=====Magnetic Force on Moving Charge=====
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?

===Facts===
  * 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 $\vec{v} = -10 \text{ m/s } \hat{y}$.

===Lacking===
  * $\vec{F}_B$

===Approximations & Assumptions===
  * The magnetic force on the charge contains no unknown contributions.
  * The charge is moving at a constant speed (no other forces acting on it)

===Representations===
  * We represent the magnetic force on a moving charge as
$$\vec{F}= q \vec{v} \times \vec{B}$$
  * We represent the two situations below.

{{ 184_notes:10_moving_charge.png?500 |Moving Charge in a Magnetic Field}}
====Solution====
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. You may remember from the [[184_notes:math_review#Vector_Multiplication|math review]] that there are a couple ways to do the cross product. 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, 4\cdot 10^{-3} \rangle \text{ T} \cdot \text{m/s}
\end{align*}

Alternatively, we could use the whole vectors and the angle between them. We find that we obtain the same result for the cross product. One would need to use the [[184_notes:rhr|Right Hand Rule]] to find that the direction of the cross product is $+\hat{z}$. The magnitude is given by

$$\left|\vec{v} \times \vec{B} \right|= \left|\vec{v}\right| \left|\vec{B} \right| \sin\theta = (10 \text{ m/s})(4\cdot 10^{-4} \text{ T}) \sin 90^{\text{o}} = 4\cdot 10^{-3} \text{ T} \cdot \text{m/s}$$

We get the same answer with both methods. Now, for the force calculation:

$$\vec{F}_B = q \vec{v} \times \vec{B} = 1.5 \text{ mC } \cdot 4\cdot 10^{-3} \text{ T} \cdot \text{m/s } \hat{z} = 6 \mu\text{N}$$

Notice that the $\sin 90^{\text{o}}$ is equal to $1$. This means that this is the maximum force that the charge can feel, and we get this value because the velocity and the B-field are exactly perpendicular. Any other orientation would have yielded a weaker magnetic force. In fact, in the second case, when the velocity is parallel to the magnetic field, the cross product evaluates to $0$. See below for the calculations.

\begin{align*}
  \vec{v} &= \langle 0, -10, 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 \rangle
\end{align*}

Or, with whole vectors:

$$\left|\vec{v} \times \vec{B} \right|= \left|\vec{v}\right| \left|\vec{B} \right| \sin\theta = (10 \text{ m/s})(4\cdot 10^{-4} \text{ T}) \sin 0 = 0$$

When the velocity is parallel to the magnetic field, $\vec{F}_B=0$.