The electric field of a plane electromagnetic wave is given by E = E_0 (i^ + j^)/sqrt(2) cos(kz + ωt

Q: Why is the wave propagating in the negative z-direction?

The phase of the wave is given by $(kz + \omega t)$. For a constant phase, $kz + \omega t = \text{constant}$, which implies $\frac{dz}{dt} = -\frac{\omega}{k}$. The negative sign indicates propagation in the $-z$ direction.

Q: How did we find the direction of the magnetic field?

In an electromagnetic wave, the electric field, magnetic field, and direction of propagation are mutually perpendicular. The direction of propagation is given by the cross product $\hat{\mathbf{E}} \times \hat{\mathbf{B}}$. Knowing $\hat{\mathbf{E}}$ and the propagation direction, we can deduce $\hat{\mathbf{B}}$.

Q: Why do we evaluate the fields at $t=0$ and $z=\frac{\pi}{k}$?

The problem asks for the instantaneous force on the particle at $t=0$. Since the particle is located at $z=\frac{\pi}{k}$ at that instant, we must evaluate the fields exactly at that point in space and time to find the forces acting on it.

Q: Does the magnitude of the velocity matter?

No, the magnitude $v_0$ only affects the magnitude of the magnetic force, not its direction. Since both the electric and magnetic forces point in the exact same direction, the net force will always point in that direction regardless of $v_0$.

Q: What if the charge was negative?

If the charge were negative, both the electric force $\mathbf{F}_e = q\mathbf{E}$ and the magnetic force $\mathbf{F}_m = q(\mathbf{v} \times \mathbf{B})$ would reverse their directions. The net force would then be parallel to $\frac{\hat{\mathbf{i}} + \hat{\mathbf{j}}}{\sqrt{2}}$.

Question

The electric field of a plane electromagnetic wave is given by
$\mathbf{E} = E_0 \frac{\hat{\mathbf{i}} + \hat{\mathbf{j}}}{\sqrt{2}} \cos(kz + \omega t)$
At $t = 0$, a positively charged particle is at the point $(x, y, z) = \left(0, 0, \frac{\pi}{k}\right)$.
If its instantaneous velocity at $(t = 0)$ is $v_0 \hat{\mathbf{k}}$, the force acting on it due to the wave is

Options:

(A) zero

(B) antiparallel to $\frac{\hat{\mathbf{i}} + \hat{\mathbf{j}}}{\sqrt{2}}$

(C) parallel to $\frac{\hat{\mathbf{i}} + \hat{\mathbf{j}}}{\sqrt{2}}$

(D) parallel to $\hat{\mathbf{k}}$

Sigma Percentile · Accepted Answer

The problem of finding the net force on a charged particle in an electromagnetic wave is a beautiful application of the Lorentz force law. Let's break down the physics step by step.

### Analyzing the Setup

We are given an electromagnetic wave with its electric field described by the equation:
$$ \mathbf{E} = E_0 \frac{\hat{\mathbf{i}} + \hat{\mathbf{j}}}{\sqrt{2}} \cos(kz + \omega t) $$

We need to find the instantaneous force on a positively charged particle located at $z = \frac{\pi}{k}$ at time $t = 0$. The particle is moving with a velocity $\mathbf{v} = v_0 \hat{\mathbf{k}}$.

Whenever a charged particle moves through a region containing both electric and magnetic fields, it experiences the **Lorentz force**, which is the vector sum of the electric and magnetic forces:
$$ \mathbf{F} = q\mathbf{E} + q(\mathbf{v} 	imes \mathbf{B}) $$

To find this net force, we must evaluate both the electric and magnetic fields exactly at the particle's position and time.

