# Magnetic field due to semi-circular wire

1. Jun 1, 2014

### ks_wann

So, I've got a wire with a semicircle about the origin, my task is to find the magnetic field at M=(0,0,z).

I want to use Biot-Savart, so I find dl', r and r-hat. (r being the vector from the source point to field point)

r^{2}=R^{2}+z^{2}\\
\mathbf{r}=R\cos{\theta}\mathbf{\hat{i}}+R\sin{\theta}\mathbf{\hat{j}}+z\mathbf{\hat{k}}\\
d\mathbf{l^{\prime}}=Rd\theta \mathbf{\hat{\theta}}

With these identified, I use biot-savart:

\mathbf{B}(\mathbf{r})=\frac{\mu_{0}I}{4\pi}\int\frac{d\mathbf{l^{ \prime}}\times\mathbf{\hat{r}}}{r^{2}}

What troubles me is the cross product. I get

d\mathbf{l^{\prime}}\times\mathbf{\hat{r}}=\frac{R}{\sqrt{R^{2}+z^{2}}} \mathbf{\hat{\theta}}\times(R\cos\theta \mathbf{\hat{i}}+R\sin\theta \mathbf{\hat{j}}+z\mathbf{\hat{k}})
,

and I'm not sure what to make of it as the unit vector would change all the way through the semicircle.

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Last edited: Jun 1, 2014
2. Jun 1, 2014

### nrqed

This is a case where the angle between $\vec{dl'}$ and $\hat{r}$ is always the same so it is much better to simply use the right hand rule to find the direction and use $dl' \, \sin \theta$ to find the magnitude. If you absolutely want to use a cross product, write instead $\vec{dl'}$ as $dl' \, \hat{\theta}$.

3. Jun 1, 2014

### ks_wann

Okay. So I basically just state that the field points downwards in the case of z=0, and get
$$\frac{\mu_0I}{4\pi} \int_{\theta_{1}}^{\theta^{2}} \frac{R}{(R^2+z^2)^{3/2}} \sin{\theta} d \theta$$?

What about the change in direction with increasing z?

4. Jun 1, 2014

### unscientific

Consider a small segment of the wire of length $d\vec l$, producing a small magnetic flux density $d\vec B$ at position $\vec r$ away from the wire.

Hint: what can you say about the direction of $d\vec l$ and $\hat r$?

Also, expressing $dl = R \space d\phi$ will make your integral much easier. Exploit symmetries instead of diving straight into Cartesian cross products..

Last edited: Jun 1, 2014
5. Jun 2, 2014

### ks_wann

Well, $d\vec l$ and $\hat r$ are perpendicular at all times. I see that for a full circle, all the horizontal components will cancel, thus giving me $B(z)$. Am I missing something?

I'm sadly not very good at identifying and exploiting symmetries..

6. Jun 2, 2014

### unscientific

Yes horizontal components cancel for a full circle, but not for a semi-cricle. Do the integration for horizontal and components separately.