A particle of mass m and charge q is moving in a region where uniform, constant electric and magneti

Q: Why can we treat the electric and magnetic motions independently?

Because the forces act in perpendicular directions. The electric force is along the y-axis, affecting only $v_y$. The magnetic force is in the x-z plane, affecting only $v_x$ and $v_z$. They don't interfere with each other's respective components.

Q: Why doesn't the magnetic field change the speed of the particle in the x-z plane?

The magnetic force is always perpendicular to the velocity vector. Since work is force dot displacement, and the angle is 90 degrees, the magnetic field does zero work. Therefore, it cannot change the kinetic energy or the speed of the particle, only its direction.

Q: How did we get the unit vector $\hat{\mathbf{k}}$?

In a right-handed coordinate system, $\hat{\mathbf{k}} = \hat{\mathbf{i}} \times \hat{\mathbf{j}}$. Since $\hat{\mathbf{i}}$ is along $\mathbf{v}_0$ and $\hat{\mathbf{j}}$ is along $\mathbf{B}$, their cross product gives the direction of the z-axis. We divide by their magnitudes to ensure it's a unit vector.

Q: What does "increasing pitch" mean?

The pitch of a helix is the distance between consecutive loops. Because the particle is accelerating along the y-axis due to the electric field, it covers more distance in the y-direction during each rotation. Thus, the loops get stretched further apart.

Q: What if the initial velocity was not perpendicular to the fields?

If $\mathbf{v}_0$ had a component parallel to $\mathbf{E}$ and $\mathbf{B}$, that component would simply add to the initial y-velocity. The circular motion in the x-z plane would remain exactly the same, but the particle would start with a non-zero velocity along the y-axis.

Question

A particle of mass $m$ and charge $q$ is moving in a region where uniform, constant electric and magnetic fields $\mathbf{E}$ and $\mathbf{B}$ are present. $\mathbf{E}$ and $\mathbf{B}$ are parallel to each other. At time $t = 0$, the velocity $\mathbf{v}_0$ of the particle is perpendicular to $\mathbf{E}$ (Assume that its speed is always $<< c$, the speed of light in vacuum). Find the velocity $\mathbf{v}$ of the particle at time $t$. You must express your answer in terms of $t$, $q$, $m$, the vector $\mathbf{v}_0$, $\mathbf{E}$ and $\mathbf{B}$ and their magnitudes $v_0$, $E$ and $B$.

Sigma Percentile · Accepted Answer

### Setting the Stage: The Coordinate System

To solve this problem elegantly, we must first establish a smart coordinate system that aligns with the physical vectors given to us. We are told that the uniform electric field $\mathbf{E}$ and magnetic field $\mathbf{B}$ are parallel. Let's align our $y$-axis with these fields.

We are also given that the initial velocity $\mathbf{v}_0$ is perpendicular to $\mathbf{E}$. Let's align our $x$-axis with $\mathbf{v}_0$.

With this setup, we can define our standard unit vectors $\hat{\mathbf{i}}$, $\hat{\mathbf{j}}$, and $\hat{\mathbf{k}}$ entirely in terms of the problem's given vectors:

$\hat{\mathbf{j}} = \frac{\mathbf{E}}{E} = \frac{\mathbf{B}}{B}$

$\hat{\mathbf{i}} = \frac{\mathbf{v}_0}{v_0}$

$\hat{\mathbf{k}} = \hat{\mathbf{i}} 	imes \hat{\mathbf{j}} = \frac{\mathbf{v}_0 	imes \mathbf{B}}{v_0 B}$

This clever substitution will allow us to seamlessly convert our final algebraic components back into the required vector format.

### The Electric Push: Motion Along the Y-Axis

One of the most beautiful principles in physics is the independence of perpendicular motions. The electric field $\mathbf{E}$ acts solely along the $y$-axis. It exerts a constant force $\mathbf{F}_e = q\mathbf{E}$ on the particle.

According to Newton's second law, this results in a constant acceleration along the $y$-axis:

$$a_y = \frac{qE}{m}$$

Using basic kinematics for constant acceleration, the velocity component along the $y$-axis at any time $t$ is simply $v_y = a_y t$. We can express this as a vector:

$$\mathbf{v}_y = \left(\frac{qE}{m}tight) \hat{\mathbf{j}} = \left(\frac{qE}{m}tight) \frac{\mathbf{E}}{E} = \left(\frac{q}{m}tight)\mathbf{E}$$

### The Magnetic Spin: Motion in the X-Z Plane

Now, let's look at the magnetic field $\mathbf{B}$. The magnetic force $\mathbf{F}_m = q(\mathbf{v} 	imes \mathbf{B})$ is always perpendicular to the velocity. Because $\mathbf{B}$ is along the $y$-axis, this force acts entirely within the $x-z$ plane.

This perpendicular force does no work; it only changes the direction of the velocity, resulting in uniform circular motion in the $x-z$ plane. The angular frequency of this rotation (the cyclotron frequency) is:

$$\omega = \frac{qB}{m}$$

At time $t$, the particle has rotated through an angle $	heta = \omega t$. Since the initial velocity $v_0$ was entirely along the $x$-axis, we can resolve the velocity in the $x-z$ plane into its components using trigonometry:

$$v_x = v_0 \cos(\omega t) = v_0 \cos\left(\frac{qB}{m}tight)$$

$$v_z = v_0 \sin(\omega t) = v_0 \sin\left(\frac{qB}{m}tight)$$

### Bringing It All Together: The Final Velocity

The total velocity vector $\mathbf{v}$ is simply the vector sum of its three independent components:

$$\mathbf{v} = v_x \hat{\mathbf{i}} + v_y \hat{\mathbf{j}} + v_z \hat{\mathbf{k}}$$

Substituting our derived expressions and replacing the unit vectors with their original definitions, we get:

$$\mathbf{v} = v_0 \cos\left(\frac{qB}{m}tight) \left(\frac{\mathbf{v}_0}{v_0}ight) + \left(\frac{q}{m}tight)\mathbf{E} + v_0 \sin\left(\frac{qB}{m}tight) \left(\frac{\mathbf{v}_0 	imes \mathbf{B}}{v_0 B}ight)$$

Simplifying the scalar magnitudes, we arrive at our elegant final answer:

$$\mathbf{v} = \cos\left(\frac{qB}{m}tight)\mathbf{v}_0 + \left(\frac{q}{m}tight)\mathbf{E} + \sin\left(\frac{qB}{m}tight)\left(\frac{\mathbf{v}_0 	imes \mathbf{B}}{B}ight)$$

### The Physical Picture: A Stretching Helix

What does this mathematical expression actually look like in the real world? The magnetic field forces the particle to constantly spiral in circles within the $x-z$ plane. Simultaneously, the electric field is relentlessly accelerating the particle upwards along the $y$-axis.

The combination of these two independent motions creates a **helix with an increasing pitch**. As time goes on, the particle completes its circular loops at a constant rate, but it travels further and further along the $y$-axis during each loop due to the constant electric acceleration. It is a beautiful dance of classical electromagnetism!

Visualized Solution

$C oor d ina t e S e t u p$

$E l ec t r i c F or ce an d A cce l er a t i o n$

$V e l oc i t y A l o n g Y - A x i s$

$M a g n e t i c F or ce an d C i r c u l a r M o t i o n$

$V e l oc i t y in X - Z P l an e$

$T o t a l V e l oc i t y V ec t or$

$T h e H e l i c a l P a t h$

The Sigma Insight: Motion of a Charge in Magnetic Fields

Setting the Stage

The Electric Push

The Magnetic Spin

Bringing It All Together

The Physical Picture

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