gabbagabbahey

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**Re: "paradox" regarding energy of dipole orientation**

My previous post was in error (I hadn't fully read all of the posts to see that you were in fact having a very different issue); Sam's resolution of the "paradox" is correct. For a detailed discussion of it, see R.H. Young,Sam,

I thought you could work out problems with that involve M in two ways:

Method 1:

Given M and the free current:

- Calculate the bound current

- Calculate the magnetic field using these current densities

- Calculate whatever you need using the magnetic field now

method 2:

Given M and the free current:

- Calculate H

- Calculate whatever you need using B and H now

I thought these two methods (use the bound current, or use H) were equivalent. In particular, when I had to take an EM course, I remember them requiring us to do many problems using both methods to apparently drive home that either way worked. So shouldn't I get the same energy either way?

What you brought up seems to clearly answer no. It seems obvious now, but I will need to read more to jive these two thoughts in my head. This however seems to raise another issue since using the relations already derived, and given that H is defined as [itex]H= \frac{1}{\mu_0} B - M[/itex] we arrive at the relation between B_external and M inside the sphere as [itex]H= \frac{1}{\mu_0} B_{ext} + (\frac{2}{3} M) - M = \frac{1}{\mu_0} B_{ext} - (\frac{1}{3} M)[/itex]. This seems to raise another "magnitude of the energy" issue.

Thanks for the input. I think you've sketched together enough that I can see all the answers are in your suggestions and examples there ... I just need to sort through this some more. At least now it is abundantly clear how the magnetized and non-magnetized situations arrive at different signs.

*Am. J. Phys*

**66**, 1043 (1998) --- In which Young answers Griffiths' question on why the Hamiltonian is [itex]-\vec{\mu}\cdot\vec{B}[/itex]--- and references cited therein.