How does the radial current density affect the area element dA in Ampere's Law?

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In the discussion, the user seeks clarification on how the radial current density affects the area element dA in Ampere's Law, specifically for a solid cylindrical wire with a radially varying current density defined as J=J0*r^2. The user correctly identifies that for a cylindrical shell at radius r, the area element is dA=2*pi*r*dr, which accounts for the cylindrical geometry. The question arises about the implications of current density varying radially versus along the length of the cylinder, with the user wondering if dA would change if the current density varied in that direction. The consensus emphasizes that in this scenario, dA remains defined as 2*pi*r*dr due to the cylindrical symmetry, and the current density is specified to vary radially only. Understanding this relationship is crucial for applying Ampere's Law correctly in this context.
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Ampere's Law question (need quick answer test tomorrow)

Using Ampere's Law on a solid cylindrical wire with radius R and a current density in the direction of the symmetry axis of the wire. The current density varies radially. J=J0*r^2. What is the magnitude of the the magnetic field when r>R, outside the wire?

Using the formula B*2*pi*r= U0* \intJ dA. Which equals B*2*pi*r = U0 * \int J0* r^2 dA.

Now why does dA = 2*pi*r dr?
Can you guys give me an explanation? How does the information that the current density varies radially show that dA= 2*pi*r*dr? I am having trouble visualizing dA.

If the current density varied along the length of the cylinder wouldn't dA= pi*r^2dx? Would the current density vary along the length of the cylinder or will the current density always vary radially?

Thanks for the help.
Stephen
 
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