• Support PF! Buy your school textbooks, materials and every day products Here!

Quick question about current density

  • Thread starter hahaha158
  • Start date
  • #1
80
0

Homework Statement





Homework Equations



J=I/A

The Attempt at a Solution


I was just looking at the solution for this, and I was confused because they seemed to use different ways of finding the area. For finding J2, they took the surface area of the end of the cable, pi(r3^2-r2^2), but for finding J2, they took the area to be ∫2pir dr, which i think is the surface area along the inner cable. Why is one using the surface area of the end, and one using the surface area across the outside of an entire cable"?

thanks for any help
 
Last edited:

Answers and Replies

  • #2
TSny
Homework Helper
Gold Member
12,537
2,956
... but for finding J2, they took the area to be ∫2pir dr, which i think is the surface area along the inner cable.
∫2pir dr is actually the area of the end of the inner cable, ##\pi##R12, as you can see by carrying out the integral from 0 to R1. The integral is summing the areas of concentric rings that fill up the entire area. This is necessary since the current density is different for each ring.
 

Attachments

Related Threads on Quick question about current density

Replies
0
Views
908
Replies
6
Views
742
Replies
1
Views
935
Replies
0
Views
1K
  • Last Post
Replies
10
Views
2K
Replies
4
Views
1K
Replies
1
Views
1K
Replies
1
Views
872
Replies
2
Views
1K
  • Last Post
Replies
2
Views
643
Top