- #1

- 117

- 15

## Homework Statement

The problem is to show that the potential energy of a dipole (whether electric or magnetic) is given by ##U = -\vec{m}\cdot\vec{B}## in the magnetic case and ##U = -\vec{p}\cdot\vec{E}## in the electric case assuming the respective magnitudes of the dipole moment remain fixed. Since the argument is identical in both cases, I will stick with the magnetic dipole throughout.

## Homework Equations

Torque on a magnetic dipole: ##\vec{N} = \vec{m} \times \vec{B}##

Force on a magnetic dipole: ##\vec{F} = \nabla(\vec{m}\cdot\vec{B})##

## The Attempt at a Solution

First of all, the potential energy of a dipole is the work required to bring it in from infinity and place it in its final position. The question now is how should the dipole be brought in. I propose to move the dipole in from infinity when it already points in its final direction. Thus, I will not have to do any rotational work and the only work that is left is translational. From here the argument is straight-forward: [tex]U = -\int\vec{F}\cdot\vec{dl} = -\int_\infty^r \nabla(\vec{m}\cdot\vec{B})\cdot\vec{dl} = -\vec{m}\cdot\vec{B}|_\infty^r = -\vec{m}\cdot\vec{B}[/tex] where I have used the fundamental theorem for gradients and the assumption that the energy at ##\infty## is zero. However, virtually all of the websites I have seen derive the equation via bringing the dipole so that it is perpendicular to the field and thus, do no translational work. Then they calculate only the rotational work. (See for example http://hyperphysics.phy-astr.gsu.edu/hbase/electric/diptor.html and http://www.chem.ox.ac.uk/teaching/Physics%20for%20CHemists/Electricity/Dipoles.html [Broken]).

Is there anything flawed with my argument that causes everyone to apply the rotational one? As far as I can see, my argument is simpler and I can detect no problems with it.

Any comments/corrections will be greatly appreciated.

Last edited by a moderator: