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Introductory Physics Homework Help
Mutual inductance of a solenoid wrapped around part of a toroid
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[QUOTE="WJSwanson, post: 3624748, member: 361047"] *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. [h2]Homework Statement [/h2] "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: [i]a[/i] and [i]b[/i], respectively. Edge length the toroid bore: [i]h[/i] Number of turns in the solenoid and the toroid: [i]N[sub]s[/sub][/i] and [i]N[sub]t[/sub][/i], respectively. [i]Now here's the work I had in my notes:[/i] 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] [h2]Homework Equations[/h2] 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] [h2]The Attempt at a Solution[/h2] 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. :) [/QUOTE]
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Mutual inductance of a solenoid wrapped around part of a toroid
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