Download 2.3 Gyromagnetic Ratio - McMaster Physics and Astronomy

More generally, the magnitude of the magnetic dipole moment is simply the current times
the area of the region enclosed by the current,
µ
~ = A I L̂
(2.16)
~ is pointing in
In the configuration we are considering µ
~ is pointing in the x̂ direction and B
the ẑ. Since x̂ ⇥ ẑ = ŷ, we could have written the resulting torque, Eq. (2.14) as,
~⌧ = x̂ ⇥ ẑ ⇡r2 I B = µB µ̂ ⇥ B̂
(2.17)
Let’s summarize what we have found. When the rotation axis is along the magnetic field as
~ and the resulting torque is zero.
in Fig. (2) the magnetic dipole moment µ
~ is parallel to B
When the rotation axis is perpendicular to the magnetic field as in Fig. (3) the magnetic
~ and the resulting torque is simply µB pointing in the
dipole moment is perpendicular to B
direction of µ̂ ⇥ B̂ as in Eq. (2.17). It seems natural to guess that in the general case when
the magnetic dipole moment is at some angle with the magnetic field the torque is given by,
~
~⌧ = µ
~ ⇥B
(2.18)
I prove in the appendix that this formula indeed gives the correct torque for any orientation
of the rotor with respect to the magnetic field.
2.3
Gyromagnetic Ratio
The magnetic dipole moment is always pointing in the direction of the angular momentum
and so very generally we have the relation,
~ .
µ
~= L
(2.19)
Here the factor is known as the gyromagnetic ratio. We know from dimensional analysis1
that has dimensions of charge over mass. For good reasons, people often prefer to talk
about dimensionless quantities and so they introduce a dimensionless factor, g, known as
the g-factor such that
=
gq
.
2m
(2.20)
The factor of two is by convention. To make things confusing, you will sometimes see g (the
g-factor) being referred to as the gyromagnetic ratio as well. As we shall see in a moment, the
1
The magnetic moment has units of area times current, as in Eq. (2.16), and thus as units of
[length]2 [charge]/[time]. Angular momentum, being the product of distance and momentum has units of
[length]2 [mass]/[time].
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gyromagnetic ratio can be thought of as a coupling constant that determines how strongly
the magnetic dipole responds to an external magnetic field.
But, before we do so, let’s work out what is the gyromagnetic ratio in the case of a charged
rotor. From Eq. (2.15) we see that the magnitude of µ
~ can be written as,
|~µ| = ⇡r2 I = ⇡r2
qv
qvr
q
q ~
=
=
mvr =
|L| .
2⇡r
2
2m
2m
(2.21)
Here I have again made use of the relation between the velocity and the current qv = 2⇡rI
as in Eq. (2.1). Thus, we can write the magnetic dipole moment as,
q ~
L
2m
µ
~=
(2.22)
Thus, in the particular case of the rotor we have for the gyromagnetic ratio,
=
q
2m
or
g=1
(2.23)
In general the gyromagnetic ratio has to be calculated for a given configuration or extracted
from experiment (but, like many dimensionless quantities in physics the g-factor is typically
an order unity number). It is an extremely important quantity that appears in both classical
physics, but especially in quantum mechanics and the interaction of atoms and fundamental
particles with magnetic fields.
2.4
Interaction Energy
F~
d~r
~r d✓
rotation axis
Figure 4: A force applied on a mass with a rigid moment arm ~r result in a rotation by an
angle d✓. This motion corresponds to a distance dr = rd✓ perpendicular to the moment
arm.
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As you know from your elementary courses on mechanics, torques do work and so there
is a potential energy associated with the interaction between the magnetic field and the
magnetic dipole. Let’s first work out the general case. The work done by a force is as always
the integral over the distance from one position to another,
Z rf
W =
F~ · d~r
(2.24)
ri
Now, in general a force would result in motion along the position vector ~r as well as motion
transverse to ~r. It is the second case that we are interested in since that corresponds to
rotation, as shown in Fig. (4). Thus, we are only interested in motion where,
d~r = rd✓ ✓ˆ
(2.25)
where ✓ˆ is a vector perpendicular to the moment arm ~r and along the direction of rotation,
as shown in the figure. Therefore the infinitesimal work is
F~ · d~r = F? r d✓
(2.26)
where F? is the component of the force perpendicular to the moment arm, ~r. You will easily
recognize this as the magnitude of the torque, |~⌧ | = F? r, and so the work done in rotating
about a fixed axis is2 ,
Z ✓
Wrotation =
|~⌧ | d✓
(2.27)
0
This is an entirely general result. Let’s now apply to our particular case of rotating the
magnetic dipole moment by an external magnetic field.
The torque in our case is given by Eq. (2.18). Let’s denote the angle between the magnetic
field and the magnetic dipole moment by ✓. Then the magnitude of the torque is,
~ = µ B sin ✓
|~⌧ | = µ
~ ⇥B
Performing the integral we obtain the work,
Z ✓
W =
µ B sin ✓d✓ = µB(1
(2.28)
cos ✓)
(2.29)
0
We can therefore associate a potential with the interaction between an external magnetic
field and a dipole moment. Dropping the irrelevant constant term we write the potential as,
V (✓) =
µB cos ✓ =
~
µ
~ ·B
(2.30)
This is an extremely important relation that underlies the dynamics of magnetic dipole
moment, both classical and quantum. I am going to use it very frequently in what follows.
