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Homework Help: Magnetic field due to semi-circular wire

  1. Jun 1, 2014 #1
    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)

    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. jcsd
  3. Jun 1, 2014 #2


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    This is a case where the angle between [itex] \vec{dl'} [/itex] and [itex]\hat{r}[/itex] is always the same so it is much better to simply use the right hand rule to find the direction and use [itex] dl' \, \sin \theta [/itex] to find the magnitude. If you absolutely want to use a cross product, write instead [itex] \vec{dl'} [/itex] as [itex] dl' \, \hat{\theta} [/itex].
  4. Jun 1, 2014 #3
    Okay. So I basically just state that the field points downwards in the case of z=0, and get
    \begin{equation} \frac{\mu_0I}{4\pi} \int_{\theta_{1}}^{\theta^{2}} \frac{R}{(R^2+z^2)^{3/2}} \sin{\theta} d \theta \end{equation}?

    What about the change in direction with increasing z?
  5. Jun 1, 2014 #4
    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
  6. Jun 2, 2014 #5
    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..
  7. Jun 2, 2014 #6
    Yes horizontal components cancel for a full circle, but not for a semi-cricle. Do the integration for horizontal and components separately.
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