# Magnetic Flux Through a Tilted Medium

1. Jun 1, 2015

### magnetpedro

Imagine a ferromagnetic medium shaped as a cylinder (a ferromagnetic fiber) with a magnetic relative permeability of μr, tilted with an angle a, as shown in the picture.

I would like to prove analytically that the sum of the inductances measured along the x axis (angle is a) and y axis (angle is π/2 - a) is independent of the angle a. (Lsum = Lx + Ly independent of a).

Each case, Lx and Ly is also the sum of the inductance of the projections of the fiber in the x and y axis.
My approach consists in using Hopkinson's Law and first determining the Reluctance for each case, and then calculating the Inductance using L= N2/Reluctance, where N is a constant number of turns of a supposed magnetomotive force.

Do you think it's possible?

Thank you very much.

2. Jun 1, 2015

### marcusl

Don't understand your post. What is the inductance of a cylinder?

3. Jun 1, 2015

### magnetpedro

It's the Inductance of a ferromagnetic medium shaped as a cylinder, that is crossed by a flux produced by a magnetomotive force.

4. Jun 1, 2015

### marcusl

Inductance is a quantity that relates energy stored in a magnetic field to the currents producung the field. It also relates an induced emf to a changing current. You have no currents, hence no inductance.

5. Jun 1, 2015

### magnetpedro

My mistake, forgot to mention that the magnetomotive force is produced by a current I (constant). That's my current.

6. Jun 1, 2015

### marcusl

Your problem doesn't make sense as stated and suggests that you don't understand inductance. Have you had an undergrad E&M course?

7. Jun 1, 2015

### magnetpedro

Yes I have. I'll try to explain this model.
Imagine that a magnetomotive force is produced by a number of turns N and a current I, being Fmm=N*I.
That magnetomotive force creates a magnetic flux ∅ that only crosses the ferromagnetic cylinder.
The inductance of the ferromagnetic medium/fiber can be determined using the following expressions:

L = N× ∅ / I

∅ = Fmm/R

L = N2/R

where R is the reluctance of the ferromagnetic medium, given by:

R= l/(μ0r*A)

where l is the lenght of the path of the magnetic flux, μ0 is the magnetic constant (vacuum's permeability), μr is the relative magnetic permeabilty of the ferromagnetic cylinder and A is the area that is crossed by the flux.

8. Jun 1, 2015

### marcusl

These formulas are usually written in terms of the number of turns per unit length n, in which case $L=\frac{n^2l}{R}$. Furthermore, this applies to long solenoids. Since your core is short, this will be an approximation at best.

To your original question, however, inductance is a scalar quantity so you can't break it into x and y components.

9. Jun 1, 2015

### magnetpedro

Yes, inductance is a scalar quantity but I can make a projection of the ferromagnetic fiber along x and y, with a length and cross area also projected, calculate both Inductances Lx and Ly, and then the "final" inductance would be L = sqrt(Lx^2 + Ly^2).

10. Jun 1, 2015

### marcusl

Well yes, you could do that, but I don't see the value since L is already independent of your angle, by definition.