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arXiv:2509.15528v1 [physics.class-ph] 19 Sep 2025

Pedagogical study of the image of a magnetic dipole in front of a superconducting sphere

Hemansh Alkesh Shah 1 , Kolahal Bhattacharya 2 ‡

1 Indian Institute of Science, Bangalore, KA-560012 (India) 2 St. Xavier’s College (autonomous), Kolkata, WB-700016 (India) E-mail: 1 [email protected], 2 [email protected] Abstract. The method of images to solve certain electrostatic boundary-value problems is taught worldwide in undergraduate-level physics courses. Though it is also possible to employ this technique for solving magnetostatic boundary value problems, examples of this usage are rarely found in textbooks or physics pedagogy literature. In particular, the problem of finding the field due to a magnetic dipole kept in front of a superconducting sphere is an interesting one, because (i) it helps the students to compare with the grounded conducting sphere image problem in electrostatics, and (ii) offers a greater degree of difficulty since the source is a dipole (vector), rather than an electric charge (scalar). The present work demonstrates an intuitive way of solving the problem. The case in which the source dipole is oriented radially with respect to the sphere is solved with a single dipole image. In the case of the transverse orientation of the source dipole, we model the dipole as a current loop. Then, we find the image of the radial and transverse current elements that satisfy the boundary conditions. Then, we show that this method can be used to deduce the form of the image dipoles when the dipole is oriented in the transverse direction. This method is very much intuitive and accessible for undergraduate-level students. Keywords: Method of images, superconductivity.

1. Introduction

The method of images is an elegant technique to solve electrostatic boundary value
problems. One can calculate the field or potential when a known charge distribution is
placed near a surface (for example, a conducting body). The boundary condition on the
surface is given, but the charge induced on the surface is unknown. The tricky job is
accomplished by asserting that the field can be expressed as a sum of the field due to the
known charge distribution, as well as that due to a simplified image charge distribution
which effectively plays the role of the induced charge. We do not need to worry about
the surface any more. The uniqueness theorem guarantees that the potential found
by this technique is unique. The method was introduced by Lord Kelvin in 1848 in
‡ Author to whom any correspondence should be addressed: [email protected]

2 one paper published in Cambridge and Dublin Mathematical Journal [1]. It is taught nowadays in the undergraduate level course on electrostatics [2, 3]. Apart from that, many interesting pedagogic applications [4, 5] and research-level applications have been discussed in the literature [6–8]. In the realm of magnetostatics, the method of images has been used to explain magnetic levitation near the surface of a superconductor [9]. Smythe [10] (1950) used the concept of image currents to solve boundary value problems in magnetostatics (section 7.23: Current Images in Plane Face). In recent times, a few pedagogical articles used the concept to throw light on magnetic levitation near planar superconducting surfaces [11, 12]. However, one does not find pedagogical examples of the method in magnetostatics in common textbooks. This is perhaps due to the fact that compared to the conductors, superconductivity is a relatively recent development in physics. The recent book by Matsushita [13], however, contains a few simple examples of the applications of the method of images involving superconductors. Here let us mention that a formal introduction to superconductivity can be found in the textbook by Tinkham [14]. For temperatures T below the critical temperature TC (T < TC ), the magnetic field B vanishes inside the superconductor, i.e. B = 0. This is called the Meissner effect [15]. When an external magnetic field H is applied, a superconductor expels the magnetic field from the interior by setting up an electric current at the surface. The surface current creates a magnetization that cancels the external magnetic field inside the bulk of the superconductor in the case of type-I superconductors. The magnetic field does not abruptly reduce to zero at the surface of a real superconductor. It does so over a small length scale called the London penetration depth. For its calculation, the interested reader is referred to the book by Tinkham [14]. In this work, we assume an ideal, Type-I superconductor with a vanishingly small London penetration depth. Coming back to the discussion of magnetic images, a pedagogically more interesting case would be to calculate the magnetic field at the test point, when a magnetic dipole (or an equivalent current distribution) is placed in front of a spherical surface as shown in the following figure.

