How would you find the vector potential, A, a distance s from an infinite straight wire carrying a current I?

Answer 1

[Edit: This is a complete rewrite, replacing the note I posted last night.]

As I suspected, this problem is in the physics book, and there are two solutions that give the same answer.

You have a horizontal infinite wire carrying a current I from right to left, and you want to find the magnetic field B at some point p located at a distance s above the wire.

There's "east-west symmetry" here, so the vector at point p will be directed the page.

The Biot-Savart law says that

dB = (mu0/ 4 pi) [(i dx sin t) / r^2]

where the permeability constant mu0 = 4 pi x 10^(-7) T m/A
(tesla-meter/ampere); i is current, dx is the differential element of wire length; t (theta) is the angle formed by the wire and the point p; and r is the distance between p and the differential element.

Then, B = (mu0 / 4 pi) int[(i dx x r) / r^3]
where the x indicates a vector cross product.

In the case of the long straight wire, we can use the symmetry to reduce the cross product to a scalar integral,

B = (mu0 / 4 pi) int[(sin t dx) / r^2]

from negative to positive infinity, and where

r = sqrt(s^2 + x^2)
sin t = s/sqrt(s^2 + x^2)

When you make those substitutions and integrate, you end up with

B = mu0 i / (2 pi s) (Answer)

In the book, they did it an easier way using Ampere's law, another name for one of Maxwell's equations. Ampere's law is: the line integral over a closed loop of B dot ds equals mu0 times i.

Applying this to your long straight-wire problem (right from the book), we again use symmetry to simplify because we know that the B vector is directed the page.

The line integral of B dot dx becomes the scalar line integral of B dx cos t = B times the line integral of dx. But that line integral simply equals 2 pi s, so from Ampere's law,

B(2 pi s) = mu0 i

B = mu0 i / (2 pi s) (Answer)

which is the same result that was obtained from the more difficult Biot-Savart law.