Suppose you have a moving charge ($q=1.5 \text{ mC}$) in a magnetic field ($\vec{B} = 0.4 \text{ mT } \hat{y}$). The charge has a speed of $10 \text{ m/s}$. What is the magnetic force on the charge if its motion is in the $+x$-direction? The $+y$-direction?

• The charge is $q=1.5 \text{ mC}$.
• There is an external magnetic field $\vec{B} = 0.4 \text{ mT } \hat{y}$.
• The velocity of the charge is $\vec{v} = 10 \text{ m/s } \hat{x}$ or $\vec{v} = 10 \text{ m/s } \hat{y}$.

• $\vec{F}_B$

• The magnetic force on the charge contains no unknown contributions.

• We represent the magnetic force on a moving charge as

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

• We represent the two situations below.

Let's start with the first case, when $\vec{v}=10 \text{ m/s } \hat{x}$.

The trickiest part of finding magnetic force is the cross-product. One can always use the Right Hand Rule, but we will go through the math here to be sure. You may remember from the math review that there are a couple ways to do this. Below, we show how to use vector components, for which it's helpful to rewrite $\vec{v}$ and $\vec{B}$ with their components.

$$\begin{align*} \vec{v} &= \langle 10, 0, 0 \rangle \text{ m/s} \\ \vec{B} &= \langle 0, 4\cdot 10^{-4}, 0 \rangle \text{ T} \\ \vec{v} \times \vec{B} &= \langle v_y B_z - v_z B_y, v_z B_x - v_x B_z, v_x B_y - v_y B_x \rangle \\ &= \langle 0 - 0, 0 - 0, 4\cdot 10^{-3} - 0 \rangle \text{ T} \cdot \text{m/s} \end{align*}$$