- #1

WJSwanson

- 81

- 0

*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.

"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:

Edge length the toroid bore:

Number of turns in the solenoid and the toroid:

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]

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]

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. :)

## 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*and_{s}*N*, respectively._{t}*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. :)

Last edited: