# Vector Potential

1. Mar 7, 2005

### heman

How to find the Vector Potential of an infinite solenoid with n turns per unit length,radius R and current I.
since here current extends to infinty..
How will it be done
Pls Help

2. Mar 7, 2005

### Astronuc

Staff Emeritus
Heman, please try to write the equation for vector potential. Then we can better point you in the right direction.

You can use LateX - see the thread on Math applications of LateX - https://www.physicsforums.com/showthread.php?t=8997 (General physics forum).

3. Mar 7, 2005

### Tom Mattson

Staff Emeritus
The infinite solenoid has a uniform magnetic field parallel to the solenoid axis. Let that axis be the z axis, for concreteness. So $\mathbf {B}=B \mathbf {k}$. Now the vector potential is $\mathbf {B}=\nabla \times \mathbf {A}$. The vector potential is not unique, so any vector-valued function $\mathbf {A}$ whose curl gives you $B \mathbf{k}$ will fit the bill.

4. Jan 23, 2010

### jimmy neutron

You may consider using the following:
then use Stoke's theorem to write surface int(gradXA.dS) = line int (A.dl), with the closed curve chosen suitably.

This gives line int(A.dl) = int(B.ds).

The surface integral is straight forward while a reasonable choic eof the closed curve makes the line integral staright forward.

Of help?

5. Jan 23, 2010

### GRDixon

The line integral of A around a closed path is equal to the flux of B through the path. For an infinitely long solenoid, B = (n)(I)/((eps0)(c^2)) at internal points. (B = 0 at all points outside of the solenoid.) Thus the flux of B inside the solenoid is (pi)(R^2)(n)(I)/((eps0)(c^2)). At a distance r>R from the solenoid's axis, (2)(pi)(r)(A)=flux of B. That is,

A=(n)(I)(R^2)/((2)(eps0)(c^2)(r))

6. Jan 23, 2010

### fluidistic

I don't think so. In fact the thread is 5 years old. It might be of help for further people though.