Multipole expansion for Magnetic Vector Potential

 Posts: 378
 Joined: Wed Mar 26, 2008 9:14 am
Multipole expansion for Magnetic Vector Potential
I am studying Griffiths and I have the following doubt. For the multipole expansion of the magnetic vector potential Griffiths simply writes the formula for a loop of wire and then ofcourse $$\oint d\mathbf{l}$$ is 0 around a loop (eq 5.78 and 5.80) and then he concludes that there is no monopole, but think about a finite segment of line charge that is moving at v(<<c) in that case the integral will not be a closed one, how do we then account for the fact that there are no monopoles then? Also why has he put a closed integral in the first place?
Re: Multipole expansion for Magnetic Vector Potential
You are probably overthinking this.
The Griffiths derivation must include the closed loop integral by the setup of the problem.
In the case of a line charge, think of the magnetic field produced. It must curl around the line charge and close in on itself. Thus, like all magnetic fields, the field produced has 0 divergence and therefore no monopole moment.
I don't have Griffiths in front of me but I'm sure if you work it out for a line charge you will conclude, if you do it correctly, that there is no monopole term.
The Griffiths derivation must include the closed loop integral by the setup of the problem.
In the case of a line charge, think of the magnetic field produced. It must curl around the line charge and close in on itself. Thus, like all magnetic fields, the field produced has 0 divergence and therefore no monopole moment.
I don't have Griffiths in front of me but I'm sure if you work it out for a line charge you will conclude, if you do it correctly, that there is no monopole term.