2
Magnetic Dipole Moment
In this section I will work out the classical dynamics of a rigid charged rotor in the presence
of an external magnetic field. Such an object has associated with it a quantity known as the
magnetic dipole moment, which determines the dynamics. The manipulation of magnetic
dipoles by external magnetic fields, and more generally the interaction of magnetic dipoles
with magnetic fields, is of extreme importance in modern physics and current day research.
2.1
Charged rotor
Consider a rigid rotor of length r, rotating with a speed v, carrying a charge q and a mass
m, as shown in Fig. (1). It is rigid in the sense that it does not get deformed under external
forces nor does it slow down. Some mechanism must of course provide this rotation, but we
will not worry about that here, and just consider the system as is. Moreover, we will assume
that the rotation is extremely rapid, much more rapid than the rest of the dynamics. So for
all intent and purposes we basically have a current carrying ring, with a current given by
the charge times the frequency of rotation,
I=q
v
2⇡r
)
qv = 2⇡rI
(2.1)
We will make good usage of this relation in a moment.
ẑ
~
L
x̂
m, q
ŷ
~r
~v
Figure 1: A particle of mass m and charge q at position ~r rotates around a fixed centre with
velocity v.
The rotation axis is along the angular momentum vector associated with the rotor, which is
of course
~ = m~r ⇥ ~v
L
(2.2)
~ define a cartesian (rotating) coordinate system. In the configThus, the vectors ~r, ~v and L
uration shown in Fig. (1) the angular momentum is pointing in the ẑ direction. By taking
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the cross product with ~r on both sides we can write this relation as,
~ = m~r ⇥ (~r ⇥ ~v ) = m (~r(~r · ~v )
~r ⇥ L
~v (~r · ~r))
(2.3)
~ ⇥ (B
~ ⇥ C)
~ = B(
~ A
~ · C)
~
~ A
~ · B).
~
where in the last equality I used the vector identity A
C(
Now I can use the fact that ~r is always perpendicular to ~v to set ~r · ~v = 0 in Eq. (2.3) and
obtain,
~ =
~r ⇥ L
m~v r2
)
~v = v L̂ ⇥ r̂
(2.4)
~ = mvrL̂ and ~r = rr̂ and reversed the order of the
In the last step I used the fact that L
cross product in account of the minus sign. Thus, the velocity is always perpendicular to
the rotation axis and radial vector, as is clear from Fig. (1).
2.2
Charged rotor in a magnetic field
~ As you
Now, let’s consider what happens when we place the rotor in a magnetic field, B.
know from your introductory courses, the Lorentz force on a moving charged particle due to
a magnetic field is,
~
F~ = q ~v ⇥ B
(2.5)
~
B
~
L
~v
ẑ
F~
x̂
~v
ŷ
F~
F~
~v
Figure 2: A charged rotor in the presence of a magnetic field (in red). In this configuration
where the rotation axis is along the direction of the magnetic field, the Lorentz force is
everywhere outwards. Thus, there is no torque on the rotor.
The magnetic field will not distort the rotor, because we assumed it to be rigid, but it can
rotate the rotation axis. Therefore, I will be interested in computing the torque on the
rotor. Deriving the general case is a bit lengthy and so I leave it to the appendix. Instead,
I will guess the general result by considering two simple configurations where things are
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visibly simpler. First, consider the case where the rotation axis L̂ is in the same direction as
~ as shown in Fig. (2). Using the right-hand rule, you can convince yourself
magnetic field B,
that the force is always pointing radially outwards. You can also see it by plugging Eq. (2.4)
into the Lorentz force law,
⇣
⌘
⇣
⌘
~
~
~
~
F = qv L̂ ⇥ r̂ ⇥ B = qv r̂(L̂ · B) L̂(r̂ · B)
= qvBL r̂ .
(2.6)
As promised, the force is pointing radially outwards. Here I have again used the vector
~ ⇥ (B
~ ⇥ C)
~ = B(
~ A
~ · C)
~
~ A
~ · B)
~ as well as the fact that L̂ and B
~ are parallel in
identity A
C(
this case and both are perpendicular to r̂. Thus, the torque vanishes since the moment arm
is in the same direction as the force,
~⌧ = ~r ⇥ F~ = qvBL ~r ⇥ r̂ = 0
(2.7)
On the other hand, when the angular momentum is exactly perpendicular to the magnetic
field, the situation is as depicted in Fig. (3). You can use the right-hand rule with the
Lorentz force to convince yourself that the forces are such as to try to align the rotor back
with the magnetic field as in Fig. (2). When the particle is in the upper part of the circle
delineating its motion the force is into the page, anti-parallel with the angular momentum.
In the lower part of the circle the force is out of the page, parallel to the direction of the
angular momentum. Let’s see how this comes about explicitly.
~
B
F =
ẑ
~v
qvB sin
x̂
ŷ
F =0
~⌧
~
L
~v
F = qvB
~ over the lower half of
Figure 3: The force is pointing out of the page, in the direction of L
~ over the upper part of the circle. This
the circle, and into the page in the direction of L
result in a torque ~⌧ (in green) pointing to the left that tries to align the rotor back with the
magnetic field as in Fig. (2).
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The position vector is in the ŷ
as,
ẑ plane and can be written in terms of the radius and angle
~r = ŷ (r cos ) + ẑ (r sin ) .
Here
(2.8)
is the angle from the positive ŷ axis, as shown in Fig. (2).
The velocity vector is also in ŷ
ẑ plane and is given by,
~v =
ŷ (v sin ) + ẑ (v cos )
(2.9)
(you can obtain this by di↵erentiating Eq. (2.8) or by simply inspecting Fig. (3)). The force,
which is given by the Lorentz force Eq. (2.5), can then be worked out to be,
~ = q ( ŷ (v sin ) + ẑ (v cos )) ⇥ (B ẑ)
F~ ( ) = q ~v ⇥ B
=
x̂ (qvB sin )
(2.10)
where in the last step I used ŷ ⇥ ẑ = x̂ and ẑ ⇥ ẑ = 0. Using the expression we just found
for the force together with the position vector Eq. (2.8) we can work out the torque,
~⌧ ( ) = ~r ⇥ F~ = (ŷ (r cos ) + ẑ (r sin )) ⇥ ( x̂ qvB sin )
= qvBr
✓
ẑ(cos sin )
ŷ(sin
2
)
◆
(2.11)
Let’s now average over one cycle, to get the net torque acting on the loop in this configuration,
Z 2⇡
Z 2⇡
1
qvBr
~⌧ =
~⌧ ( )d =
ẑ(cos sin ) ŷ(sin2 ) d
(2.12)
2⇡ 0
2⇡
0
=
ŷ
qvBr
2
(2.13)
Using the relation between the velocity and the current, qv = 2⇡rI (see Eq. (2.1)) we finally
arrive at,
~⌧ =
ŷ ⇡r2 I B
(2.14)
The combination ⇡r2 I is the magnitude of what is known as the magnetic dipole moment.
It is actually a vector that points in the same direction as the angular momentum and so I
will write it as,
µ
~ = ⇡r2 I L̂
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(2.15)
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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