# Find magnetic flux density(B) via vector potential(A)

1. Apr 22, 2014

### baby_1

Hello
As you know we can find the flux density (B) via vector potential with this equation

So with this equation and solution at first i try to find vector potential completely then use above equation

Problem:

Solution:

my approach:

now we use curl A to find B

we find that

but it is different form the book soltuion

what is my problem and wrong?

Thanks

Last edited: Apr 22, 2014
2. Apr 22, 2014

### TSny

Note how you wrote R in terms of x, y, z. Is the 3/2 power correct?

3. Apr 25, 2014

### baby_1

Thanks Dear Tsny.you'r correct
as i want to solve the new integral

a new problem appears

for example if i want to calculate the inner integral the integral going to be infinite.

Ln(infinite)=infinite

now how can solve this problem?
Thanks

4. Apr 25, 2014

### TSny

Yes, the integral is infinite. It's similar to trying to finding the potential V of an infinite line charge by integrating dV = k dq/r. Here you will also get a divergent integral. This is due to the charge distribution extending to infinity while at the same time trying to take the potential to be zero at infinity. (Note kdq/r goes to zero at infinity.)

You can avoid the problem by not demanding that the potential be zero at infinity. For example, in the infinite line charge example, you can pick an arbitrary point in space to be zero potential. The potential at other points is then defined relative to this chosen point.

You can try a similar trick for the vector potential. Thus define A as the following (with constants left out)

$$\int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{\sqrt{x^2 +y^2+z^2}} dx dy - \int_{0}^{\infty} \int_{0}^{\infty} \frac{1}{\sqrt{x^2 +y^2+z_0^2}} dx dy$$

$z_0$ is just an arbitrarily chosen point on the z axis where you are taking A to be 0.

Each integral individually diverges. But if you combine the integrals before letting the upper limits go to infinity, you get a finite result that you can use for A.