0.1. POTENTIAL ENERGY OF A MAGNETIC DIPOLE 215
Quantity Electric Dipole Magnetic Dipole
Dipole Moment ~p=q~ℓ ~m=N IAnˆ
Force in Uniform Field F~= 0 F~= 0
Torque in Uniform Field ~τ=~p×E~ ~τ=~m×B~
Potential Energy U=−~p·E~ U=−~m·B~
Force in Non-Uniform Field F~=−∇~U F~=−∇~UTable 3: Similarity of results for the electric and magnetic dipoles in (orlater, as the source of) their
respective fields.
Example 6.3.3: The Magnetic Moments of Rotating Charged Objects
Not all current carrying wires or current densities will have magnetic dipole moments that are easy
to compute. In fact,mostcurrent densities will have moments that aretoo difficultto compute with
anything less than a computer! Imagine a spool of wire tangled up likefishing line with a current
running through it – this is only one of the infinity of arbitrary shapesto consider, most of which
cannot even be expressed as a simple function of three dimensionalcoordinates! Still, our plane figure
result above appears to bevery usefulbecause when we as humans design a magnetic apparatus
(say, a motor) we can certainlychooseto wrap our coils in a plane (at least approximately). Also,
we can see how to at leastformulatethe problem for arbitrary currents. We are therefore done (for
this level of instruction) with current loops.
There is one more generic distribution of moving charge that we verymuch need to consider before
quitting. A surprisingly common occurrence in physics is to have a “particle” that is microscopically
more or less a ball with a mass and a charge that isrotatingabout some axis. A proton, for example,
can be modelled as a ball of some radiusrp≈ 10 −^15 meters, containing a massmpand a charge
e. The proton also has aspin– an intrinsic angular momentum – ofLz=~/2 where~is Planck’s
constant over 4π(a number that need not concern us in this course – it is very small in macroscopic
terms but islargeas far as the proton’s physical dimensions are concerned).
If we imagine the proton to be a uniform ball of charge with radiusR, total (uniformly distributed)
chargeQ=e, (uniformly distributed) massM, spinning about some axis through its center at an
angular velocity~ωso that:
~L=I~ω=(
2
5
M R^5
)
~ω (463)then it is clear that the proton willalsohave amagnetic momentparallel toL~. What isnotso
obvious is that this magnetic moment will be directly proportional to the angular momentum in a
way that is independent of the shape of the proton (or even that itis a proton), so that
~m=Q
2 M
~L=μL~ (464)foranysymmetric spinning particle with identically distributed charge and mass, where I have
defined the ratio:
μ=Q
2 M
(465)
as the classical equivalent of the “Bohr magneton” in the quantum physics of the electron^62. Let us
understand this.
Suppose we have a ring of chargeQ, massM, and radiusRspinning at angular speedωabout
its axis of symmetry as drawn in figure 74
(^62) The Bohr magneton of the electron isμB= 2 em~
to make the remaining parameter a dimensionless quantum nume, which we recognize as ourber. μ, but with the units of~appended