Ampere's Law: Determining magnetic fields of a shell conductor

[Renaldo]

Homework Statement

A current of constant density, J₀, flows through a very long cylindrical conducting shell with inner radius a and outer radius b. What is the magnetic field in the regions r < a, a < r < b, and r > b? (Use any variable or symbol stated above along with the following as necessary: μ₀.)

Homework Equations

Ampere's Law
[itex]\oint B \bullet ds[/itex] = μ₀i_enc

The Attempt at a Solution

[itex]\oint B \bullet ds[/itex] = μ₀i_enc

Solving for B:

B[itex]\oint ds[/itex] = μ₀i_enc
B2∏r = μ₀i_enc

B = μ₀i_enc/2∏r

At r < a:
B = 0 because i_enc at this point = 0

At a < r < b:

B = μ₀i_enc/2∏r

I don't know how to get i_enc. I know it has something to do with the current density.

At r > b:

i_enc = i_total, but I would need an expression for the volume of the cylinder.

V = ∏b²h - ∏a²h
V = ∏h(b²-a²)
I_enc = J₀V [itex]\Rightarrow[/itex] I_enc = J₀∏h(b²-a²)

B = μ₀i_enc/2∏r

B = μ₀J₀h(b²-a²)/2r

However, by including h, I am introducing variables that the problem hasn't allowed me to use.

 

[rude man]

Homework Statement

A current of constant density, J₀, flows through a very long cylindrical conducting shell with inner radius a and outer radius b. What is the magnetic field in the regions r < a, a < r < b, and r > b? (Use any variable or symbol stated above along with the following as necessary: μ₀.)

Homework Equations

Ampere's Law
[itex]\oint B \bullet ds[/itex] = μ₀i_enc

The Attempt at a Solution

[itex]\oint B \bullet ds[/itex] = μ₀i_enc

Solving for B:

B[itex]\oint ds[/itex] = μ₀i_enc
B2∏r = μ₀i_enc

B = μ₀i_enc/2∏r

At r < a:
B = 0 because i_enc at this point = 0

At a < r < b:

B = μ₀i_enc/2∏r

I don't know how to get i_enc. I know it has something to do with the current density.

Current = current density times cross-sectional area. What is the cross-sectional area of your cylinder at a < r < b?

At r > b:

i_enc = i_total, but I would need an expression for the volume of the cylinder.

No. Volume does not enter the picture. Cross-section does.

 

[Renaldo]

That makes sense.

So at a < r < b:

B = μ₀J₀(r²-a²)/2r

At r > b:

B = μ₀J₀(b²-a²)/2r

That worked. Thanks for your help.