Using ampere's law with current density

[terminator88]

"What is the ? part? I don't know since I don't know the area of the sheet. You seem to be taking the whole area enclosed by the sheet of 4R (wrongly though, it should be ), but (as long as I understood the question well) the current is only running through the (infinitely thin) sheet at 4R, not anywhere < 4R... Right?"
Isn't the area of the sheath (pi)(2R)^2...so I multiply tht by J which will give the current of the sheath which is (4)(pi)(R^2)
Then the current of the wire is (pi)R^2...agree?
so the net current is the difference of both (3)(pi)(r^2)

 

[Nick89]

Why would the area of the sheet be \pi (2R)^2? I don't understand, why?

The area of a wire with a radius of 2R is \pi (2R)^2.
The area of a sheet with inner radius 2R and outer radius 3R (thus, thickness R) is \pi[ (3R)^2 - (2R)^2] (for example)
The area of a sheet with inner radius = outer radius = 4R (and thus thickness 0) is 0.

If I understand the question correctly, the last example is what you are dealing with, an sheet without any thickness. If this is not correct please tell me.

Also, you keep forgetting to add the J in "the current of the wire is (pi)R^2...". This is an area, not a current. Multiply it with a current density J to make it a current.

Finally about what I said a few posts earlier, that the current density J may be a current per unit length. This is often called a surface current density. I'm not sure if this is what we're dealing with because you haven't specified the question clearly.EDIT
You may not understand the link between current density, area and current correctly.
In the equation I = JA, the area A is the area that the current runs through. In this case, the current does not run through and area of \pi (4R)^2, the current only runs through the infinitely thin sheet (right?) which has no area and thus no current.

If the current density J is a current per unit length, then the current through the sheet is simply JL where L is the length of the wire. Since this length is not given and does not cancel out anywhere I still think the question as you have asked it is wrong.What I want to know is, is the 'sheet' you are talking about infinitely thin or does it have a thickness? Maybe an infinitely small thickness dr or something?

 

[terminator88]

Oh I got an idea...maybe the area between the smaller and bigger circle is the thickness.Acctually foorget abt thiss qquestion..its getting very messy and I have alrredy done my AP(not very well though!)...thank you very much...and appreaciate it

 

[Nick89]

If that is the case I get a magnetic field of B = \frac{14}{8} \mu_0 J R

Your original answer specified a magnetic field containing pi however. I can't see any way how the magnetic field can contain pi if you are given the current per area J. The pi will always cancel out...