Two Coaxial Linear Magnetic Materials

[ApoI1o]

Homework Statement

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.

Homework Equations

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

The Attempt at a Solution

[/B]
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(μ₁/μ₀ - 1) / (2πs) in the inner material
M= I(μ₂/μ₀ - 1) / (2πs) in the outer material

Plugging M into equation 5, I calculate that J_b = 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?

 

[Charles Link]

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 $\oint B \cdot dl$.

 

[Charles Link]

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 $\oint B \cdot dl$.

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.