You are a collector of magnetic field detectors. A fellow detector collector is trying to trim down her collection, and so it's your job to see if an old detector is still working properly, in which case it's yours! Today, you are a magnetic field detector collector, inspector, and hopefully a selector. First, you run a test in which a charged particle ($q = 15 \text{ nC}$) is sent through the detector, and you look at the detector's readings to see if the detector seems to be working properly. The particle is travelling at $2 \text{ m/s}$ and you need to predict the magnetic field's magnitude and direction at the points shown below. Be careful! – You'll be using vectors.

• $\vec{v} = 2 \text{ m/s } \hat{x}$
• The locations of interest are indicated above (Locations 1, 2, and 3), if the particle is at the origin.
• $\vec{r}_1 = -0.5 \text{ m } \hat{x}$ (vector pointing from q to Location 1)
• $\vec{r}_2 = 0.5 \text{ m } \hat{y}$ (vector pointing from q to Location 2)
• $\vec{r}_3 = -0.5 \text{ m } \hat{x} + 0.5 \text{ m } \hat{y}$ (vector pointing from q to Location 3)
• $q = 15 \text{ nC}$

• $\vec{B}_1$, $\vec{B}_2$, $\vec{B}_3$

• The particle can be treated as a point particle..
• We are only interested in the $B$-field at this specific moment in time.

• We represent the Biot-Savart Law for 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 diagram given above.

We begin by cracking open the Biot-Savart Law. In order to find magnetic field, we will need to take a cross product $\vec{v}\times \vec{r}$. Notice that $\vec{r}$ indicates a separation vector, directed from source (point particle) to observation (location 1, 2, or 3). Since our source is at the origin of our axes, then the separation vectors are actually just $\vec{r}_{sep} = \vec{r}_{obs} - 0 = \vec{r}_{obs}$, which are just the r-vectors listed in our “Facts”. Below, we have calculated the cross product for each of our three locations:

Note that for the first cross-product, Location 1 is situated perfectly in line with the straight-line path of the particle. This means that the $\vec{r}_1$ points in the same direction as $\vec{v}$ (so the angle between them is 0 degrees). When we take the cross product, then: $$\vec{v}\times \vec{r}_1 = |\vec{v}||\vec{r}_1|sin(0)=0$$ This means there is no magnetic field at Location 1 at all!

For Location 2, the $\vec{r}_2$ is perpendicular to the velocity vector $\vec{v}$ (the angle between them is 90 degrees). When we take this cross product, then: $$\vec{v}\times \vec{r}_2 = |\vec{v}||\vec{r}_2|sin(90)=2*0.5*1= 1 \text{ m}^2\text{s}^{-1}$$ Now we just need to figure out the direction using the right hand rule.

\begin{align*} \\ \vec{v}\times \vec{r}_2 &= 1 \text{ m}^2\text{s}^{-1} \hat{z} \\ \vec{v}\times \vec{r}_3 &= 1 \text{ m}^2\text{s}^{-1} \hat{z} \end{align*} The other two cross-products are the same because locations 2 and 3 are equidistant from this path. Though they are difference distances from the particle itself, the $\hat{x}$ portion of location 3's separation vector does not contribute at all to the cross-product.

Next, we find the magnitudes of $r^3$, since that is another quantity we need to know in the Biot-Savart Law. \begin{align*} {r_2}^3 &= 0.125 \text{ m}^3 \\ {r_3}^3 &= 0.354 \text{ m}^3 \end{align*}

We don't including location 1 above since we already know the magnetic field is 0 at that location! Below, we give the magnetic field at all three locations.

\begin{align*} \vec{B}_1 &= \frac{\mu_0}{4 \pi}\frac{q\vec{v}\times \vec{r}_1}{{r_1}^3} = 0 \\ \vec{B}_2 &= \frac{\mu_0}{4 \pi}\frac{q\vec{v}\times \vec{r}_2}{{r_2}^3} = 12 \text{ nT } \hat{z} \\ \vec{B}_3 &= \frac{\mu_0}{4 \pi}\frac{q\vec{v}\times \vec{r}_3}{{r_3}^3} = 4.2 \text{ nT } \hat{z} \end{align*}