Finding H,B, and M in Infinite Slab w/ Weird Free Current

[NeedPhysHelp8]

Homework Statement

There is an infinite slab of material with magnetic susceptibility $$\chi_m$$ parallel to xy plane and between z=-a and z=a. There is a free current with density $$J=J_0 \frac{z}{a}$$ in the x direction, so it's positive for z>0, and negative for z<0.
What is the H field inside and outside the material? also asks for B and M but I figure those should be easy once I get H.

Homework Equations

I'm not quite sure if we can use this:
$$\int H \cdot dl = I_f_e_n_c$$

The Attempt at a Solution

Ok I'm not quite sure how to set up this question, or if I can use Amperes law because the current switches direction at the xy plane. I tried putting a rectangular ampere loop sitting on z=0 and going up to z=r with r<a with the enclosed current being $$I_f_e_n_c= \int J \cdot da$$. With this I got $$H= \frac{J_0 r^2}{2a}$$
My only problem is I'm not sure that H equals 0 on the xy plane because the current isn't symmetric.
Any help to get me started is much appreciated thanks!

 

[anathema]

Hello! Thank you for your post. Let me try to help you with this problem.

First, you are correct in using Ampere's law to solve this problem. However, since the current density is not constant throughout the material, we need to split the integration into two parts: one for z>0 and one for z<0.

For z>0, we can choose a rectangular loop with sides parallel to the z-axis and centered at z=r, where r<a. The enclosed current will be J_0(z/a)da and the length of the loop will be 2r. So, applying Ampere's law, we get:

∫H⋅dl = J_0(z/a)da
H(2r) = J_0(z/a)da
H = J_0(z/2a)

For z<0, we can choose a rectangular loop with sides parallel to the z-axis and centered at z=-r, where r<a. The enclosed current will be -J_0(z/a)da and the length of the loop will be 2r. So, applying Ampere's law, we get:

∫H⋅dl = -J_0(z/a)da
H(2r) = -J_0(z/a)da
H = -J_0(z/2a)

So, we can see that the H field inside the material is given by H = J_0(z/2a) for z>0 and H = -J_0(z/2a) for z<0.

Now, for the H field outside the material, we can choose a rectangular loop with sides parallel to the z-axis and centered at z=r, where r>a. The enclosed current will be J_0, since there is no current inside the material. So, applying Ampere's law, we get:

∫H⋅dl = J_0
H(2r) = J_0
H = J_0/2r

Therefore, the H field outside the material is given by H = J_0/2r.

To find the B field, we can use the relation B = μH, where μ is the permeability of the material. Since the material has a magnetic susceptibility χ_m, we can write μ = μ_0(1+χ_m), where μ_0 is the permeability of