Ampere's Law - Showing that B at center is due only to wire

Join the discussion
Registration is free. Start your own thread to ask a follow-up.
1 reply · 4K views
whoareyou
Messages
162
Reaction score
2

Homework Statement



In Fig. 29-71, a long circular pipe with outside radius R = 2.6 cm carries a (uniformly distributed) current i =
8.00 mA into the page.A wire runs parallel to the pipe at a distance of 3.00R from center to center. Find the (a) magnitude and (b) direction (into or out of the page) of the current in the wire such that the net magnetic field at point P has the same magnitude as the net magnetic field at the center of the pipe but is in the opposite direction.

http://i.imgur.com/b5aCEF0.png

Homework Equations



Ampere's Law - ##\displaystyle\oint\vec{B}\cdot d \vec{s} = \mu_{0}i_{enc}##

The Attempt at a Solution



I know how to do this question already. Conceptually, we know that the magnetic field at the center of the pipe is due only to the wire. Set that equal to the total field at P and solve. What I'm having trouble understanding is how to show that the magnetic field at the center of the pipe is due only to the wire mathematically. If you draw an Amperian Loop with the center at the wire and passing through the center of the pipe (so radius 3R), the enclosed currents are from the wire and the little piece of arc from the pipe which does not imply that the magnetic field at the center is from the wire alone. So how can one show, mathematically, that at the center of the pipe, the magnetic field is only from the wire?
 
Physics news on Phys.org
You don't want to use that Amperian loop when considering both the wire and the pipe because you don't have symmetry to take advantage of.

Think about using the principle of superposition to calculate the field at the inside the pipe.