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Amperes law for circular arc

  • Thread starter Uku
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  • #1
Uku
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Homework Statement



When I use Biot-Savart law to calculate the magnetic field at a distance from an arc, I get the well known

[tex]B=\frac{i\mu_{0}\phi}{4 \pi R}[/tex]

Why cant I use Amperes law for this? I could imagine that the arc instead to be a complete loop of current. Then I could find the B inside the loop using Amperes law, and then multiply this simply by the [tex]\frac{arc angle}{360}[/tex], since I see superposition used for B fields.

I smell a rat in the "I could find the B inside the loop using Amperes law" part though... Things cancel out that should not cancel out. But I don't get it intuitively...
 

Answers and Replies

  • #2
collinsmark
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Homework Statement



When I use Biot-Savart law to calculate the magnetic field at a distance from an arc, I get the well known

[tex]B=\frac{i\mu_{0}\phi}{4 \pi R}[/tex]

Why cant I use Amperes law for this? I could imagine that the arc instead to be a complete loop of current. Then I could find the B inside the loop using Amperes law, and then multiply this simply by the [tex]\frac{arc angle}{360}[/tex], since I see superposition used for B fields.

I smell a rat in the "I could find the B inside the loop using Amperes law" part though... Things cancel out that should not cancel out. But I don't get it intuitively...
Ampère's law can be expressed in this form:

[tex] \oint _P \vec B \cdot \vec{dl} = \mu_0 I_{inc} [/tex]

Ampère's law can be useful in many situations involving magnetostatics (particularly if you know what B already is). However, there's only a few situations where it can be useful it you are attempting to find B.

[Edit: deleted most of my original post. I want to make sure I'm not giving too much away.]

The reason has to do with symmetry.

Uku, is this question for a homework problem in particular? Or are you just curious?
 
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  • #3
Uku
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I don't think "giving too much away" is really possible here. The problem created conflict in my thought process. You can apply superposition with Biot-Savart and find the field in a point due to a arc-current, or for exampe the field of a magnetic dipole, but not with Ampere's law. The problem arose in calculating the field by a current arc, I found it with Savart and I wondered why does it not work out with Amperes.

Or, well you can apply superposition with the Amperes, even more so than with Biot actually. For example when finding the field in a solenoid or toroid. So... yeah, why not with a circular arc.
 
  • #4
collinsmark
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I don't think "giving too much away" is really possible here. The problem created conflict in my thought process. You can apply superposition with Biot-Savart and find the field in a point due to a arc-current, or for exampe the field of a magnetic dipole, but not with Ampere's law. The problem arose in calculating the field by a current arc, I found it with Savart and I wondered why does it not work out with Amperes.

Or, well you can apply superposition with the Amperes, even more so than with Biot actually. For example when finding the field in a solenoid or toroid. So... yeah, why not with a circular arc.
Okay, as long as the homework problem wasn't asking "why can't you use Ampère's law in this situation," then I can give you more detail.

In order to use Ampère's law to find B, you need to set up the closed path such that the magnetic field strength is a constant at any point on the path
(technically, you must ensure that B · dl = constant for all dl on the closed path).

Typically, this relies on symmetry. You need to find some path such that
B · dl doesn't change when you move around the path. If you're successful at doing so, the left side of the equation reduces to (notice how the vector notation goes away):

[tex] \oint _P \vec B \cdot \vec{dl} \rightarrow \int _P Bdl [/tex]

[tex] = B \int_P dl [/tex]

But of course that assumes that the parallel component B is a constant over the path.

In the problem in the arc, it's not possible to maintain symmetry, such that B is a constant over any given closed path. So even though Ampère's law still applies in this situation, you just cant use it to find B because B is not constant over any conceivable closed path.
 
  • #5
Uku
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Hello!

Yes, but why can't you assume that the arc is a ring and then multiply the result by arcangle/360?
For example when calculating the field in a solenoid, you draw a loop and add four B quantities to find the net field.
 
  • #6
collinsmark
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Hello!

Yes, but why can't you assume that the arc is a ring and then multiply the result by arcangle/360?
But you can't use Ampère's law to find B for a single ring of wire either. For a single ring of wire, even if it does form a complete circle, you're still stuck using Biot-Savart law to find B.
For example when calculating the field in a solenoid, you draw a loop and add four B quantities to find the net field.
Yes, but in the case of a straight solenoid, you must assume that the solenoid is infinitely long in order to use Ampère's law. Why is that? Because its the only case where the magnetic field B is constant, and independent of the position within the solenoid. In other words, its the only case where B · dl is constant for the straight section of the path within the solenoid (and the only case where B · dl of the other sections are zero).

If you have a torus shaped solenoid, you can draw a closed path completely within the torus shaped solenoid. And in that case, due to symmetry, B · dl is constant.

If you have a finite length, straight solenoid, you can use Ampère's law to calculate the approximate magnetic field, as long as the solenoid is long. But realize that is only an approximation. The approximation falls apart when the length of the solenoid is short -- and doesn't apply at all for a single ring of wire.

Don't get me wrong, Ampère's law still applies. And if you already happened to know B for all space, you could use it to find the current in the ring of wire. But you won't be able to use it to find B, because B · dl is not constant over a simple, path (of course, if you could find some path where B · dl was constant over some oddly shaped path, then you could use Ampère's law to find B, but in this case it's not obvious how such a path would be shaped).
 
  • #7
Uku
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Right! Thanks for the replies! I was thinking offbeat a bit, the [tex]B \bullet ds[/tex] needs to be constant around the wire.

Sincerely,
Uku
 

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