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Mutual indutance of two long parallel wires

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Two long parallel wires whose centres are a distance d apart carry equal currents in opposite directions. If the flux within the wires is neglected, the inductance of such arrangement of wire of length l and radius a will be?

I am not able to relate this question to the definition of mutual inductance. In mutual inductance, we basically change the current in one component so as to induce emf in the other component but here both of the components are carrying currents in opposite directions.

I know that:

$\phi = MI$

Please help me with the physical interpretation of the question and provide a hint on how to solve it.

asked Aug 10 in Physics Problems by Reststack (404 points)

1 Answer

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It is not clear what you are asking. Mutual inductance is not mentioned in the problem statement. Perhaps you are making the problem more complicated than it is by asking about mutual inductance.

Current-carrying wires cannot exist on their own without some circuit. I think the key to understanding what is going on in such questions is to ask "Where is/are the circuit/s?"

If identical parallel wires AB and DC carry the same current in opposite directions (so that both currents increase/decrease together) then they can be considered to be part of the same circuit, eg the loop ABCD of circuit I in the upper diagram below. If so you are really asking for the self-inductance of this circuit.

On the other hand if the two parallel wires AB and A'B' are part of separate circuits, such as the loops ABCD and A'B'C'D' in circuits I and II below, then you are indeed asking for the mutual inductance of these two circuits. This depends on the relative orientation of the circuits.

Self Inductance of Circuit I

The sides AB, CD are so long that the magnetic fields due to the short sides AD, BD and corners effects are assumed to be negligible.

Assume that the current is distributed uniformly over the cross-section of the wire in the circuit. Using Ampere's Law the magnetic field around each wire has the form $$B(r)=\frac{r^2}{a^2}\frac{\mu_0 I}{2\pi r} \text{ for } r \le a$$ $$B(r)=\frac{\mu_0 I}{2\pi r} \text{ for } r \gt a$$ where $a$ is the radius of the wire and $r$ is the distance from each wire.

The magnetic flux through a unit length of circuit in the direction AB is $$\Phi=2\int_0^d B(r)dr=\frac{\mu_0 I}{\pi}(\int_0^a \frac{r}{a^2}dr+\int_a^d \frac{1}{r}dr)=\frac{\mu_0 I}{\pi}[\frac12+\ln\frac{d}{a}]$$ The self inductance per unit length $L$ of the circuit is defined by $$\Phi = LI$$ therefore $$L=\frac{\mu_0}{\pi}[\frac12+\ln\frac{d}{a}]$$

For comparison see Inductance of two parallel wires.

Mutual Inductance between Circuits I & II

Assume that the 2 circuits lie in the same plane.

The flux through unit length of circuit I due to circuit II is

$$\Phi_{12}=\Phi_{A'B'}-\Phi_{C'D'}=\int_x ^{d+x} B(r)dr+\int_{2d+x} ^{d+x} B(r)dr=\frac{\mu_0 I}{2\pi}\ln\frac{(d+x) ^2}{(2d+x)x}$$

The mutual inductance per unit length $M_{12}$ is defined by

$$\Phi_{12}=M_{12} I_2$$

therefore

$$M_{12}=\frac{\mu_0}{2\pi}\ln\frac{(d+x)^2}{(2d+x)x}$$

This means that if the current in circuit II is changing then an emf is induced in circuit I of magnitude

$$E=\frac{d\Phi_{12}}{dt}=M_{12}\frac{dI_2}{dt}$$

answered Aug 11 by sammy gerbil (26,096 points)
edited Aug 14 by sammy gerbil
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