Magnetic field inside a ring

In summary, the wire is reeled on a ring with radii R_{1} and R_{2} and a current I passes through the wire. The ring is made up of N waps reeled tightly in only 1 layer. The wire follows the circular shape of the ring and the current inside the ring is given by I. The magnetic field in the center of the ring can be calculated using the Biot-Savart law and is given by d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{Id\mathbf{l}\times \mathbf{\hat{r}}}{r^2}. The wire can also be formed into a torus, in which case
  • #1
mtr
21
0

Homework Statement


A wire is reeled on a ring (radii [tex]R_{1}[/tex] and [tex]R_{2}[/tex]). 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


[tex]H=\frac{IN}{l}[/tex]
[tex]B=\mu H[/tex]

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.
 
Physics news on Phys.org
  • #2
The current inside the ring (or, more clearly, along the circumference) is given in your problem statement - it's [tex]I[/tex]. The magnetic field in center can be calculated from Biot-Savart law:

[tex]d\mathbf{B} = \frac{\mu_0}{4\pi} \frac{Id\mathbf{l}\times \mathbf{\hat{r}}}{r^2}[/tex]
 
  • #3
How exactly is it reeled? Does the wire follow the circular shape of the ring or does it pass through the hole repeatedly?
 
  • #4
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 [tex]I[/tex]. The magnetic field in center can be calculated from Biot-Savart law:

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

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:
[tex] d\vec B = K_m \frac{ \vec j \times \mathbf{\hat r}} {r^2} dV [/tex]?
I think it might be important, because in the task there are given [tex]R_{1}[/tex] and [tex]R_{2}[/tex], 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.
 
  • #5
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.
 
  • #6
Well if the wire is reeled in that manner doesn't that reduce the setup to one of a current carrying solenoid?
 
  • #7
The wire forms a torus.
 
  • #8
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.
 
  • #9
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:

[tex]\oint \mathbf{B} \cdot d\mathbf{l} = \mu_0 I[/tex]

where [tex]d\mathbf{l}[/tex] 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.
[tex]R_{1}[/tex] is the radius of a hole inside a ring and [tex]R_{2}[/tex] is the radius of the loop of current.
 
  • #13
In Ampere's law the current [tex]I[/tex] is the total current pierced by integration. In your case, the current in one loop is [tex]I[/tex], so the total is [tex]NI[/tex]. Just change in the law [tex]I \rightarrow NI[/tex].
 
  • #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 [tex]I[/tex] is the total current pierced by integration. In your case, the current in one loop is [tex]I[/tex], so the total is [tex]NI[/tex]. Just change in the law [tex]I \rightarrow NI[/tex].

So actually in this case I should use [tex]B=\frac{\mu _{0} N I}{l}[/tex] ?
 
Last edited:
  • #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:
[tex]x = (1+a\cos(wt))\cos t[/tex]
[tex]y = (1+a\cos(wt))\sin t[/tex]
[tex]z = a\sin (wt) [/tex]

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.
 

1. What is a magnetic field?

A magnetic field is a region in space where a magnetic force can be detected. It is produced by moving electric charges, such as the movement of electrons in a wire or the rotation of the Earth's core.

2. How is a magnetic field created inside a ring?

A magnetic field inside a ring is created by a current flowing through the ring. This current produces a magnetic field that is perpendicular to the plane of the ring, with the strength of the field depending on the amount of current flowing through the ring.

3. What affects the strength of the magnetic field inside a ring?

The strength of the magnetic field inside a ring is affected by the amount of current flowing through the ring, the diameter of the ring, and the material the ring is made of. Materials with higher magnetic permeability, such as iron, will have a stronger magnetic field.

4. What is the direction of the magnetic field inside a ring?

The direction of the magnetic field inside a ring follows the right-hand rule. If you point your right thumb in the direction of the current flow, your curled fingers will indicate the direction of the magnetic field inside the ring.

5. How does the magnetic field inside a ring affect objects placed inside it?

Objects placed inside the magnetic field of a ring will experience a force if they have an electric charge or if they are magnetic. The direction and strength of the force will depend on the direction and strength of the magnetic field inside the ring.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
139
  • Introductory Physics Homework Help
Replies
2
Views
584
  • Introductory Physics Homework Help
Replies
1
Views
666
  • Introductory Physics Homework Help
Replies
8
Views
917
  • Introductory Physics Homework Help
Replies
25
Views
1K
  • Introductory Physics Homework Help
Replies
20
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
212
  • Introductory Physics Homework Help
Replies
1
Views
267
  • Introductory Physics Homework Help
2
Replies
51
Views
7K
  • Introductory Physics Homework Help
Replies
8
Views
2K
Back
Top