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Consider a current-carrying loop of radius R, and we want to find the field at the center of the loop on the same plane as the loop. Biot-Savart law tells me that it is $B_{center}=\frac{\mu_0I}{2R}$.

With Ampere's Law, I tried to consider the loop as a superposition of many infinitesimal arc segments of the loop , and each arc segment contributes to a field B towards the loop center. Then I applied Ampere's Law to find the B field due to the arc toward the loop center as $B=\frac{\mu_0I}{2\pi R}$. Further, as these arc segments combine to form the loop, I imagined that the contribution of each arc sums up. Then I did an integration over all these arc segments as $B_{center}=\int_0^{2\pi} \frac{\mu_0I}{2\pi R} Rd\theta=\mu_0I$ which looks very wrong. Indeed, I just redid the Amphere's Law.

I believe I applied the Amphere's Law in a wrong way and it has been bugging me, would you mind giving me some hints? It seems to me that it wasn't correct to say the arc segments have a circular field distribution, or superposition can't be done this way.

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3 Answers 3

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For a short wire of length $||dx||$, with current flowing in the direction of $d\vec{x}$, there will be a field at displacement $\vec{r}$ away from the wire, ($k\equiv \tfrac{\mu_0}{4\pi}$):

$$d\vec{B}= kI \frac{d \vec{x} \times \vec{r}}{||r||^3} $$

If we integrate for a full circle around its center, then $d\vec{x}$ and $\vec{r}$ will always be perpendicular, $d \vec{x} \times \vec{r}=rdx$, and all field contributions $d\vec{B}$ will be in the same direction by the right hand rule, crossing $d\vec{x}$ with $\vec{r}$. Therefore they will simply sum up (integrate). Each small length $dx$ will be $dx=rd\theta$, so that $$d \vec{x} \times \vec{r}=rdx = r^2d\theta$$

We can integrate around the circle:

$$d\vec{B}= kI \frac{d \vec{x} \times \vec{r}}{||r||^3} $$ $$ dB= kI \frac{d \theta}{r} $$

$$\implies B = kI \int_{0}^{2\pi} \frac{d\theta}{r} =\frac{2\pi k I}{r} =\frac{\mu_0 I}{2r}$$

As $r$ is a constant. I don’t know what the answer is, but as you can see was pretty meticulous, and think I did it correctly. Edit: Oh I see is correct, misunderstood initial statement.

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  • $\begingroup$ This is the first time Ive ever been irritated at bot getting a thanks or vote. Lol! $\endgroup$
    – Al Brown
    Sep 1, 2021 at 7:34
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The issue is that the formula for $B$ from Ampere's law is valid only for infinite wires, not infinitesimal ones. Even parts of an infinite wire far from some point $P$ contribute to the field at $P$.

See, for instance, the first answer to Ampère's law applied on a "short" current-carrying wire

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  • $\begingroup$ Thanks for the link, I was reading it also. For a broken current path, I could relate that to the complication in apply Stroke’s theorem, where there will be some surface integral (a deep dome) that does not cross any current at all, so it is not well defined. Apply that back to a loop, however, they will always cross each other, and that confuses me. I will read it again to see if I missed something. $\endgroup$
    – Sandbo
    Mar 20, 2018 at 4:50
  • $\begingroup$ @Sandbo, I think the issue is that a "broken current path" would break conservation of charge, this is more completely discussed in the first four paragraphs of the linked answer. $\endgroup$
    – Munthe
    Mar 20, 2018 at 22:16
  • $\begingroup$ Ampere's law (including a displacement current term) applies in all circumstances. Perhaps you could rephrase your answer? $\endgroup$
    – ProfRob
    Nov 10, 2019 at 7:35
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Ampere's law uses a line integral of the B field around a closed loop. For it to be useful you need a function for the component of the B field which is parallel to the loop at each point and which can conveniently integrated. In this problem, if you could find the B field at each point on any chosen loop, you would not need ampere's law to find the B on the axis.

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