Proving Equal Proportionality Constant for Mutual Induction in Two Coils

In summary, the professor did not care to address the issue of the proportionality constant of two coils being equal, saying it should be assumed. However, using the vectors from the source to the field point and the fact that they are interchangeable, it is easy to see that the proportionality constants are equivalent.
  • #1
DWigs87
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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
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.
 
  • #3
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
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.
 
Last edited:

1. What is mutual induction?

Mutual induction is a phenomenon in which a changing magnetic field in one circuit induces an electromotive force (EMF) in a nearby circuit. This is due to the interaction between the magnetic fields of the two circuits.

2. How is mutual induction different from self-induction?

Self-induction refers to the production of an EMF in a single circuit due to changes in its own magnetic field. Mutual induction involves the interaction between two separate circuits, resulting in an EMF in one circuit due to the changing magnetic field in the other.

3. What is the mathematical relationship between mutual inductance and the number of turns in a coil?

The mutual inductance between two circuits is directly proportional to the number of turns in each coil. This means that increasing the number of turns in one coil will result in a higher mutual inductance between the two circuits.

4. Can mutual induction be used to transfer energy between two circuits?

Yes, mutual induction can be used to transfer energy between two circuits. This is the principle behind devices such as transformers, which use mutual induction to increase or decrease the voltage in a circuit.

5. How is mutual induction used in practical applications?

Mutual induction has many practical applications, such as in transformers, motors, generators, and wireless power transfer systems. It is also used in electronic devices to reduce interference between circuits and in wireless communication systems to transmit and receive signals.

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