Ampere's Law with a coaxial cable

  • Thread starter Ignitia
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  • #1

Homework Statement

A portion of a long, cylindrical coaxial cable is shown
in the accompanying figure. A current I flows down the
center conductor, and this current is returned in the outer
conductor. Determine the magnetic field in the regions (a)
R ≤ r1, (b) r2 ≥ R ≥ r1, (c) r3 ≥ R ≥ r2, and (d)
R ≥ r3. Assume that the current is distributed uniformly
over the cross sections of the two parts of the cable.


Homework Equations

B∫dl = μI

Inner current = I1
Outer Current = I2

I = J*A
J = Current Density
A = Area

The Attempt at a Solution

(a) R ≤ r1
B∫dl = μI
B*2πR = μ*J*A
B*2πR = μ*(I/πr12)*πR2
B = (μ*I*R)/2πr12
The answer given states it's only B = (μ*I*R)/r12 Why is that?

(b) r2 ≥ R ≥ r1

My thinking was that since this is outside the conductor, B = 0T since I = 0. The answer says the enclosed charge would be I, not 0, but doesn't go into any further detail. Can someone please explain why?

(c) r3 ≥ R ≥ r2
My answer was the same as a and was wrong, and likely related to B. the book stated that I = I1 - I2. Why is this?

(d) R ≥ r3
Since it's outside the conductor, then I = 0 and B = 0T This was correct.

Thanks for the help.


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Answers and Replies

  • #2
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It seems you have some problems understanding what the enclosed current in Ampere's law is. That there is no current where the loop is does not matter. The only thing that matters is how much current passes through an area that has the loop as its boundary. So for example, in the case of ##r_1 < r < r_2##, the area could be a disk with radius ##r##, what would be the current through this disk? This problem affects both (b) and (c).

For (a), there definitely should be a ##2\pi## in the denominator. The field is ##B = \mu_0 I_{\rm enc}/(2\pi r)## and ##I_{\rm enc} = I r^2/r_1^2##.
  • #3
Okay, I see what my confusion is now. I was thinking the area was shaped like a washer, and only integrated from r1 to r2, not from 0 to r2. I got it now!