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3 3 case of uniform external magnetic field. To show this point, let us note that the vector potential of the form U0 = dV J A = A dV J (4) 1 Z k k k Z k A = r × B produces a uniform magnetic field. Since X=1 X=1 2 k k the vector potential in this expression is of the first order ∂2A in the coordinate x , it is obvious that [ k ]0 is equal and i ∂xi∂xj to 0, which means that the quadrupole term, as well as 3 all remaining higher order terms vanish. We have shown that the dipole term in the expansion U1 = dV Jkr · [∇Ak]0 (5) Z of the magnetic is expressed as +m · B, kX=1 whereas the dipole term in the electric energy expansion which are the monopole and dipole terms respectively. has the −p · E form, which is an unexpected result con- For a stationary current system, the monopole term U0 sidering that the electric dipole and the magnetic dipole vanishes since JkdV = 0. Now, if the constant vector have so many similar properties. Although this result [∇Ak]0 is temporarilyR denoted as c, the dipole term U1 is rather unfamiliar, such ’wrong’ sign of the magnetic can be expressed as, dipole energy has already been discussed by D. J. Grif- fiths [4]. In that paper, the author derives the expression

3 U = +m · B using the the magnetic charge model and provides a rather lengthy explanation on why such oppo- U1 = U1 (6) k site sign result should be obtained. X=1 k The fact that the energy of a magnetic dipole under the influence of an external field should be expressed as +m· where U1k = dV Jk(r · c). U1k term can now be ex- pressed as, R B could be confusing to many students, since the energy of a magnetic dipole is usually given as −m · B in most textbooks[1, 5]. Such sign difference in magnetic dipole energy does not make any difference when the magnitude U1 = dV x (J · c) − dV [c × (r × J)] (7) k Z k Z k of force on the magnetic dipole only is of interest. But it results in the opposite direction of the force or torque on if the vector identity c × (r × J) = r(J · c) − J(r · c) the dipole, and a good understanding on this situation is used. It could be noted here that the integral of the may be important pedagogically. second term on the right hand side can be expressed as 2[c × m]k where m is the magnetic dipole moment which 1 III. Discussion on the Force and Torque on the is defined as m = 2 r × JdV . Magnetic Dipole Let us now considerR the first term on the right hand Before discussing the force and torque on the dipole, c side of Eq.(7). Since is a just constant vector, this let us make sure that +m · B is the right expression of term can be written as i ci xkJi. Therefore, using the magnetic dipole energy. To show that point, con- the relation dV (xkJi +PxiJk)R = 0 which is satisfied for sider situations where a current loop is placed in an uni- a stationaryR current system[8], this term can now be ex- form external magnetic field, in which magnetic dipole is r c pressed as − i ci dV xiJk, or, as − dV Jk( · ), which directed (1)parallel (2)anti-parallel and (3)perpendicular means that theP firstR term reduces to −RU1k. Therefore we to the magnetic field direction. Considering the super- get the expression, posed field of the external field and of the field due to the current loop and calculating the magnetic field en- ergy dV |B|2, it is obvious that the energy is increased U1 = −[c × m]k (9) k for caseR (1) and is decreased for case (2). It is also ob- vious that there is no change in the magnetic energy for in which c stands for [∇Ak]0. Therefore, using the Levi-Civita symbol and Einstein case (3). This example shows that the energy variation summation convention, we finally obtain the expression for each situation is consistent with the magnetic energy expression U =+m · B. for the magnetic dipole energy as U1 = −εkij (∂iAk)mj , or Now let us consider the force and torque on the mag- netic dipole. If one attempt to apply the F = −∇U relation between the force and potential energy using the U1 = εjik(∂iAk)mj (10) energy expression U =+m·B, the force on the magnetic dipole becomes F = −∇(m · B) which is wrong by a sign using the εkij = −εjik property . Now since εjik∂iAk = from the right expression[5]. We again do not get the Bj from the relation B = ∇× A, we find that the mag- right expression for the torque if U =+m · B is used in netic dipole energy can be written as +m·B(0), in which connection with the common expression τ(θ)= −dU/dθ, B(0) is the magnetic field at an arbitrary point inside the since we get the result τ(θ)=+mB sin θ, which shows current distribution. We may note that this dipole term that the torque is exerted in the wrong direction, i.e. the 3 dipole tends to orient anti-parallel to the magnetic field. be expressed as F = ∇U rather than F = −∇U, when a This situation seems to indicate that the expression current loop is placed under the influence of external field U = +m · B is wrong by a sign, and suggests that the due to all other circuits. Using that argument, it could be right force expression ∇(m · B) is obtained if the expres- shown that the energy supplied by the external source to sion U = −m · B is is used. In fact, the exact reverse a circuit loop is exactly twice the potential energy change steps are followed in the textbook by J.D. Jackson[5], of the circuit[9, 10]. The mechanism of energy exchange in which the author derives the dipole energy expression between a current loop and all others in this situation U = −m · B using the calculated force expression, but is found to be the electromagnetic induction. For ex- with a careful remark that this is the case for the ”per- ample, if a displacement is made on a current loop, a manent magnetic dipole” only[5]. In that textbook, the change of magnetic flux in the loop is accompanied and author also provides a comment that ”U = −m · B is not induced electromagnetic effect arises. This means there the total energy of the in the external will be an exchange of energy between the current loop field”, which is intended to emphasize that there could and all others. Of course, no such exchange of energy be an exchange of energy between the dipole and the occurs between a permanent magnetic dipole and the ex- external source. This implies that if the magnetic mo- ternal magnetic field, or between an electric dipole and ment does not have a fixed magnitude as for the case of the external electric field. spin magnetic moment, then exchange of energy could oc- cur between the magnetic dipole and the external source. IV. Acknowledgement This work was supported Such situation is nicely discussed by J.R. Reitz et al.[9]. by the Inha University Research fund. In that textbook, a rigid circuits system is used to ex- plain why the force and magnetic energy relation should

[1] D.Halliday, R.Resnick, J.Walker, Fundamentals of [7] ”The force on a magnetic dipole”, Am. J. Phys. 56, 688- Physics (J. Wiley and sons, New York,2003), 6th ed. 692 p.678 [8] This relation can be proved by employing the vector iden- [2] R.P.Feynman, R.B.Leighton, M.Sands,The Feynman tity ∇ · (xixj J)= Jixj + Jj xi. Lectures on Physics, vol. 2 (Addison-Wesley, Lon- [9] J.R.Reitz, F.J.Milford, R.W.Christy, Foundations of don,1964), p.15-1 Electromagnetic Theory (New York,1993), 4th ed. p.294 [3] E.N.Purcell and Magnetism, Berkeley Physics [10] It is interesting to note that the exactly similar situation Course, vol. 2 (Mcgraw-Hill, New York,1965), p.367 occurs in the electrostatic system, in which a [4] D.J.Griffiths Introduction to Electrodynamics (Prentice- slab is inserted inside a parallel plate connected Hall, New York,1989), 2nd ed. to a battery. In that case, it could be shown that the [5] J.D.Jackson Classical Electrodynamics (Wiley, New energy supplied by the battery is exactly twice the mag- York,1999), 3rd ed. pp. 188-190 nitude of the change of the capacitor stored energy. [6] L.D.Landay, E.M.Lifshitz, Electrodynamics of Continu- ous Media (Pergamon Press, New York,1960) p.143