The Magnetic Field of Coaxial Cables

Click For Summary
SUMMARY

The discussion focuses on determining the magnetic induction B in a coaxial cable with a thin-walled cylindrical inner conductor carrying current I. Using Ampere's Law, the participants analyze three scenarios: when the radius r is less than a, between a and b, and greater than b. The key equation applied is \(\oint(B.dl) = \mu_0 I\), where I represents the net current through the Amperian loop. Clarifications on the value of I for each case are essential for accurate calculations.

PREREQUISITES
  • Understanding of Ampere's Law in magnetostatics
  • Knowledge of coaxial cable structure and current flow
  • Familiarity with magnetic induction concepts
  • Basic calculus for evaluating integrals
NEXT STEPS
  • Study the application of Ampere's Law in different geometries
  • Learn about magnetic fields in coaxial cables
  • Explore the concept of enclosed current in Amperean loops
  • Review the principles of magnetostatics and magnetic induction
USEFUL FOR

Students studying electromagnetism, electrical engineers, and anyone interested in the magnetic properties of coaxial cables.

Ghassan_2k
Messages
1
Reaction score
0

Homework Statement


A coaxial cable of length l, has a thin wall cylindrical inner conductor of radius a carrying current I. All current returns along the outer conductor whose inner radius is b.

Assuming that the current flows only in a thin layer on each conductor, determine the magnitude and direction of the magnetic induction B at radius r.

a) when r<a
b) when a<r<b
c) when r>b

Homework Equations


\oint(B.dl) = \mu0 I


The Attempt at a Solution


Ive just started studying magnetostatics and am a little bit confused. I think using Ampere's Law in each of the three cases is the right thing to do, but I am unsure on what the value for I is in each case.

For part a), i think B * 2 * pi * r = \mu0 * I
but then i don't know how it differs for part b).

Thanks for any help
 
Physics news on Phys.org
The I_{enc} in Ampere's Law is the net current flowing through your Amperean loop. So, what are you using for your Amperian loop?
 

Similar threads

Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K
Replies
2
Views
2K
  • · Replies 16 ·
Replies
16
Views
3K
Replies
10
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K