Theory

Consider a circular current loop in an external magnetic field . The force on an infinitesimal section of this loop of length dl is $$dF=IBdl\sin \theta$$

This force produces a torque about the axis AC of magnitude

$$d\tau =dFr\sin \theta$$

Figure 1: Torque on a current loop

The torque on each half of the loop is found by integrating , using the relation , to be

$$\tau =\int _{0}^{\pi }IBr^{2}\sin ^{2}\theta d\theta =IB\frac{\pi }{2}r^{2}$$

since the torque on each half of the loop acts in the same direction, the total torque is
 
 

$$\tau =I\pi r^{2}B=IAB$$

This result, although derived for a circular loop, actually applies to loops of arbitrary shape. The quantity IA is known as the magnetic moment of the loop, and is generally denoted by the symbol . It is convenient to define  as a vector quantity with magnitude IA which points in the direction normal to the plane of the loop. To see why, note that the above result has been derived assuming that the magnetic field lies in the plane of the loop. If it does not, then the torque becomes

$$\tau =\mu B\sin \phi$$ (1)

where  is the angle between the direction normal to plane of the loop and the magnetic field . Furthermore, the direction of the torque is perpendicular to both  and . We can therefore express Eq. 1 in vector form as

$$\overrightarrow{\tau }=\overrightarrow{\mu }\times \overrightarrow{B}$$

The results for a current loop may be used to find the appropriate expression for a charge moving around a fixed path. A charge q moving at velocity v around an circular orbit of radius r represents an average current

$$I=\frac{q}{t}=\frac{q}{\frac{2\pi r}{v}}=\frac{qv}{2\pi r}$$

The expression for  now becomes

$$\mu =\frac{qvr}{2}$$

Using the relation L=mvr, this expression can be rewritten

$$\overrightarrow{\mu }=\frac{q}{2m}\overrightarrow{L}$$ (2)

This expression must be modified to account for the fact that the electron is not a point object but a charge distribution. To do this, the charge q is multiplied by a dimensionless factor g, which is known as the gyromagnetic ratio. If the angular momentum of the electron is due to spin only, then g = 2.0023. Equation 1 then becomes

$$\overrightarrow{\mu }=\frac{gq}{2m}\overrightarrow{L}$$ (3)

The next step toward resonance is to consider the effect of a static magnetic field on . Equating torque to the time rate of change of L, we obtain the expression

$$\frac{d\overrightarrow{L}}{dt}=\overrightarrow{\mu }\times \overrightarrow{B}$$

If we multiply this expression by , it follows from Eq. 3 that

$$\frac{d\overrightarrow{\mu }}{dt}=\frac{gq}{2m}\overrightarrow{\mu }\times \overrightarrow{B}$$ (4)

Figure: Precession of  around 

From the properties of the vector cross product, we know that the rate of change of  is perpendicular to both and . As a result, precesses around  (see Figure 2). To see how this comes about, we expand the equations of motion (Eq. 4) to obtain

$$\begin{array}{cccc}\frac{dL_{x}}{dt}= & \frac{d(L\sin \theta......c{dL_{x}}{dt}= & \frac{d(L\cos \theta )}{dt}=0 & &\end{array}$$

does not change magnitude, since the torque is perpendicular to it. It follows from the third equation that  is a constant of the motion. Also, from the first equation we obtain

$$\frac{dL_{x}}{dt}=L\sin \theta \cos \phi \frac{d\phi }{dt}=\frac{gq}{2m}LB\sin \theta \cos \phi$$ (5)

The quantity  is called the Lamor frequency. From Eq. 5, it is equal to

$$\omega _{0}=\frac{gq}{2m}B$$ (6)

The Lamor frequency can be measured by driving the system with a weak oscillating magnetic field B₁. When the drive frequency is equal to the Lamor frequency, resonance occurs and the system absorbs maximum energy from the driving field. Once this value is known, Eq. 6 can be solved for the gyromagnetic ratio.

1999-09-14