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

AI Thread Summary
A current of constant density, J0, flows through a long cylindrical conducting shell with inner radius a and outer radius b, prompting the need to determine the magnetic field in different regions. Using Ampere's Law, the magnetic field is found to be zero for r < a, B = μ0J0(r² - a²)/2r for a < r < b, and B = μ0J0(b² - a²)/2r for r > b. The discussion highlights the importance of understanding the relationship between current density and cross-sectional area to derive the enclosed current, ienc. The final expressions for the magnetic field in each region were confirmed to be correct. The problem-solving process emphasized the need to focus on cross-sectional area rather than volume for this scenario.
Renaldo
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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
\oint B \bullet ds = μ0ienc

The Attempt at a Solution



\oint B \bullet ds = μ0ienc

Solving for B:

B\oint ds = μ0ienc
B2∏r = μ0ienc

B = μ0ienc/2∏r

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

At a < r < b:

B = μ0ienc/2∏r

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

At r > b:

ienc = itotal, but I would need an expression for the volume of the cylinder.

V = ∏b2h - ∏a2h
V = ∏h(b2-a2)
Ienc = J0V \Rightarrow Ienc = J0∏h(b2-a2)

B = μ0ienc/2∏r

B = μ0J0h(b2-a2)/2r

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

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
\oint B \bullet ds = μ0ienc


The Attempt at a Solution



\oint B \bullet ds = μ0ienc

Solving for B:

B\oint ds = μ0ienc
B2∏r = μ0ienc

B = μ0ienc/2∏r

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

At a < r < b:

B = μ0ienc/2∏r

I don't know how to get ienc. 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:

ienc = itotal, but I would need an expression for the volume of the cylinder.
No. Volume does not enter the picture. Cross-section does.
 
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.
 
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