Why would a magnetic monopol's field be 1/r^2?

[Yonah]

Hi,
I was reading Purcell's Electricity and Magnetism. It says that if there were a magnetic charge, its field would behave like an electric charge (field pointing out radially and magnitude decreasing as 1/r^2). Why? Could this be deduced from magnetic fields produced by electric charges? If so, how?

I'd appreciate your help!

 

[Dale]

The only reason to think that magnetic monopoles exist is because it would make things like Maxwells equations more symmetrical. If they didn't behave like an electric monopole then that would make the equations asymmetrical, which would defeat the whole point.

 

[Yonah]

Ok, so you are saying that there would be no contradiction among maxwell's equations necesarily (except of course ∇⋅B=0) if a magnetic monopol existed which decreased in a way other than 1/r^2. Things just wouldn't be symetrical in that case. Is that correct?

 

[jtbell]

If the $\vec B$ field from a magnetic monopole did not go like $1/r^2$, what would replace $\vec \nabla \cdot \vec B = 0$ and its integral equivalent $\oint {\vec B \cdot d \vec a } = 0$?

 

[Dale]

Ok, so you are saying that there would be no contradiction among maxwell's equations necesarily (except of course ∇⋅B=0) if a magnetic monopol existed which decreased in a way other than 1/r^2. Things just wouldn't be symetrical in that case. Is that correct?

No, that isn't what I am saying. I am saying that to my knowledge nobody even considers that. The idea is to make Maxwell's equations more symmetrical, not less. I don't know how a change like the one you describe could wind up with a set of equations even remotely similar to Maxwells equations. It would be the complete opposite of what monopole proponents want to accomplish.

 

[Yonah]

Ok, I think I understand what your getting at. Thanks!