Mutual Induction Proof

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I am currently a sophomore in college, and as of now, my calculus-based physics class is studying the phenomena of mu tual induction. However, the actual text does not display a proof showing that the proportionallity constant of two coils are equal (Msub(2 1) = Msub(1 2) = M, which is then used in the equation EMF = -M(dIsub(1)/dt) and EMF = -M(dIsub(2)/dt) where I is the steady current in the coil(s)), and the professor did not care to address it, saying it should be assumed.

Does anyone know this how to go about doing the proof for proving that the proportionallity constant is equal for both coils? I have math background up to and including calculus III and differential equations. Any input would be appreciated. Thanks a lot.
 

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  • #2
Galileo
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The flux through loop 2 (as a result of B1) is:

[tex]\Phi_2=\int \vec B_1 \cdot d\vec a_2[/tex]

Now use the vector potential [itex]\vec \nabla \times \vec A_1=\vec B_1[/itex], apply Stokes and insert the integral form expression for the vector potential.

That's it. You can read off the result.
 
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I see.

I understand what you are doing, and why you have done it as such, but I am still unable to arrive at the conclusion that the proportionality constants are equivalent even after I applied Stokes (I am coming up with a very odd integral expression, so I may have made an error there). Do you mind walking me through it a bit more? I really appreciate it.
 
  • #4
Galileo
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Yeah, the integral looks odd and it's horrible for doing any calculation. But from its form it's easy to see [itex]M_{12}=M_{21}[/itex].

For the vector potential of loop 1, you can use:
[tex]\vec A_1=\frac{\mu_0}{4\pi}I_1\oint \frac{d\vec l_1}{\mathcal{R}}[/tex]

(I wanted to use script-r for the vector from the source to the field point, but it turned out funky)
And remember you can just interchange the orders of integration. Can you show your work.
 
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