Note that the lowest energy state corresponds to a magnetic dipole aligned with the magnetic
field.
2
The limits of integration changed from position to angles where I defined the angle associated with ri
as 0 and that associated with rf as ✓.
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2.5
Dipole Dynamics
The first consequence of the expression for the energy of a dipole in a magnetic field,
Eq. (2.30), is that in the absence of dissipation energy is conserved and the angle between
~ does not change. Hence ✓ is a constant. So, how does
the dipole µ
~ and the magnetic field B
the torque a↵ect the dipole moment?
To figure this out, let me first quickly derive something you might already know, the relation
between the torque and the rate of change of the angular moment. Newton’s second law
states that,
d~p
F~ =
dt
By taking the cross-product with ~r on both sides we obtain,
(2.31)
d~p
d
~r ⇥ F~ = ~r ⇥
=
(~r ⇥ p~)
(2.32)
dt
dt
In the last equality I used the fact that the moment arm does not change with time. Thus,
as you probably already know from an elementary course in mechanics, we have a relation
~ = ~r ⇥ p~
between the torque ~⌧ = ~r ⇥ F~ and the rate of change of the angular momentum L
with respect to time,
~⌧ =
~
dL
dt
(2.33)
Using the relation between angular momentum and the magnetic moment, Eq. (2.19) and
the expression for the torque, Eq. (2.18), this can be written as,
~ =
µ
~ ⇥B
1 d~µ
dt
d~µ
=
dt
)
~
µ
~ ⇥B
(2.34)
The equation of motion is telling us that the change in the vector µ
~ is always perpendicular
~
to that vector and the magnetic field B. Thus, as shown in Fig. (5), the vector µ
~ will simply
precess around the magnetic field, maintaing a constant angle ✓ with it and a constant rate
of rotation. Let’s see this more explicitly and derive the rate of rotation.
~ = 0),
By taking the dot product of Eq. (2.34) with µ
~ itself we immediately find (since µ
~ ·(~µ⇥B
d~µ
d
d
= (~µ · µ
~ ) = (µ2 )
(2.35)
dt
dt
dt
Thus, the magnitude of the dipole, µ, is conserved. Let’s then write down the dipole explicitly
in cartesian coordinates,
✓
◆
✓
◆
✓
◆
µ
~ = µ sin ✓ cos '(t) x̂ + µ sin ✓ sin '(t) ŷ + µ cos ✓ ẑ
(2.36)
0=µ
~·
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~
B
~µ
Figure 5: The torque on the rotor will result in a precession of the magnetic dipole moment
about the external magnetic field. The direction of precession shown assumes a positive
~ and µ
charge so that L
~ are parallel (rather than anti-parallel).
Both µ, the magnitude, and ✓, the angle with the magnetic field do not depend on time. Only
the angle in the x̂ ŷ plane has possible time dependence. Plugging this expression into the
equation of motion, Eq. (2.34) we can cancel the dependence on µ and equate components,
x̂ component:
d cos '
= B sin '
dt
(2.37)
ŷ component:
d sin '
= B cos '
dt
(2.38)
~ = B ẑ and the usual relations x̂ ⇥ ẑ =
Where I have used B
These two equations are in fact a single relation,
d'
=
dt
B
)
'(t) =
ŷ, ŷ ⇥ ẑ = x̂, and ẑ ⇥ ẑ = 0.
Bt.
(2.39)
In the last step I assumed that t = 0 is chosen to coincide with ' = 0. The negative sign is
just telling us that the dipole precesses clockwise, as you can see directly from Fig. (5). Thus
we proved that indeed the dipole precesses around the magnetic field, and with an angular
frequency of
!= B
(2.40)
This is an incredibly important relationship and is known as Larmor frequency. Experiments
typically have good control over two of these quantities and this relation can be used to
measure the third. For example, by placing a known magnetic field and observing a specific
angular frequency one can determine the gyromagnetic ratio to be,
=
!
.
B
(2.41)
Incidentally, this explains the origin of the name, is indeed the ratio of the rate of gyration
to the magnetic field, hence gyromagnetic ratio. Alternatively, if is known (from previous
measurements) it can be used to determine the magnetic field present if the angular frequency
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can be detected. This is one of the principles behind Magnetic Resonance Imaging, where
the strength of the inferred magnetic field is then correlated with position to produce an
image in space.
Most systems do dissipate and energy is slowly lost to the environment. The dipole continues to precess, but the angle ✓ with the magnetic field slowly diminishes. After some
characteristic time, the system will relax to its ground state and the magnetic dipole will
align with the magnetic field.
2.6
Magnetic field of a dipole moment
While we will not need it for anything that follows, the physical picture I presented would
not be complete without recognizing that our little charged rotor itself creates a magnetic
field, as does in fact every magnetic dipole moment. The magnetic field of the charged rotor
is shown in Fig. (6). Since it is outside the scope of our analysis, I will not try to prove it to
you, but it is a rather straightforward exercises in applying the basic laws of electrodynamics
(in particular the Biot-Savard law). The behavior of the magnetic field lines far from the
charged rotor are universal in that every other magnetic dipole would result in the same
behavior.
~ dipole
B
~µ
I
Figure 6: Magnetic field associated with a charged rotor. The behavior of the field lines far
from the rotor are universal and would behave the same for any magnetic dipole.
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