• Support PF! Buy your school textbooks, materials and every day products Here!

Find the mutual inductance

  • #1
198
0

Homework Statement



Find the mutual inductance between an infinitely long wire and a single loop of wire separated by a distance d. (radius = a)

Homework Equations





The Attempt at a Solution



I've adjusted the magnetic flux density to a coordinate system to make everything easier
[tex]B =\frac{ \mu_0 I }{2\pi x}[/tex]
[tex]L_{12} = N\phi / I [/tex]

My main problem is finding the flux through the circle.

[tex]\int_{d-a}^{d+a} \int_{-\sqrt(a^2-x^2)}^{\sqrt(a^2-x^2} \frac{1}{x} dydx[/tex]
So I'm basically integrating the B field over a circle (i removed constants from the integral)

This unfortunately yields a result that isn't quite what it's supposed to be, but it seems close. Does anyone know where I went wrong?
 
Last edited:

Answers and Replies

  • #2
1,169
5
Before I say anything, is the answer for the magnetic flux 2*u0*I*a*ln((d+2a)/d) so that the mutual inductance is 2*u0*a*ln((d+2a)/d)?
 
  • #3
198
0
The mutual inductance should be [tex]\mu_0 (d - \sqrt(d^2-a^2) )[/tex]

the integral I have yields something slightly similar but it doesn't take out the pi
 
  • #4
1,169
5
heh...my integration looks nothing like the answer...I sort of see what you did there, but how did you get that domain for the first integral?
 
  • #5
198
0
Thanks for your replies,
Here's a pic of the problem so it's not confusing
http://img240.imageshack.us/img240/8214/woahsa4.th.png [Broken]

so I was thinking back to calc 3 and I made my limits like [d-a, d+a] for x and then for y I decided to make a circle equation and solve for y.
x^2 + y^2 = a^2
y = +- sqrt(a^2 - x^2)

But then when I eval the integral I get
Four terms of ln's and a sqrt, but unfortunately the ln's don't go away
 
Last edited by a moderator:
  • #6
alphysicist
Homework Helper
2,238
1
Hi jesuslovesu,

One good way to check that everything is okay with the limits on your integrals is to calculate the area; in other words, evaluate the integral without the B field contribution (1/x) and see if you get (pi a^2).

I think there is a problem with the limits on your integrals. The limits on the x integral indicate that the x=0 point is on the wire itself. If that is that case, then the equation of the circle is not x^2 + y^2 =a^2, since the center of the circle is not at the origin.

If the distance d that they give is the distance from the wire to the center of the circle, then the equation would be

[tex](x-d)^2 + y^2 = a^2[/tex]

and you could solve that for y to get your limits on the y integral.

However, I would suggest putting the origin at the center of the circle and making the adjustment to the magnetic field formula; it seems to make the integrals a bit nicer (at least to me).
 
  • #7
198
0
Hey, thanks for your reply, I think I've got it now but I've run into another problem!

Like you said, I checked this
[tex]
\int_{-a}^{a} \int_{-\sqrt(a^2-x^2)}^{\sqrt(a^2-x^2} \frac{1}{1} dydx = \pi a^2
[/tex]
P.S. I used my calculator to eval that, how does one do that in Cartesian coords? I thought I used to be able to do that by hand, but I can't remember quite how.


Then I tried modifying my B equation
[tex]B = \frac{\mu_0 I }{2\pi(x+d)}[/tex]
Which I think is correct if I am saying that my origin is defined at the center of the circle.

But then I tried to evaluate this integral with both my TI-92 and MATLAB but I can't get them to evaluate it, heh
[tex]
\int_{-a}^{a} \int_{-\sqrt(a^2-x^2)}^{\sqrt(a^2-x^2} \frac{1}{x+d} dydx = ?
[/tex]

Is there something I'm still doing wrong?
I know in another book, there is the same problem and it says the integral is supposed to be very unusual but I am surprised I can't solve it
 
Last edited:
  • #8
alphysicist
Homework Helper
2,238
1
For the area part, you can first evaluate the y integral which is straightforward:

[tex]
\int\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} dy = y \big|\limits_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}} = 2 \sqrt{a^2 - x^2}
[/tex]

Then the x integral is

[tex]
\int\limits_{-a}^a 2 \sqrt{a^2-x^2} dx
[/tex]

I looked this up in a table and it gave:

[tex]
2 \frac{1}{2} \left[ x\sqrt{a^2-x^2} + a^2 \sin^{-1}\frac{x}{|a|}\right]_{-a}^{a}
[/tex]

which gave [itex]\pi a^2[/itex].

I personally don't use Matlab; maybe if you did the y integral by hand and only gave it the x integral? Does that give the answer?
 
  • #9
1,169
5
So this right here is the answer: [tex]\mu_0 (d - \sqrt(d^2-a^2) )[/tex]...hmm...all the models I've come up with have integrals that produce ln's...none of them seem to go away (unless I'm just not seeing it clearly).
 
  • #10
alphysicist
Homework Helper
2,238
1
Yes, trying to do it in Cartesian coordinates looks to be quite difficult. It did seem to give me the answer, but it would be much better here to write the integral in polar coordinates.

Instead of da=dx dy, you have da = r dr d(theta). Instead of x+d you have d + (x component of the r coordinate), so in the integral coordinates the magnetic field will be

[tex]
B = \frac{\mu_0 I}{2 \pi (d + r \cos\theta)}
[/tex]

if theta=0 is the direction directly away from the wire. With the appropriate limits I think this should give you the answer much faster than with Cartesian coordinates.
 

Related Threads for: Find the mutual inductance

  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
2
Views
457
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
10
Views
690
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
17
Views
3K
Top