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

Ampere's Law: Toroids

  • Thread starter Knissp
  • Start date
  • #1
75
0
1. Homework Statement
The question gives a coil of N turns carrying a current of I Amperes wound on a ring with rectangular cross section of inner radius r1 and outer radius r2 and height h. The ring has magnetic permeability mu. What is the flux in webbers?


2. Homework Equations
Ampere's Law:
Closed Integral (B * ds) = mu * I


3. The Attempt at a Solution

My problem is mostly a conceptual one since I have all the equations for sure.

flux = closed integral (B * dA)
integral (B * ds) = B integral (ds) = B (2 pi r) = mu * N * I
since it has N loops

B is not constant, so for flux we must integrate:
flux = integral (B dA) = integral (mu*N*I/(2 pi r) dA)

This is where I'm confused:
the solution says:
"The area element dA can be expressed as dA = h dr where h is the height of the rectangular cross section."
Then it proceeds to integrate from r1 to r2 to get: flux = N*I*h/(2 pi)*mu*ln(r2/r1) and the question is solved. But I don't understand how that substitution can be made. Doesn't that imply dA/dr = h so A is a function of r, or A = hr + c??

Any help will be appreciated greatly.
 
Last edited:

Answers and Replies

Related Threads for: Ampere's Law: Toroids

  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
2
Views
800
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
4
Views
1K
Top