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Question 6.21: Place the object in a region of zero magnetic field, and hea...

(a) Show that the energy of a magnetic dipole in a magnetic field B is

U = –m · B.                                    (6.34)

[Assume that the magnitude of the dipole moment is fixed, and all you have to do is move it into place and rotate it into its final orientation. The energy required to keep the current flowing is a different problem, which we will confront in Chapter 7.] Compare Eq. 4.6.

U = –p · E                                  (4.6)

(b) Show that the interaction energy of two magnetic dipoles separated by a displacement r is given by

U=\frac{\mu_{0}}{4 \pi} \frac{1}{r^{3}}\left[ m _{1} \cdot m _{2}-3\left( m _{1} \cdot \hat{ r }\right)\left( m _{2} \cdot \hat{ r }\right)\right]                            (6.35)

Compare Eq. 4.7.

U=\frac{1}{4 \pi \epsilon_{0}} \frac{1}{r^{3}}\left[ p _{1} \cdot p _{2}-3\left( p _{1} \cdot \hat{ r }\right)\left( p _{2} \cdot \hat{ r }\right)\right]                                     (4.7)

(c) Express your answer to (b) in terms of the angles \theta_{1} \text { and } \theta_{2} in Fig. 6.30, and use the result to find the stable configuration two dipoles would adopt if held a fixed distance apart, but left free to rotate

(d) Suppose you had a large collection of compass needles, mounted on pins at regular intervals along a straight line. How would they point (assuming the earth’s magnetic field can be neglected)? [A rectangular array of compass needles aligns itself spontaneously, and this is sometimes used as a demonstration of “ferromagnetic” behavior on a large scale. It’s a bit of a fraud, however, since the mechanism here is purely classical, and much weaker than the quantum mechanical exchange forces that are actually responsible for ferromagnetism .^{13}]

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(a) The magnetic force on the dipole is given by Eq. 6.3; to move the dipole in from infinity we must exert an opposite force, so the work done is

F = ∇(m · B)                                  (6.3)

U=-\int_{\infty}^{ r } F \cdot d l =-\int_{\infty}^{ r } \nabla( m \cdot B ) \cdot d l =- m \cdot B ( r )+ m \cdot B (\infty)

(I used the gradient theorem, Eq. 1.55). As long as the magnetic field goes to zero at infinity, then, U = –m·B. If the magnetic field does not go to zero at infinity, one must stipulate that the dipole starts out oriented perpendicular to the field.

\int_{ a }^{ b }(\nabla T) \cdot d l =T( b )-T( a )                           (1.55)

(b) Identical to Prob. 4.8, but starting with Eq. 5.89 instead of 3.104.

B _{ dip }( r )=\frac{\mu_{0}}{4 \pi} \frac{1}{r^{3}}[3( m \cdot \hat{ r }) \hat{ r }- m ]                         (5.89)

E _{\text {dip }}( r )=\frac{1}{4 \pi \epsilon_{0}} \frac{1}{r^{3}}[3( p \cdot \hat{ r }) \hat{ r }- p ]                             (3.104)

(c)  U=-\frac{\mu_{0}}{4 \pi} \frac{1}{r^{3}}\left[3 \cos \theta_{1} \cos \theta_{2}-\cos \left(\theta_{2}-\theta_{1}\right)\right] m_{1} m_{2} . \text { Or, using } \cos \left(\theta_{2}-\theta_{1}\right)=\cos \theta_{1} \cos \theta_{2}+\sin \theta_{1} \sin \theta_{2} ,

U=\frac{\mu_{0}}{4 \pi} \frac{m_{1} m_{2}}{r^{3}}\left(\sin \theta_{1} \sin \theta_{2}-2 \cos \theta_{1} \cos \theta_{2}\right) .

Stable position occurs at minimum energy:  \frac{\partial U}{\partial \theta_{1}}=\frac{\partial U}{\partial \theta_{2}}=0

 

\left\{\begin{array}{l}\frac{\partial U}{\partial \theta_{1}}=\frac{\mu_{0} m_{1} m_{2}}{4 \pi r^{3}}\left(\cos \theta_{1} \sin \theta_{2}+2 \sin \theta_{1} \cos \theta_{2}\right)=0 \Rightarrow 2 \sin \theta_{1} \cos \theta_{2}=-\cos \theta_{1} \sin \theta_{2}; \\\frac{\partial U}{\partial \theta_{2}}=\frac{\mu_{0} m_{1} m_{2}}{4 \pi r^{3}}\left(\sin \theta_{1} \cos \theta_{2}+2 \cos \theta_{1} \sin \theta_{2}\right)=0 \Rightarrow 2 \sin \theta_{1} \cos \theta_{2}=-4 \cos \theta_{1} \sin \theta_{2}.\end{array}\right.

 

\text { Thus } \sin \theta_{1} \cos \theta_{2}=\sin \theta_{2} \cos \theta_{1}=0 .\left\{\begin{aligned}\text { Either } \sin \theta_{1}=\sin \theta_{2}=0: \overset{(1)}{\rightarrow \rightarrow }or\overset{(2)}{\rightarrow \leftarrow } \\\text { or } \quad \cos \theta_{1}=\cos \theta_{2}=0:\overset{\uparrow \uparrow }{(3)}or\overset{\uparrow \downarrow }{(4)} \end{aligned}\right.

Which of these is the stable minimum? Certainly not ;\text { (2) or (3)-for these } m _{2} \text { is not parallel to } B _{1} , whereas we know  m _{2} will line up along B _{1} .

It remains to compare  \text { (1) (with } \theta_{1}=\theta_{2}=0 \text { ) and (4) (with } \theta_{1}=\pi / 2, \theta_{2}=-\pi / 2 \text { ) } :

U_{1}=\frac{\mu_{0} m_{1} m_{2}}{4 \pi r^{3}}(-2) ; U_{2}=\frac{\mu_{0} m_{1} m_{2}}{4 \pi r^{3}}(-1) . U_{1} is the lower energy, hence the more stable configuration.

Conclusion: They line up parallel, along the line joining them: → →

(d) They’d line up the same way:  →  →  →  →  →  →

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