How to Set Lower Limit for Intergration on Vector Magnetic Potential Problem

[yungman]

Homework Statement

My question is mainly on the set up of the limits of the integral.

The original question is:

Find vector magnetic potential A distance s from a infinite long straight wire carrying DC current I.

The Attempt at a Solution

Let wire on z-axis and \vec I = \hat z I

\nabla \times \vec A = \vec B = \hat {\phi} \frac {\mu_0 I}{2\pi r}

We know A is same direction as I and is function of r in this case.

\nabla \times \vec A = \nabla \times \hat z A_{(r)}} = \hat {\phi} \frac {\partial A_{(r)}}{\partial r} = \hat {\phi} \frac {\mu_0 I}{2\pi r}

\Rightarrow\; \vec A_{(s)} = \frac {\mu_0 I}{2\pi} \int \frac 1 r dr

The book use

\vec A_{(s)} = \frac {\mu_0 I}{2\pi} \int^s_b \frac 1 r dr

Where b is a constant.

My question is how do you justify using b as the lower limit. I understand b cannot be zero. But how do you justify using b constant?

My main question is how do you set the lower limit in this case.

Thanks

 

[Mike Pemulis]

I think it's just a case of the familiar idea that you can always add an arbitrary constant to the potential. The lower limit on the integral adds a constant to A, and since the derivative of a constant is 0, that doesn't affect the B field. So your last integral is the potential at radius s relative to some point b.

 

[yungman]

I think it's just a case of the familiar idea that you can always add an arbitrary constant to the potential. The lower limit on the integral adds a constant to A, and since the derivative of a constant is 0, that doesn't affect the B field. So your last integral is the potential at radius s relative to some point b.

Thanks a lot, now I remember that I can choose any A as long as I can satisfy

\nabla \times \vec A =\vec B \;\hbox { and } \nabla \cdot \vec A = 0

and any b will satisfy this!

Thanks

Alan

 

[ideasrule]

b is just an arbitrary constant. You can easily show that whatever b happens to be, the equation curl A=B is still satisfied.

Incidentally, the book's solution is not the most general possible. You can add any curl-free field to it and get an expression that still satisfies curl(A)=B.

 

[ideasrule]

Oops, Mike already answered.

 

[Antiphon]

Your equation (and therefore your integral) are wrong.

Where is s?

The integral limits must run from - to + infinity along the wire.

 

[ideasrule]

Your equation (and therefore your integral) are wrong.

Where is s?

They don't. There's no s because the OP decided to use r to represent distance from the wire. His solution would be perfectly fine if he used s instead.

The integral limits must run from - to + infinity along the wire.

They don't. The indefinite integral solves curl(A)=B; a definite integral is equivalent to a constant, and would always have curl(A)=0.