# Find the current in the copper wire

1. Mar 4, 2013

### Maiq

1. The problem statement, all variables and given/known data

a) What is the current contained in the cylindrical shell of wire defined by R/2<r<R?

b) What is the total current running through the wire?

c) Given your answer to (b) what potential difference V must be applied between the ends of the wire?

2. Relevant equations

J=ar
i=JA
A=$\pi$r$^{2}$
R=ρL/A

3. The attempt at a solution

For part a I found the area of the shell by subtracting the area of two circles with radius R and R/2. I then used that to find the current using equations i=JA and J=aR/2. For part b i did the same thing except used A=$\pi$r$^{2}$ for the area and J=aR. As for part c I found the resistance using the equation R=ρL/A with ρ=1.68x10$^{-8}$ Ωm and the area the same as from part a. Then I used the answer for b with ohm's law to get the voltage. A classmate of mine did this problem differently and had different answers for part a and c (b was the same).

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Last edited: Mar 4, 2013
2. Mar 4, 2013

### rude man

You realize you need to do some integrations here.

3. Mar 4, 2013

### Maiq

Well my classmate used integrations for his and like I said we had the same answer for b but for a our answers were different by a multiply of two. At the time I hadn't completed c so i couldn't compare those answers.

4. Mar 5, 2013

### rude man

The important thing is not to get the right answer, it's to understand what to do. You need to integrate. If you got the right answer without integrating you just lucked out.

5. Mar 5, 2013

### haruspex

J=aR/2 gives the current density only at r=R/2, the least of the range of values for J. And it won't help to take the average of aR/2 and aR, because there's more area outside 3a/4 than inside 3a/4. This is why you need integration.

6. Mar 6, 2013

### Maiq

Alright I reworked part a using integration and checked the professors solution and I got it right. I see now why my answer was wrong thanks a lot for the help guys.