# Two Coaxial Linear Magnetic Materials

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1. Oct 27, 2016

### ApoI1o

1. The problem statement, all variables and given/known data
Consider a two-layered cylindrical wire with inner-layer permeability μ1 and outer-layer permeability μ2. A line current I runs through the center in the z direction. Calculate the bound currents and the magnetic field produced by the bound currents.

2. Relevant equations

[1] ∫ B⋅dl = Iμ0
[2] ∫ H⋅dl = Ifree
[3] B = μ0 (H+M)
[4] B = μH
[5] Jb = ∇x M
[6] Kb = M x n

3. The attempt at a solution

Using equation 2 and symmetry, I come up with
H = I/(2πs)

Using equation 4, I found the inner and outer material B fields. These point in the φ direction.
B=Iμ1/(2πs) in the inner material
B=Iμ2/(2πs) in the outer material.

Plugging B and H into equation 3, I found the inner and outer material M fields. These point in the φ direction.
M= I(μ10 - 1) / (2πs) in the inner material
M= I(μ20 - 1) / (2πs) in the outer material

Plugging M into equation 5, I calculate that Jb = 0 in both the inner and outer material.

I expected some bound volume current, so this result is strange to me. If there is no bound volume current and I draw an amperian loop within the inner material, the enclosed total current must be equal to I. If that's the case, I should be able to use equation 1 to find that
B = Iμ0/(2πs)
but I already calculated a different inner B field above.

How can I reconcile the different B values in this inner material?

2. Oct 27, 2016

For a little bit, it seemed you had found a case where $\oint M \cdot dl$ is non-zero and where $\nabla \times M=0$. The $I_{free}$ is uniformly distributed through the $\mu_1$ material. Thereby, the current enclosed in your $B \cdot dl$ loop and $H \cdot dl$ loop is proportional to $r^2$ and thereby you will not have a vanishing curl and you will get a finite $J_b$. $\\$ If you did work the problem as the case where you have a very thin inner inner conductor (of radius "a" )down the center carrying the entire current, I think you would find there is a magnetic current at the boundary where the $\mu_1$ material begins that supplies the necessary difference between $\mu_o \oint H \cdot dl$ and $\oint B \cdot dl$.

Last edited: Oct 27, 2016
3. Oct 28, 2016

By magnetic current, I mean "magnetic surface current". In general it is magnetic surface currents (and magnetic currents $J_b$ ) that cause the magnetic field $B$ in materials, along with the $H$ from free currents.