Suppose you have a moving charge $q=16 \text{ mC}$ in a magnetic field $\vec{B} = 15 \text{ mT } \hat{y}$. The charge has a velocity of $\vec{v} = (3\hat{x} + 2\hat{y}) \text{ m/s}$, and a mass of $m = 1 \text{ g}$. What does the motion of the charge look like?

• There is a charge $q = 16 \text{ mC}$.
• The charge has velocity $\vec{v} = (3\hat{x} + 2\hat{y}) \text{ m/s}$.
• The charge has a mass $m = 1 \text{ g}$.
• The charge is in a field $\vec{B} = 15 \text{ mT } \hat{y}$.

• The motion of the charge.

• The B-field is constant.

• We represent the magnetic force on a moving charge as

$$\vec{F}= q \vec{v} \times \vec{B}$$

• We represent the situation below.

We can recall an earlier example where we had to find the motion of a charge in a magnetic field. You should be able to convince yourself based on that example that if our particle were moving in the $\hat{x}$ direction, then the $\hat{y}$-directed magnetic field would cause circular motion in the $xz$-plane. The radius of this motion would be, as before, $$r = \frac{mv}{qB}$$

However, in this example the motion of the particle is a little different. There is a $\hat{y}$ component to the velocity, which is parallel to the magnetic field – to be clear, the velocity as a whole is not parallel to the magnetic field, but the existence of the $\hat{y}$ component guarantees that they are not perpendicular. To see why we care about this nuance, consider this: Since the magnetic field is directed completely in the $\hat{y}$-direction, we know that $\vec{v}\times\vec{B}$ will always be perpendicular to the $\hat{y}$-direction. In connection with this, since $\vec{F}= q \vec{v} \times \vec{B}$, we know that the $\hat{y}$ component is zero, $F_y = 0$. Since the particle never experiences a force in the $\hat{y}$-direction, the $\hat{y}$ component of the velocity will never change.

So it seems like the motion here will not be circular. The magnetic force still plays a role, though. You should be able to convince yourself that $$\vec{F}= q \vec{v} \times \vec{B} = q v_x B \hat{z}$$

So when we use the Right Hand Rule, we point our fingers in the direction of $\vec{v}$, which is $\hat{x}$. When we curl our fingers towards $\vec{B}$, which is directed towards $-\hat{z}$, we find that our thumb end up pointing in the $\hat{y}$ direction. Since $q$ is positive, $\hat{y}$ will be the direction of the force. The math should yield: $$\vec{F}= q (v\hat{x}) \times (-B\hat{z}) = qvB\hat{y}$$

So, the force on the charge is at first perpendicular to its motion. This is pictured above. You can imagine that as the charge's velocity is directed a little towards the $\hat{y}$ direction, the force on it will also change a little, since the cross product that depends on velocity will change a little. In fact, if you remember from the notes, this results in circular motion if the charge is in a constant magnetic field.

Finding the radius of this circular motion requires recalling that circular motion is dictated by a centripetal force. This is the same force that we computed earlier – the magnetic force – since this is the force that is perpendicular to the particle's motion. Below, we set the two forces equal to find the radius of the circular motion.

$$\begin{align*} \left| \vec{F}_B \right| &= \left| \vec{F}_{cent} \right| \\ qvB &= \frac{m v^2}{r} \\ r &= \frac{mv}{qB} \end{align*}$$

Notice that the units work out when we check:

$$\frac{\text{kg}\cdot\text{m/s}}{\text{C}\cdot\text{T}} = \frac{\text{kg}\cdot\text{m/s}}{\text{C}\cdot\frac{\text{kg}}{\text{C}\cdot\text{s}}} = \text{m}$$

So now we know the radius. You can imagine that if the particle entered from outside the region of magnetic field, it would take the path of a semicircle before exiting the field in the opposite direction. Below, we show what the motion of the particle would be in each situation.