# Inductance and coils

1. Dec 18, 2006

### 6Stang7

1. The problem statement, all variables and given/known data
Coil 1 has L=25mH and N=100 turns. Coil 2 has L=40mH and N=200. The coils are rigidly positioned with respect to each other, and their mutual inductance is 3.0mH. A 6.0mA current in coil 1 is changing at 4.0 A/s. (a)What magnetic flux Phi12 links coil 1 to coil 2, and what is the self-induced emf that appears in coil 1? (b) What magnetic flux Phi21 links coil 2 to coil 1, and what is the mutually induced emf that appears in coild 2?

2. Relevant equations

I know I need E=-L(dI/dt) for the self induced emf and E=-M(dI/dt) for the emf induced in coil2.

3. The attempt at a solution
Solving for both emfs easy and I understand it. What I don't get is what is ment by the magnetic flux that links coil 1 to coil 2 and visea-versa. Would Phi12=L1I1+MI2 and Phi21=L2I2+MI1?
Given that I dont have length (but since I do have number of total turns and would assume that I can assume a length of 1m?) or cross-sectional area, I dont see how you could ever calculate the induced current. I think I have a conceptual problem here and my book is sort of famous for this. Anyone?

2. Dec 18, 2006

### vanesch

Staff Emeritus
There is always something confusing about "the flux through a coil".
We know that a closed loop spans a surface, and this is not different for a coil. However, this is a "multi-layered" surface. The flux through this "multi-layered" surface will be approximately N times the flux through a simple surface spanning the coil "only once", when N is the number of windings.
The simple surface would be a disk, and the "multi-layered" surface would be some kind of helicoidal surface with a very small step (the step of the windings).

If we call $$\phi_{multi}$$ the correct, multi-layered surface attached to the coil, then we have that the EMF is the time variation of this flux. However, we usually call the flux through a coil, the flux through the simple surface $$\phi_{simple}$$. So we have:

$$EMF = \frac{d \phi_{multi}}{dt} = N \frac{d \phi_{simple}}{dt}$$

We also know that $$EMF_1 = L \frac{d i_1}{dt} + M \frac{d i_2}{dt}$$

From this, it follows that
$$\phi_{simple} = \frac{L i_1 + M i_2}{N}$$

Last edited: Dec 18, 2006
3. Dec 18, 2006

### 6Stang7

why do you have emf=-M(dI2/dt)? isnt it emf=-M(dI1/dt)? And what is N in this case? If I am lokking for the flux of 1 onto 2, then would I set N to be coil 1's turns or coil 2? I don't know the current in coil 2 either; is there a way to solve for that?

4. Dec 18, 2006

### 6Stang7

I guess what I am not getting here is the conceptual difference between the two fluxes.

5. Dec 18, 2006

### vanesch

Staff Emeritus
$$L i_1$$ is the flux seen by the (multilayered) surface on coil 1 by the current in coil 1. $$M i_2$$ is the flux seen by the multilayered surface on coil 1 by the current in coil 2.
Now, we usually call the flux, not the flux through the multilayered surface, but by the simple surface, which is then N times less, with N the number of windings of coil 1.

EDIT: you can put in minus signs. That depends on some conventions.

EDIT 2: I was taking coil 1 here as an abstract example. Of course something similar can be written about coil 2.

Last edited: Dec 18, 2006
6. Dec 19, 2006

### 6Stang7

so for this case, the flux that links coil 1 to coil 2 is MI2 + LI1 and the flux that links coil 2 to coil 1 is LI2 + MI1.

now, is there any way to solve for the induced current in coil 2? Wouldn't I need more information such as length and area of coil 2?

7. Dec 19, 2006

### 6Stang7

anyone got an idea?

8. Dec 20, 2006

### vanesch

Staff Emeritus
No. The flux (on the helicoidal surface) induced in coil 1 by the current in coil 2 is M I2.
The flux induced in coil 1 by the current in coil 1, is L1 I1.
The total flux is the sum of both contributions.

And if you want the flux in a single "disk" surface, you have to divide by the number of windings of coil 1.

In the same way:

The flux (on the helicoidal surface) induced in coil 2 by the current in coil 1 is M I1.
The flux induced in coil 2 by the current in coil 2, is L2 I2.
The total flux is the sum of both contributions.

And if you want the flux in a single "disk" surface, you have to divide by the number of windings of coil 2.

No: obviously, what you need are the currents (you have them), the coefficients of induction (L1, L2 and M) and the number of windings. All the geometry is already included in the coefficients of induction.