You have spotted an unidentified flying object! Naturally, you wish to find its charge. You have a compass, a good sense of direction, and keen eyesight. You notice that it is flying due south on a course that will pass directly overhead, and it is $15 \text{ m}$ above you, travelling at $200 \text{ m/s}$. You observe the dial on your properly aligned compass as the object passes overhead. As a function of time, this is what you see:

• We know $B_{\text{earth}} = 32 \mu\text{T}$.
• If we align our coordinate axes according to the representation below, $\vec{B}_{\text{earth}} = 32 \mu\text{T } \hat{y}$.
• $h = 15 \text{ m}$.
• $\vec{v} = -200 \text{ m/s } \hat{y}$.
• You have the graph of $\theta$ versus $t$.

• $\vec{B}_{text{UFO}}$
• $q$

• The UFO can be approximated as a moving point charge.
• $q$, $\vec{v}$, $h$, and $\vec{B}_{\text{earth}}$ are all constants.
• Your sense of direction and eyesight can be trusted.

• We represent the Biot-Savart Law for the magnetic field from a moving point charge as

$$\vec{B}= \frac{\mu_0}{4 \pi}\frac{q \vec{v}\times \vec{r}}{r^3}$$

• We represent the situation with the following pictures. Coordinate axes and cardinal directions are specified.

We can find the magnetic field from the UFO based on the compass measurement. Below we show the geometric argument.

We know that the UFO is closest when it is directly overhead, which is also when it will have the largest contribution to the magnetic field at our observation point, resulting in the largest displacement of the compass needle. On our graph, this would be $20^\circ$. We can also reason that the magnetic field of the UFO at our observation point will point in the $+x$-direction, due to the Right Hand Rule, assuming the charge of the UFO is positive. If the charge of the UFO were negative, we would expect the B-field to be directed in the $-x$-direction. This is not the case, so we can conclude that the UFO is positively charged.

$$\vec{B}_{\text{UFO}} = B_{\text{earth}} \tan \theta \hat{x}$$

When the UFO is directly overhead, it should not be hard to see (pictured below) that the separation vector is $\vec{r}=-h\hat{z}$ and the velocity of the UFO is $\vec{v}=-v\hat{y}$. Their cross product is then

$$\vec{v} \times \vec{r} = hv \hat{x}$$

At this point, we have two ways to express the magnetic field of the UFO, which should help us find an expression for $q$, the charge of the UFO.

$$\begin{align*} \vec{B}_{\text{UFO}} &= \frac{\mu_0}{4 \pi}\frac{q \vec{v}\times \vec{r}}{r^3} \\ B_{\text{earth}} \tan \theta \hat{x} &= \frac{\mu_0}{4 \pi}\frac{qhv}{h^3}\hat{x} \end{align*}$$

Solving for $q$, we have

$$q=\frac{4 \pi h^2 B_{\text{earth}} \tan \theta}{\mu_0 v} = 131 \text{ C}$$