### The Electric Force

Let's start by finding the electric field at the particle's location. We substitute $t = 0$ and $z = \frac{\pi}{k}$ into our wave equation:
$$ \mathbf{E} = E_0 \frac{\hat{\mathbf{i}} + \hat{\mathbf{j}}}{\sqrt{2}} \cos\left(k\frac{\pi}{k} + 0ight) = E_0 \frac{\hat{\mathbf{i}} + \hat{\mathbf{j}}}{\sqrt{2}} \cos(\pi) $$

Since $\cos(\pi) = -1$, the electric field vector at this instant is:
$$ \mathbf{E} = -E_0 \frac{\hat{\mathbf{i}} + \hat{\mathbf{j}}}{\sqrt{2}} $$

Because the particle has a positive charge $q$, the electric force $\mathbf{F}_e = q\mathbf{E}$ points in the exact same direction as the electric field. Therefore, the electric force is directed along $-(\hat{\mathbf{i}} + \hat{\mathbf{j}})$.

### The Magnetic Force

To find the magnetic force, we first need to determine the magnetic field $\mathbf{B}$. In an electromagnetic wave, the direction of propagation is given by the cross product of the electric and magnetic field unit vectors: $\hat{\mathbf{E}} 	imes \hat{\mathbf{B}} = \hat{\mathbf{c}}$.

Looking at the phase term $(kz + \omega t)$, we can deduce that the wave is propagating in the **negative z-direction** ($-\hat{\mathbf{k}}$).

We know that $\hat{\mathbf{E}}$ is along $\frac{-\hat{\mathbf{i}} - \hat{\mathbf{j}}}{\sqrt{2}}$. Let's set up the cross product to find $\hat{\mathbf{B}}$:
$$ \left( \frac{-\hat{\mathbf{i}} - \hat{\mathbf{j}}}{\sqrt{2}} ight) 	imes \hat{\mathbf{B}} = -\hat{\mathbf{k}} $$

Solving this cross product reveals that the magnetic field must point along $\frac{-\hat{\mathbf{i}} + \hat{\mathbf{j}}}{\sqrt{2}}$.

Now, we can calculate the magnetic force using $\mathbf{F}_m = q(\mathbf{v} 	imes \mathbf{B})$. The velocity is $\mathbf{v} = v_0 \hat{\mathbf{k}}$.
$$ \mathbf{F}_m = q \left( v_0 \hat{\mathbf{k}} 	imes B_0 \frac{-\hat{\mathbf{i}} + \hat{\mathbf{j}}}{\sqrt{2}} ight) $$

Evaluating the cross products $\hat{\mathbf{k}} 	imes \hat{\mathbf{i}} = \hat{\mathbf{j}}$ and $\hat{\mathbf{k}} 	imes \hat{\mathbf{j}} = -\hat{\mathbf{i}}$, we get:
$$ \mathbf{F}_m = q v_0 B_0 \frac{-\hat{\mathbf{j}} - \hat{\mathbf{i}}}{\sqrt{2}} = -q v_0 B_0 \frac{\hat{\mathbf{i}} + \hat{\mathbf{j}}}{\sqrt{2}} $$

Fascinatingly, the magnetic force also points in the direction of $-(\hat{\mathbf{i}} + \hat{\mathbf{j}})$.

### The Net Lorentz Force

We have found that both the electric force $\mathbf{F}_e$ and the magnetic force $\mathbf{F}_m$ point in the exact same direction: $-(\hat{\mathbf{i}} + \hat{\mathbf{j}})$.

When we add them together to find the net Lorentz force, the resultant vector will naturally point in this shared direction.

Therefore, the net force acting on the particle is **antiparallel to $\frac{\hat{\mathbf{i}} + \hat{\mathbf{j}}}{\sqrt{2}}$**.

Select Answer:

Visualized Solution

Setup

Lorentz Force

Electric Field

Calculate $E$

Electric Force

Wave Propagation

Magnetic Field

Magnetic Force

Net Force

The Sigma Insight: Characteristics of Electromagnetic Waves

Analyzing the Setup

The Electric Force

The Magnetic Force

The Net Lorentz Force

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