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Mutual inductance of a solenoid wrapped around part of a toroid

  1. Nov 19, 2011 #1
    *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.

    1. The problem statement, all variables and given/known data
    "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: Ns and Nt, 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]

    2. Relevant 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]

    3. 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: Nov 19, 2011
  2. jcsd
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