The Curl of Magnetic Potential

1. Feb 3, 2013

quantumfoam

Hi guys! I recently saw on Wiki that given a magnetic potential A(r)=(u/4π)(mXR/r^3) ,( where u is the permeability of free space, m is the magnetic dipole moment of a magnetic field, R is the position vector, and r is the distance from the magnetic field ) upon taking the curl of the magnetic potential A(r), one obtains (u/4π)([3(m•r)R/r^5]-[m/r^3]). Can someone show me how they ended up with this result? I would like a step by step process please.(: the wiki article is called Magnetic Moment under the section called Effects of an External Magnetic Field on a Magnetic Moment.

2. Feb 3, 2013

SammyS

Staff Emeritus
That's not how we do things in the Homework section of PF. We're not even allowed to give such help. But we can try to help you work your way through such a problem.

One way to get the result, is to write A(r) in terms of it's Cartesian components, then take the curl. I suppose you could also take the curl directly in spherical coordinates.

3. Feb 3, 2013

quantumfoam

I'm sorry! Well I thought about doing it in spherical coordinates but I was not sure if I should use the whole three partials in the del operator.

4. Feb 3, 2013

SammyS

Staff Emeritus
Yes, use all three partials.

5. Feb 3, 2013

Dick

I don't think there is any advantage is using spherical coordinates. Try using the identity:

$$\nabla \times (a \times b)=(b \cdot \nabla) a-b(\nabla \cdot a) -(a \cdot \nabla) b+a(\nabla \cdot b)$$

If you set a to be m and b to be $\frac{\vec r}{r^3}$, the first two terms vanish because m is a constant. The last one vanishes because $\frac{\vec r}{r^3}$ is divergence free. So you only have to puzzle out what the third term means.

6. Feb 3, 2013

quantumfoam

Thank you very much! (:

7. Feb 3, 2013

quantumfoam

Is there any other identity I could use for (a(div))b? I did the calculation but I think I did it wrong. I'm missing the second term in the curl of the magnetic potential. The (m/r^3) term

8. Feb 3, 2013

Dick

Show what you did. You'll need to use the quotient rule on the dervivatives $\frac{\vec r}{r^3}$.

9. Feb 3, 2013

quantumfoam

Well, at first I used the identity (a(div))b= (div(a))b-b(div(a)). Was that a correct move?

10. Feb 3, 2013

quantumfoam

Oh! I get it! Nevermind! I didn't think to carry out the quotient rule. That makes a lot more sense! Thank you!

11. Feb 3, 2013

quantumfoam

So I should calculate using the radial term of the spherical version of the del operator?

12. Feb 3, 2013

quantumfoam

That's the only way I see to take the derivative of R/r^3.

13. Feb 3, 2013

Dick

R/r^3 is just (x,y,z)/(x^2+y^2+z^2)^(3/2). It worked out pretty ok for me in cartesian coordinates.

14. Feb 3, 2013

quantumfoam

Would it be a little easier with spherical coordinates?

15. Feb 3, 2013

Dick

Maybe. I didn't try it that way. I just know it's not hard in cartesian.

16. Feb 3, 2013

quantumfoam

I will try it! Thank you!