Relation between magnetic dipole moment and angular momentum

In classical electrodynamics, a magnetic dipole moment \mu is associated with a current loop (see diagram below), and represents the magnitude and orientation of a magnetic dipole. It is a pseudo-vector, whose direction is perpendicular to the plane of the loop and given by the right hand thumb rule.

The magnitude of a magnetic dipole moment of a current loop is defined as

\mu=IA\; \; \; \; \; \; \; \; 60

where I is current, and A is the area of the loop.

Rewriting eq60 in terms of I=\frac{q}{t}=\frac{qv_{\perp}}{2\pi r} (where q is charge, t is time and v_{\perp} is the tangential velocity of the charged particle) and using A=\pi r^{2}, we have \mu=\frac{q}{2m}rmv_{\perp}, where m is the mass of the charged particle. Substitute eq59 in \mu=\frac{q}{2m}rmv_{\perp}, we have the relation between magnetic dipole moment and angular momentum:

\boldsymbol{\mathit{\mu}}=\gamma\boldsymbol{\mathit{L}}\; \; \; \; \; \; \; \; 61

where \gamma=\frac{q}{2m} is the classical gyromagnetic ratio.

When placed in an external magnetic field , the magnetic dipole moment experiences a torque, whose energy  is

U=-\boldsymbol{\mathit{\mu}}\cdot\boldsymbol{\mathit{B}}\; \; \; \; \; \; \; \; 62



How is eq62 derived?


In order to rotate a current loop, we must do work W against the torque \tau due to the magnetic field \boldsymbol{\mathit{B}}. For a rotating system (see diagram I below), W=-\int F_{\perp}ds, where according to convention, the negative sign is added so that work on the system by the field is positive.

For small \theta, ds=rd\theta. Since \boldsymbol{\mathit{\tau}}=\boldsymbol{\mathit{r}}\times\boldsymbol{\mathit{F}} and W=-\Delta U, we have

U_f-U_i=\int F_{\perp}ds=\int_{\theta_i}^{\theta_f}Fsin\theta rd\theta=\int_{\theta_i}^{\theta_f}\tau d\theta\; \; \; \; \; \; \; \; 62a

Diagram II below shows a square current loop with side 1 parallel to side 3 and side 2 parallel to side 4 (not shown in diagram). The length of each side is b.

Since \tau=r\times F=\frac{b}{2}Fsin\theta+\frac{b}{2}Fsin\theta and F=BIb (\theta=90^{\circ}),

U_f-U_i=\int_{\theta_i}^{\theta_f}IBb^{2}sin\theta\, d\theta=IA B\left (- cos\theta_f+cos\theta_i \right)\; \; \; \; \; \; \; \; 62b

where A is the area of the loop.

Substitute eq60 in the above equation,

U_f-U_i=\mu B\left (cos\theta_i-cos\theta_f \right)\; \; \; \; \; \; \; \; 63

Comparing eq62a and eq62b, torque can also be expressed as

\boldsymbol{\mathit{\tau}}=BIAsin\theta=\boldsymbol{\mathit{\mu}}\times\boldsymbol{\mathit{B}}\; \; \; \; \; \; \; \; 64

The torque exerted by the magnetic field on the magnetic dipole tends to rotate the dipole towards a lower energy state. So, we let \theta_i=90^{\circ} with U_i=0 and eq63 becomes U_f=-\mu Bcos\theta_f, or simply:

U=-\boldsymbol{\mathit{\mu}}\cdot \boldsymbol{\mathit{B}}\; \; \; \; \; \; \; \; 65


Other than a torque, the external magnetic field may exert another force on the current loop. For a magnetic field pointing in the z-direction, \boldsymbol{\mathit{B}}=B_z(x,y,z)\boldsymbol{\mathit{k}}, where B_z(x,y,z) is a scalar function. We substitute eq65 in F_z=-\frac{\partial U}{\partial z} to give

