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Am I correct to say that a uniform conductor is cylindrical, while a non-uniform conductor can have any geometry as long as it is open at both ends.

The textbook does not seem to define these 'uniform' or 'non-uniform' conductors.

Buzz
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Nway
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1 Answers1

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The integration is performed on a path, usually a section of a circuit

$\mathbf{B}(\mathbf{r}) = - \dfrac{\mu}{4 \pi} \displaystyle \int_{\mathbf{r}' \in\ell} i \dfrac{\mathbf{r} - \mathbf{r}' }{|\mathbf{r}-\mathbf{r}'|^3} \times d \mathbf{r}'$.

So far, this is a equation of physics. When you need to evaluate this integral, you usually need to describe the path $\ell$ in a parametric way, i.e. the points of the path $\ell$ in space are function of a parameter $s$, namely $\mathbf{r}'(s)$. It's convenient to use a regular parametrization, so that each point of the path is associated with one value of the parameter $s$.

The choice of the parametrization induces the extreme of integration as well, being $\mathbf{r}_0 = \mathbf{r}(s_0)$ the starting point of the path, and $\mathbf{r}_1 = \mathbf{r}(s_1)$ the other extreme point, so that you can recast the integral as

$\mathbf{B}(\mathbf{r}) = - \dfrac{\mu}{4 \pi}\displaystyle \int_{s_0}^{s_1} i \dfrac{\mathbf{r} - \mathbf{r}'(s) }{|\mathbf{r}-\mathbf{r}'(s)|^3} \times \dfrac{d \mathbf{r}'(s)}{ds} ds$

On the other hand, you can write

$\mathbf{B}(\mathbf{r}) = \displaystyle \int_{\mathbf{r}' \in \ell} d\mathbf{B}(\mathbf{r}, \mathbf{r}')$, being

$d\mathbf{B}(\mathbf{r}, \mathbf{r}') = - \dfrac{\mu}{4 \pi} i \dfrac{\mathbf{r} - \mathbf{r}' }{|\mathbf{r}-\mathbf{r}'|^3} \times d \mathbf{r}'$

basics
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  • Thank you for your answer! I've added my attempt to include limits of integration above. Let me know if anything is wrong. – Nway Dec 14 '22 at 21:08
  • Thanks again! Sorry I have not done vector calculus or line integrals. Only really single variable. – Nway Dec 14 '22 at 21:14
  • Once you manage to write the curve in a parametric way $\mathbf{r}(s)$, it's single variable integration :). You have to do it properly, but then it's single variable integration – basics Dec 14 '22 at 21:16
  • Thanks for your edit. Sorry why do you not have limits of integration like I included? – Nway Dec 14 '22 at 21:16
  • Thanks for your comment! :) Just one question, why did you do a line integral? I thought the Bio-Savart Law was for a point? – Nway Dec 14 '22 at 21:20
  • Integration means summation, $\mathbf{r}' \in \ell$ indicates the place where we need to perform the integration, in space. Once you write $\mathbf{r}(s)$, you may write the limits of integration of your parameter. I think it doesn't have much sense to write $\int_0^{B}$. Trust me, since I did all the courses of Math and Classical Physics – basics Dec 14 '22 at 21:20
  • Thanks for your comment! I don't have too much knowledge. I'm just studying Bio-Savart Law for first year physics. I'm guessing that what they discuss in Purcell's? – Nway Dec 14 '22 at 21:21
  • $\mathbf{r}$ is the "passive" point where we are calculating the field, generated by all the "active" points $\mathbf{r}'$ of the wire. Since we need to evaluate the overall effect of all the points $\mathbf{r}' \in \ell$, we need to sum their contributions and thus performing the line integral $\int_{\mathbf{r}' \in \ell} d\mathbf{B}(\mathbf{r}, \mathbf{r}')$ – basics Dec 14 '22 at 21:23
  • Thanks again for you comments! Ok could we not just integrate each component of the B-field? (Like what is the simplest way to solve my confusion) – Nway Dec 14 '22 at 21:24
  • One of the things I'm confused on is the notation. I don't know what dB(r, r') means sorry. The only notation I know is boldface B for vector B_x for x-component and etc for each component of vector. Would you mine using my notation please? – Nway Dec 14 '22 at 21:29