(adsbygoogle = window.adsbygoogle || []).push({}); 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* [tex]\int[/tex]J dA. Which equals B*2*pi*r = U0 * [tex]\int[/tex] 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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# Homework Help: Ampere's Law question

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