Magnetic field due to semi-circular wire

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Homework Help Overview

The discussion revolves around calculating the magnetic field at a point due to a semicircular wire using the Biot-Savart law. Participants are exploring the geometric and mathematical aspects of the problem, particularly focusing on the integration of contributions from the wire segments.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the setup of the problem, including the identification of vectors and the application of the Biot-Savart law. There is a focus on the cross product involved in the calculation and the implications of the changing direction of the unit vector throughout the semicircle. Some participants suggest using the right-hand rule for direction and simplifying the integral by expressing the differential length in polar coordinates.

Discussion Status

The conversation is ongoing, with various approaches being considered. Some participants have provided hints and suggestions to simplify the calculations, while others express uncertainty about the implications of the semicircular geometry on the magnetic field direction and the integration process.

Contextual Notes

There is a noted difficulty in exploiting symmetries in the problem, particularly regarding the cancellation of components for the semicircle compared to a full circle. Participants are also grappling with the implications of the vertical and horizontal components of the magnetic field.

ks_wann
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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)



\begin{equation}
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}}

\end{equation}

With these identified, I use biot-savart:

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


What troubles me is the cross product. I get

\begin{equation}
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}})
\end{equation},

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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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}.
 
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?
 
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##?

34q42ts.png


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:
unscientific said:
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##?

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..
 
ks_wann said:
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..

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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