\boldsymbol{\mathit{F_z}}=\frac{\partial(\boldsymbol{\mathit{\mu}}\cdot\boldsymbol{\mathit{B}})}{\partial z}=\frac{\partial[(\mu_x\boldsymbol{\mathit{i}}+\mu_y\boldsymbol{\mathit{j}}+\mu_z\boldsymbol{\mathit{k}})\cdot B_z(x,y,z)\boldsymbol{\mathit{k}}]}{\partial z}=\mu_z\frac{\partial B_z(x,y,z)}{\partial z}\; \; \; \; \; \; \; \; 66

where \boldsymbol{\mathit{i}}, \boldsymbol{\mathit{j}} and \boldsymbol{\mathit{k}} are unit vectors, and \frac{\partial B_z(x,y,z)}{\partial z} is the gradient of the external magnetic field along the z-direction.

For a uniform field, B_z(x,y,z)=B_0 (i.e. a constant) and \boldsymbol{\mathit{F_z}}=0; while for an inhomogenous field, B_z(x,y,z)=B_0+\alpha z, where \alpha is a scalar representing the change in B_0 along the z-axis, and \boldsymbol{\mathit{F_z}}\neq 0.

Substituting eq61 in eq62,

U=-\gamma\boldsymbol{\mathit{B}}\cdot\boldsymbol{\mathit{L}}\; \; \; \; \; \; \; \; 67

For an inhomogenous magnetic field pointing in the z-direction, the above equation becomes

U=-\gamma(B_0+\alpha z)\boldsymbol{\mathit{k}}\cdot(L_x\boldsymbol{\mathit{i}}+L_y\boldsymbol{\mathit{j}}+L_z\boldsymbol{\mathit{k}})=-\gamma(B_0+\alpha z)L_z\; \; \; \; \; \; \; \; 68

Eq68 is used as a starting point in analysing results from the Stern-Gerlach experiment.



Does the magnetic field \boldsymbol{\mathit{B}}=(B_0+\alpha z)\boldsymbol{\mathit{k}} violate Maxwell’s 2nd equation of \nabla\cdot\boldsymbol{\mathit{B}}=0 where \nabla=\frac{\partial}{\partial x}\boldsymbol{\mathit{i}}+\frac{\partial}{\partial y}\boldsymbol{\mathit{j}}+\frac{\partial}{\partial z}\boldsymbol{\mathit{k}}?


The field should be \boldsymbol{\mathit{B}}=-\alpha x\boldsymbol{\mathit{i}}+(B_0+\alpha z)\boldsymbol{\mathit{k}}, which satisfies \nabla\cdot\boldsymbol{\mathit{B}}. However, the precession of the spin magnetic moment of a silver atom around B_0 in the Stern-Gerlach experiment is so fast that the x-component of the spin moment averages to zero, resulting in an effective field of \boldsymbol{\mathit{B}}=(B_0+\alpha z)\boldsymbol{\mathit{k}} interacting with the atom.



Why is \nabla\cdot\boldsymbol{\mathit{B}}=0?


A simple way of showing \nabla\cdot\boldsymbol{\mathit{B}}=0 is to consider a two-dimensional diagrammatic representation of the vector function \boldsymbol{\mathit{B}}=-y\boldsymbol{\mathit{i}}+x\boldsymbol{\mathit{j}} below, where each point in the two-dimensional space is associated with a vector.

When y=0, the successive points in the negative x direction and the positive x direction are associated with the vectors -\boldsymbol{\mathit{j}},-2\boldsymbol{\mathit{j}},\cdots and \boldsymbol{\mathit{j}},2\boldsymbol{\mathit{j}},\cdots respectively. Similarly, when x=0, the successive points in the negative y direction and the positive y direction are associated with the vectors \boldsymbol{\mathit{i}},2\boldsymbol{\mathit{i}},\cdots and -\boldsymbol{\mathit{i}},-2\boldsymbol{\mathit{i}},\cdots respectively. Clearly, \nabla\cdot\boldsymbol{\mathit{B}}=0.



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