric field $\mathbf{E}$. Therefore, our entire problem boils down to finding the net electric field at the origin.

The magnetic field provided in the problem is a superposition of two distinct waves:

To find the electric field, we must analyze each component separately using the fundamental relationship for electromagnetic waves: $\hat{\mathbf{E}}\times \hat{\mathbf{B}}=\hat{\mathbf{c}}$ (where $\hat{\mathbf{c}}$ is the unit vector in the direction of wave propagation) and $E_0=c{B}_0$.

Wave 1: The first term is ${\mathbf{B}}_0=B_0\cos (kz-\omega t)\hat{\mathbf{i}}$. The phase $(kz-\omega t)$ tells us this wave is propagating in the positive $z$-direction ($+\hat{\mathbf{k}}$). Applying the right-hand rule, ${\hat{\mathbf{E}}}_0\times \hat{\mathbf{i}}=\hat{\mathbf{k}}$, which means ${\hat{\mathbf{E}}}_0$ must be $-\hat{\mathbf{j}}$. Thus, the corresponding electric field is ${\mathbf{E}}_0=-c{B}_0\cos (kz-\omega t)\hat{\mathbf{j}}$.

Wave 2: The second term is ${\mathbf{B}}_1=B_1\cos (kz+\omega t)\hat{\mathbf{j}}$. Notice the plus sign in the phase $(kz+\omega t)$! This indicates the wave is propagating in the negative $z$-direction ($-\hat{\mathbf{k}}$). Applying the right-hand rule again, ${\hat{\mathbf{E}}}_1\times \hat{\mathbf{j}}=-\hat{\mathbf{k}}$, which means ${\hat{\mathbf{E}}}_1$ must be $-\hat{\mathbf{i}}$. Thus, the corresponding electric field is ${\mathbf{E}}_1=-c{B}_1\cos (kz+\omega t)\hat{\mathbf{i}}$.

Now, we need the net electric field specifically at the location of our charge, which is $z=0$. Substituting $z=0$ into our electric field equations, the spatial dependence vanishes, and we are left with purely time-varying fields:

Since $\cos (-\theta )=\cos (\theta )$, this simplifies beautifully to:

Notice that the two electric field components are perfectly perpendicular to each other. The maximum magnitude of this net electric field occurs when the cosine term hits its peak value of $1$. Using the Pythagorean theorem, the maximum electric field is:

With the maximum electric field in hand, the maximum force is simply $F_{max}=Q|{\mathbf{E}}_{net}{|}_{max}$. Let's plug in the numbers:

Since $\sqrt{904}$ is approximately $30.066$, we find $F_{max}\approx 0.902$ N.

However, the question specifically asks for the RMS value of the force. Because the force oscillates sinusoidally with time (due to the $\cos (\omega t)$ term), the RMS value is the peak value divided by $\sqrt{2}$:

Looking at our options, $0.638$ N is closest to $0.6$ N. The elegant interplay of vector cross products and wave propagation directions leads us straight to the correct answer!