# Mutual Inductance in a circuit

If there is a circuit consists of a long thin conducting shell of radius a and a parallel return
wire of radius b on axis inside. If the current is assumed distributed uniformly
throughout the cross section of the wire, must we calculate mutual inductance?

If there is a circuit consists of a long thin conducting shell of radius a and a parallel return
wire of radius b on axis inside. If the current is assumed distributed uniformly
throughout the cross section of the wire, must we calculate mutual inductance?

Yes! if the current is AC. Any conductor of any shape or form placed in any physical medium will possess self as well as mutual inductances depending on the proximity of the conductors with each other. The magnitude of this inductance will vary in accordance with the variations in the factors contributing to the said inductance like size, material property proximity, etc.

However, the mutual inductance in this case will contribute towards 'Eddy currents' induced in either conductors depending on the factors mentioned above.

First, I am not going to solve your problem, but I will give you some guidance. I will be using the mks system of units (which I recommend you use). Furthermore, there are no eddy currents, and the current is uniform throughout the inner conductor.

In general, the relation between the inductance per unit length L, and the capacitance per unit length C for long concentric or parallel conductors is (for air between the two conductors)

L C = u0 e0 where e0 is the permittivity of free space, and u0 is the permeability of free space:

u0 = 4 pi x 10-7 henrys per meter
e0 = 1/(u0 c2) = 8.85 x 10-12 farads per meter
where c = 2.9979 x 108 meters per sec (speed of light)

For two concentric cylinders of radius a and b (b<a), the capacitance per meter is

C = 2 pi e0/Ln(a/b) Farads per meter

Then if the two cylinders are shorted at the end, and the current on the inner cylinder flows back on the outer cylinder, the inductance per meter is

L = e0 u0 / C = u0 Ln(a/b) /(2 pi) = 2 x 10-7 Ln(a/b) Henrys per meter

Your problem has a solution that looks very similar to this, but because the current is flowing uniformly inside cylinder a (and not on the surface), there is additional inductance, so you need to solve the complete problem.