The Magnetic Moment
or
If I Can Figure it Out So Can You

This page uses a lot of Greek symbols. The font is SPIONIC.TTF and is available for free download here. You will also need a bit of familiarity with very basic calculus and vector algebra.

The idea of the magnetic moment is pretty simple and comes from some basic observations of the physical nature of electric currents and magnetic fields.

The most basic observation is that a charge, q, moving through a magnetic field, B, with velocity, v experiences a force, F, which is directed at right angles to both v and B.

This is, of course, a vector cross-product from which we can get the magnitude of F quite easily:

F = qvBsin(f)

where f is the angle between v and B. We can write the first vector equation in terms of infinitesimals:

dF = dq(vXB)

and since v represents the movement of charge per unit time, dl/dt, we can write:

dF = (dq/dt)(dlXB)

If you integrate over the length of the wire to get the total force of the current:

F = i(lXB)
(i = dq/dt)

Now, if l and B are at right angles to each other then we can calculate the magnitude of F:

F = ilB (f = 90o so sin(f) = 1)

where i is the current, l is the length of the wire and B is the magnitude of the magnetic field.

Now, imagine a current conducting coil in a magnetic field at an angle, f to the field, B:

The first diagram is a side view of the coil and the second is a top view. Here, the field, B, in the second diagram is directed away from the viewer into the page as indicated by the x's. The current carrying coil is rectangular in shape with sides of length a and b. The force, F, is then given by:

F = iaB

and:

F' = ibBsin(f)

Forces F' on sides b cancel each other out but forces F constitute a torque:

G = iaBbsin(a)

The torque, G, is at a minimum when the angle a is zero and a maximum when it is 90 degrees. Since a times b is the area inside the coil we can write:

G = iABsin(a)

iA is referred to as the magnetic moment and is symbolised with the greek letter m. Thus:

G = mBsin(a)

The vector version of this equation is:

G = mXB

where m is the vector version of the magnetic moment. The vector direction is, of course, at right angles to the plane defined by m and B and, using the right-hand rule, we find that in the above case the torque is directed along the axis of rotation and is colinear with the angular momentum of the rotating coil.

Note that in this case the torque and the angular momentum are colinear. When this is not the case the rotational system executes a somewhat more complex precessional motion. Consider the rotating top:

In this case the torque, G is perpendicular to angular momentum, L.

G = mgXR

So, rather than a change in the magnitude of the angular momentum as in the case of the coil, the direction of the angular momentum changes. The top precesses about the pivot point. You're familiar with this effect if you have ever played with a child's top. The top is spun up to speed, let go at which point it begins to 'wobble' or precess, due the force of gravity.

The z-component of the spin magentic moment is proportional to the z-component of the spin angular momentum:

mz = iA (i-current, A-area)

= qvpr2/2pr
(i = q/t, A = pr2, t = v/2pr)

= (q/2m)mvr

=(q/2m)Lz

(q/2m) is often replaced with g and is called the magnetogyric ratio (or the gyromagnetic ratio). The same arguement can be made for the x and y components so that, generally:

m = gL

When the spin magnetic moment of a particle such as, oh ... a nucleus let's say, finds itself in a magnetic field acts much as the child's top does in a gravitational field. The magnetic moment precesses about the magnetic field. How fast, you ask? Depends on the field strength and the nucleus.

G = dL/dt
(from basic physics ... look it up)

dL = Gdt

= mXBdt

and, using the proportionality between m and L:

dm = g(mXB)dt

=gmBsin(q)dt

Also:

dm = msin(q)df

=gmBsin(q)dt
(from above)

so that:

df/dt = gB

or:

w = gB

w is the angular Larmor frequency in 'natural units' or radians per second. To convert to hz we divide by 2p. Note that it is proportional to both the magnetic field and the gyromagnetic ratio. This means that for a given nucleus, the precessional frequency will double if the field strength doubles. It also means that for a given field strength, different nucleii with different values of g will have different precessional frequencies.

We can calculate the Larmor frequency of a proton with g=2.67522 x 108s-1T-1 in a field of 11.75 Telsla:

w = 2.67522 x 108s-1T-1 x 11.7T

= 3.143 x 109s-1

n = w/2p

= 500.3 MHz

That's how fast! A magnet of this field strength is usually referred to as a 500 MHz magnet rather than its 11.75 Tesla field strength and in general, the approximate Larmor frequency of protons in a magnet's field is refered to as the magnet's 'field'.

The energy of a magnetic moment can be worked out by considering the action of a torque on the magnetic moment, as in an external magnetic field, B. For the coil example above, the torque rotates the coil and hence, the magnetic moment towards the direction of the field and in the process does work. The potential energy of the coil depends on the angle that it's magnetic moment makes with the field. We can come up with an analytical expression for this by integrating the angular momentum expression:

dW = Gdq
(basic physics by analogy with W = Fs)
(q is the angle that m makes with B)

W = ∫Gdq

= ∫mBsin(q)dq

= mBcos(q) + C

If we arbitrarily take the q = p/2 position as the zero point for the energy then we can say:

E = -mBcos(q)

or in dot-product vector notation:

E = -m.B

Amazingly, I have never seen this explicitly derived in any textbook ... frequently the result is simply stated. Yet this is needed to build the Hamiltonian for the quantum mechanical description of the behaviour of magnetic moments!!

Also, if we integrate over the range q=0 to q=p we will obtain an energy difference of 2mB. This is the energy difference between the compass needle (which has a magnetic moment) aligned north and the needle aligned south.

The Hamiltonian is constructed from the above energy equation:

H = -ghB0Iz

where B0 is the field along the z-axis.