218 Week 6: Moving Charges and Magnetic Force
Bo
m = Lμp
θ
τ
Proton
Figure 75: A rotating proton with a magnetic moment ~m=μp~Laligned with its rotation axis
precessesaround an applied magnetic fieldB~ 0 with precession frequencyωp=μB 0 independent of
the particular angleθbetween~mandB~ 0. Note thatμp= 2 mepfrom the previous section.
The magnetic field exerts atorque~τon the magnetic dipole of the proton, out of the page at the
particular instant drawn, that is given by:
~τ=~m×B~ 0 (474)or (using the fundamental definition of the torque as the time rateof change of the angular momen-
tum):
dL~
dt =μp(~L×B~ 0 ) (475)
This is a very important first order, linear, homogeneous, ordinarydifferential equation. It
describesprecession, a common phenomenon in physics, especially when considering motionin ro-
tating frames. Itdoeshave exponential forms for its solution, but we will at first skip overa detailed
treatment of this approach and “solve” it more intuitively, graphically and algebraically.
The magnitude of the torque is given by:τ=μpLB 0 sin(θ)≈∆L
∆t(476)
The direction is out of the page. If we breakL~up into two components, one parallel and one
perpendicular toB~ 0 as shown in the simplified figure??, we can easily see that the torque only
comes from, and only changes, theL⊥component.
The torque is alwaysperpendicularto~L⊥, and hence changes itsdirectionbut not itsmagnitude.
This is a familiar situation in physics – obviously~L⊥turns in a circle of radiusL⊥where~τis always
perpendicular to it. This situation is pictured in an “overhead view” in figure 76 at an instant when
~L⊥has bothxandycomponents.
In a short time ∆t, the angular momentum changes a small amount ∆L⊥that is the length of
the arc on theL⊥circle subtended by the angle through which it turns in that time, ∆φ. That is:
∆L= ∆L⊥=L⊥∆φ (477)