What is the Magnetic Field Inside a Tightly Wound Ring?

AI Thread Summary
The discussion focuses on calculating the magnetic field inside a tightly wound ring with a current flowing through it. Participants clarify that the wire is wrapped around the ring, forming a toroidal shape, and that the magnetic field can be determined using Ampere's law. The total current is calculated as NI, where N is the number of turns and I is the current per turn. There is also a mention of using the Biot-Savart law for a more complex evaluation, but it is noted that this approach may be tedious. Ultimately, the consensus leans towards using Ampere's law for a straightforward solution.
mtr
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Homework Statement


A wire is reeled on a ring (radii R_{1} and R_{2}). Find the induction of a magnetic field in the middle of the ring, if there is a current I through the wire and there are N waps. Waps are reeled very tightly in only 1 layer.

Homework Equations


H=\frac{IN}{l}
B=\mu H

The Attempt at a Solution


I tried to think about the ring as about one wap of a coil, however I have no idea how to figure out a current inside the ring. I also do not really understand in which part of this construction I should look for induction.
 
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The current inside the ring (or, more clearly, along the circumference) is given in your problem statement - it's I. The magnetic field in center can be calculated from Biot-Savart law:

d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{Id\mathbf{l}\times \mathbf{\hat{r}}}{r^2}
 
How exactly is it reeled? Does the wire follow the circular shape of the ring or does it pass through the hole repeatedly?
 
The wire is reeled as tightly as possible and follows the circular shape of the ring.

Irid said:
The current inside the ring (or, more clearly, along the circumference) is given in your problem statement - it's I. The magnetic field in center can be calculated from Biot-Savart law:

d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{Id\mathbf{l}\times \mathbf{\hat{r}}}{r^2}

Thanks for help, but I have one more question. I read, that Biot-Savart law behaves a little bit different, when the wire's size is important. Do you think I should use:
d\vec B = K_m \frac{ \vec j \times \mathbf{\hat r}} {r^2} dV?
I think it might be important, because in the task there are given R_{1} and R_{2}, so I can actually count ring's volume. I am also not sure what to do, if the wire is isolated from the ring or the ring is not made from conductor, but e.g. from plastic etc.
 
Defennder said:
How exactly is it reeled? Does the wire follow the circular shape of the ring or does it pass through the hole repeatedly?

This is a good observation! Is the wire reeled along the circumference of the ring, or is it reeled around the ring to form a torus?

I supposed it was the former case. If the ring is of finite extent from R1 to R2, I think you should consider a plane current flowing uniformly in a strip from R1 to R2.
 
Well if the wire is reeled in that manner doesn't that reduce the setup to one of a current carrying solenoid?
 
The wire forms a torus.
 
mtr said:
I am also not sure what to do, if the wire is isolated from the ring or the ring is not made from conductor, but e.g. from plastic etc.

Usually one considers an isolated wire (metal wrapped around with plastic coating), so it doesn't matter what the ring is made of. If the wire wasn't isolated, it would short-circuit and you couldn't have N wounds.
 
Well, now you're saying it forms a torus. That corresponds to the wire passing through the ring hole repeatedly.
 
  • #10
OK, it forms a torus. For sure.
 
  • #11
In the torus case, you should use Ampere's law:

\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I

where d\mathbf{l} points around along the circumference of the torus. R1 probably is the radius of the torus, while R2 is the radius of the loop of current, or vice versa.
 
  • #12
So why do I need N - number of waps? Usually known data should be somehow used in such problems.
R_{1} is the radius of a hole inside a ring and R_{2} is the radius of the loop of current.
 
  • #13
In Ampere's law the current I is the total current pierced by integration. In your case, the current in one loop is I, so the total is NI. Just change in the law I \rightarrow NI.
 
  • #14
Another question on clarification, you want to find the B field within the solid ring itself or at the point in space corresponding the centre of the hole of the ring?
 
  • #15
Defennder said:
Another question on clarification, you want to find the B field within the solid ring itself or at the point in space corresponding the centre of the hole of the ring?

I am looking for B at the point in the centre of the hole of the ring.

Irid said:
In Ampere's law the current I is the total current pierced by integration. In your case, the current in one loop is I, so the total is NI. Just change in the law I \rightarrow NI.

So actually in this case I should use B=\frac{\mu _{0} N I}{l} ?
 
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  • #16
Well, just think of it in terms Ampere's circuital law. Try to set up the closed line integral in a way which exploits the symmetry of the torus geometrically. You should be able to get a rough idea of what the magnitude of the B field should be.

The alternative, I think would be to evaluate the B-field directly in toroidal coordinates. Unfortunately I have no idea how to do so, and it does look horribly complicated:
http://mathworld.wolfram.com/ToroidalCoordinates.html
 
  • #17
I think it's possible to do this by parametrising the space curve of the wire. From:
http://mathworld.wolfram.com/Slinky.html
you should see that the parametric path would have the form:
x = (1+a\cos(wt))\cos t
y = (1+a\cos(wt))\sin t
z = a\sin (wt)

Now what you need to do is work out the values of w and a so that it fits the parameters of your problem and evaluate the line integral using the Biot-Savart law.

Unfortunately it appears that it is extremely tedious to do this by hand. I just entered the integral expression into the Wolfram online integrator and it tells me
Sorry, The Integrator was unable to finish doing this integral in the time allotted. Try running it in Mathematica on your own computer. (Download a trial version of Mathematica [here].)

Oh well, so if you want a quick answer just use your intuition and Ampere's circuital law.
 
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