Continue to Site

Welcome to

Welcome to our site! is an international Electronics Discussion Forum focused on EDA software, circuits, schematics, books, theory, papers, asic, pld, 8051, DSP, Network, RF, Analog Design, PCB, Service Manuals... and a whole lot more! To participate you need to register. Registration is free. Click here to register now.

Mutual inductance in concentric cylinders

Not open for further replies.


Newbie level 3
Mar 29, 2006
Reaction score
Trophy points
Activity points

I have been thinking about this question for a while, so I hope you can help! Here it is:

Lets say you have two concentric, conducting cylinders.

The outer cylinder ("Cylinder 2") has the following properties: radius R2, carries a current of I2, relative magnetic permeability mu2, impedance of Z2 = R2 + jwL2.

The inner cylinder ("Cylinder 1") has the following properties: radius R1, carries a current of I1 (in the same direction as I2), relative magnetic permeability mu1, impedance of Z1 = R1 + jwL1.

My question is: how does the mutual inductance between these cylinders increase their impedance?

My guess is the following:

"Cylinder 1" creates a magnetic field B = (mu0)(mu1)(I1)/(2*pi*R1)
This creates a mutual inductance on "Cylinder 2". The mutual inductance created on "Cylinder 2" is M21 = [(mu0)(mu1)(I1)/(2*pi*R1)]*LN(R2/R1).
Thus the impedance of "Cylinder 2" increases to Z2 = R2 + jwL2 + jwM21.

"Cylinder 2" creates a magnetic field B = (mu0)(mu2)(I2)/(2*pi*R2). However, this field exists only outside the cylinder. Inside the cylinder, the magnetic field is 0. Therefore, the mutual inductance created on "Cylinder 1" by "Cylinder 2" is M12 = 0.
Thus the impedance of "Cylinder 1" remains unchanged at Z1 = R1 + jwL1.

Please let me know if this is correct or if I'm terribly mistaken! Thank you kindly!

Hi Jesteraas,

by reciprocity, it should be M12=M21. You consideration seems to violate this.
I have not a complete answer, but an idea: You should consider not just two pieces of conductor, but two closed circuits.
Different situations can arise. Fos instance, the inner circuit can be closed inside the outer cylinder or outside. The linked flux is different in the two cases.
Interesting problem.

interesting question..

to simplify this we assume that cylinders are made of copper.
sine cylinders caring AC current. it produces Eddy current on another cylinder, this Eddy current produces opposing flux that will decrease the primary field. since the resultant primary flux decreases the impedance also should decrease.

for non conducting magnetic material (like ferrite) the resultant impedance will increase because there is no eddy current..

this may not be complete (mathematical) solution, simply an idea
There's an important point missing in your considerations. How do you measure the inductance of a single conductor? It's undefined without assuming a return path. To analyze the two conductors as separate inductors, a third conductor has to be introduced. Or analyze both conductors as a single inductor, as it's usually done when calculating transmission lines.

Thank you all for your responses! I am learning much more now.

Not open for further replies.

Similar threads

Part and Inventory Search

Welcome to