Mutual inductance of a solenoid wrapped around part of a toroid

[WJSwanson]

*This isn't a problem so much as needing to fill in the gaps from a lecture where I was taking notes on painkillers due to surgery. The problem statement is my best guess as to what exactly was said by the instructor.

Homework Statement

"Find the mutual inductance (M) of a square-bore toroid that has a solenoid wrapped partially around it. Assume the toroid and solenoid behave as their respective ideals. No numerical values aside from physical constants are included; we are looking for an analytical solution."

Inner and outer radii of the square-bore toroid: a and b, respectively.
Edge length the toroid bore: h
Number of turns in the solenoid and the toroid: N_s and N_t, respectively.

Now here's the work I had in my notes:
1.
[itex]EMF_s = -M \frac{di_t}{dt}[/itex]

[itex]EMF_t = -M \frac{di_s}{dt}[/itex]

2.
[itex]N_t \frac{d\Phi_{B,s}}{dt} = M \frac{di_t}{dt}[/itex]

[itex]N_s \frac{d\Phi_{B,t}}{dt} = M \frac{di_s}{dt}[/itex]

3.
[itex]\oint \vec{B} . d\vec{s} = \mu_0 N_t i_t[/itex]

4.
[itex]\Phi_{B,1} = \int \vec{B} . d\vec{A} = \int^b_a \frac{\mu_0 N_2 i_2 h}{2\pi r} dr = \frac{\mu_0 N i_t h}{2\pi} ln(\frac{b}{a})[/itex]

5.
[itex]N_s \Phi_{B,1} = M i_2 \Rightarrow M = \frac{N_s N_t \mu_0 h}{2\pi} ln(\frac{b}{a})[/itex]

Homework Equations

Faraday-Lentz Law:
[itex]EMF = -d\Phi_B / dt[/itex]

EMF vs inductance:
[itex]EMF = -L di/dt[/itex]

Ampere's Law:
[itex]\oint \vec{B} . d\vec{s} = \mu_0 i_{encl}[/itex]

Self-inductance of a solenoid:
[itex]L = N^2_s \mu_0 A * \frac{1}{l}[/itex]

Self-inductance of a toroid:
[itex]L = \int^b_a N^2_t \mu_0 h dr / 2\pi r = \frac{N^2_t \mu_0 h}{2\pi} \int^b_a \frac{dr}{r} = \frac{N^2_t \mu_0 h}{2\pi} ln(\frac{b}{a})[/itex]

The Attempt at a Solution

So the question arises, did I write everything down correctly? That might not be the case. Assuming I did, I run into some weirdness.

Using Faraday's Law, we find that
[itex]EMF_s = -N_s \frac{d \Phi_{B,s}}{dt} \Rightarrow N_s \frac{d\Phi_{B,s}}{dt} = M \frac{di_t}{dt}[/itex]

and

[itex]EMF_t = -N_t \frac{d \Phi_{B,t}}{dt} \Rightarrow N_t \frac{d\Phi_{B,t}}{dt} = M \frac{di_t}{dt}[/itex].

So now we have:
[itex]N_s \frac{d\Phi_{B,s}}{dt} = M \frac{di_t}{dt}[/itex]
&
[itex]N_t \frac{d\Phi_{B,t}}{dt} = M \frac{di_s}{dt}[/itex].

From there, I'm not sure what to do. I suppose it would be time to pick which geometry has the easier flux calculation (probably the toroid, since we aren't given the dimensions of the solenoid) to find di/dt for its counterpart?

For the toroid:

[itex]\oint \vec{B} . d\vec{s} = \mu_0 i_{encl.} = \mu_0 N_t i_t[/itex]
[itex]d(\oint \vec{B} . d\vec{s}) / dt = \mu_0 \frac{d i_{encl.}}{dt} = \frac{\mu_0}{L} \frac{d\Phi_B}{dt}[/itex]

or at least I think. I really am not sure where any of the rest of this stuff comes from, though. I would seriously appreciate anyone who can help me figure this out. :)

 

[AvroArrow]

The solution is as follows:1. Use Faraday's Law to relate the emf of the solenoid and toroid to the mutual inductance between them:EMF_s = -M \frac{di_t}{dt}EMF_t = -M \frac{di_s}{dt}2. Use Ampere's Law to relate the enclosed current in the toroid to its self-inductance:\oint \vec{B} . d\vec{s} = \mu_0 N_t i_t3. Calculate the flux of the solenoid through the toroid, using the self-inductance equation for a solenoid. This will allow us to find the mutual inductance:\Phi_{B,1} = \int \vec{B} . d\vec{A} = \int^b_a \frac{\mu_0 N_2 i_2 h}{2\pi r} dr = \frac{\mu_0 N i_t h}{2\pi} ln(\frac{b}{a})4. Plugging this into the equation for EMF_t, we can solve for the mutual inductance:N_s \Phi_{B,1} = M i_2 \Rightarrow M = \frac{N_s N_t \mu_0 h}{2\pi} ln(\frac{b}{a})