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Using Ampere's law to find B just outside finite solenoid

by quasar_4
Tags: ampere, finite, solenoid
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quasar_4
#1
Nov28-10, 03:01 PM
P: 290
1. The problem statement, all variables and given/known data

We have a solenoid of radius a, length L, with ends at z = +/- L/2. The problem is to use Ampere's law to show that the longitudinal magnetic induction just outside the coil is approximately

[tex] B_z (\rho=a^+, z) \approx \left(\frac{2 \mu_0 N I a^2}{L^2} \right) \left(1+ \frac{12 z^2}{L^2}- \frac{9 a^2}{L^2} + \ldots \right) [/tex]

(This is part b of problem 5.5 in Jackson 3rd edition).
2. Relevant equations

Ampere's law: [tex] \oint B \cdot dl = \mu_0 I [/tex]

3. The attempt at a solution
I'm pretty sure I'm thinking about this too simplistically, which is why I'm stuck. For an infinite solenoid, the magnetic induction outside is 0. Since we've got a finite solenoid, there are obviously fringing effects of some sort and I guess we can't expect the field to be 0 outside anymore (though it should be reasonably small compared to the field on axis in the center of the solenoid).

I can't figure out what to do differently with Ampere's law though. Does my amperian loop enclose the coils over the full length L of the solenoid or just a short bit? What I was initially thinking was that if my amperian loop encloses all of them, then

[tex] \oint B \cdot dl = BL = \mu_0 I N L \rightarrow B = \mu_0 N I \hat{z} [/tex]

but this is the result you get with the infinite solenoid inside the solenoid, and is clearly not what I need to have since there's nothing to expand on... so I guess I'm not sure how to set this up. Can anyone help?
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kreil
#2
Nov29-10, 08:09 PM
kreil's Avatar
P: 544
end of page 6/beginning of page 7 is relevant.

http://www.contrib.andrew.cmu.edu/~b...2005EM1HW8.pdf
Chino_br
#3
Mar19-11, 12:35 PM
P: 2
Hey guys, how do you know that for an infinite solenoid, the magnetic induction outside is 0?


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