How to Find Vector Magnetic Potential A from Magnetic Flux Density B?

[jezse]

magnetic flux density B for an infinitely long cylindrical conductor of radius b is know:

inside: B = (Mu*R*I)/(2*pi*b^2), R<b

outside: B = (Mu*I)/(2*pi*R^2), R>=b

I need to determine the vector magnetic potential A using only the relation B = curl A (del X A)

 

[quantumdude]

Of course, the vector potential A is not unique, owing to gauge invariance of physical fields. Thus, you can use any gauge you wish, and an easy one to use is the one in which:

A=(1/2)Bxr

I got this from Goldstein, Classical Mechanics. I can't remember if I got the order of B and r correct, you should verify that curl[A] does indeed equal B.

 

[M98Ranger]

Using the definition of curl, we can write:

curl A = (1/R)*(d/dR)(R*A)

Since we know B = curl A, we can substitute in the expressions for B inside and outside the conductor:

For R<b: (Mu*R*I)/(2*pi*b^2) = (1/R)*(d/dR)(R*A)
Solving for A, we get:

A = (Mu*I*R/2)*(ln(R/b))/(pi*b^2) + C1

For R>=b: (Mu*I)/(2*pi*R^2) = (1/R)*(d/dR)(R*A)
Solving for A, we get:

A = (Mu*I*R/2)*(1/R + C2)

Where C1 and C2 are integration constants. Therefore, we can determine the vector magnetic potential A for both inside and outside the conductor using only the given relationship B = curl A.