3 Figure 1: Field lines of a magnetic dipole (red colour) of arbitrary orientation placed in front of a superconducting sphere (depicted in blue). The field lines are parallel to the interface just outside the sphere. The magnetic field inside the sphere is zero. Since a dipole of arbitrary orientation can be resolved into radial and transverse components, one can approach the problem systematically. Lindell analysed the case, where a current loop is placed in front of a material sphere, such that its axis is radially oriented relative to the sphere [16]. He used an image current loop to solve the problem. In the limit where the size of the current loop is small, one can talk about a point magnetic dipole placed in front of a sphere pointing in the radial direction. The same technique was used by Sezigner et al. [17] in the context of a superconducting sphere. Rezdic [18,19] showed that the it is possible to replace the superconducting sphere with a permeable sphere with an embedded current loop which coincides with the image current for the superconducting sphere case. Yang [20] and Coffey [21] dealt with the same problem to find levitating force, without any reference to magnetic image currents. They matched the boundary conditions of the magnetic scalar potential due to the source and the image dipoles to solve the problem. However, the case of transverse orientation of the magnetic dipole (or current loop) was not covered in these works. A systematic way to approach this problem is shown by Lin [7] who separately considers the cases of the radial and transverse orientations. He treats a magnetic dipole as two closely spaced magnetic monopoles of opposite signs. Then, he solves the image problem for a single magnetic monopole and finally uses the principle of superposition to arrive at the correct answers. Other notable works on dipoles with transverse orientation came from Coffey [22, 23]. However, like the radial orientation case, these works are also concerned with the levitating force but not with the image dipole configuration. Hence, the only consolidated solution to the problem using the method of images is the work by Lin [7]. In many ways, this problem is an instructive tool for physics students. First of all, this problem gives an option for a comparative study between the application of the image method in electrostatics and magnetostatics. It also offers a higher degree of complication since the source is represented by a vector, whereas the standard electrostatic problem involves only a scalar charge. In principle, one can also solve the problem of an electric dipole placed in front of a grounded conducting sphere

4 by the method of images [4]. This exercise, therefore, could give a great case-by-case comparison between the application of the method in electrostatics and magnetostatics. There is another simpler example of the use of the method in magnetostatics. Let us consider a straight infinite wire carrying constant current, placed at a small distance from an infinite superconducting plane; or a magnetic dipole placed over an infinite superconducting plane [11]. But these problems perhaps do not involve many interesting features of the more generalised problem of a sphere. Despite Lin [7] giving a systematic development of the application of the method of images in typical textbook-type problems and its conceptual appeal, we think this work has little pedagogical value, for many reasons. First of all, Lin uses magnetic scalar potential, a quantity that physics students are usually discouraged to use, due to the subtlety of the topological nature of the region where they could be applicable. Earlier Coffey [23] justified the usage of scalar potential “by the absence of electric current outside of the superconducting sphere”. This statement does not explicitly state the nature of the current. In this problem the magnetic scalar potential can be used, because we have a bound (induced) current on the sphere and the free current is zero. This would imply ∇ × H = Jf ree = 0. In addition, we must consistently keep ourselves in a simply-connected region. Lin considers the region outside the sphere and gives the expressions for magnetic scalar potential due to the source monopole [Eq.(2) in [7]] and that due to the image monopole’s distribution [Eq.(3) in [7]]. After that, it is stated without justification that the scalar potential due to the image monopole can be expressed as the difference between a monopole term and a line-monopole density term [Eq.(7) in [7]]. This step appears ad-hoc from a pedagogical viewpoint. Finally, we note that using magnetic monopole is a matter of convenience. One cannot adopt this as a physical theory, because monopoles have never been located in the universe, through several experiments. For these reasons, it is perhaps worthwhile to look for alternative pedagogical approaches to solve the problem. If we want to look beyond the magnetic scalar potential, a safe starting point would be to start with the boundary conditions matching the magnetic field B or the vector potential A. For example, the normal component of the magnetic field must be zero at the surface: B ⊥ = 0 (1) due to the Meissner effect. We can use this condition to deduce the image configuration when the source magnetic dipole is oriented radially. The boundary condition on A, as pointed out in Chapter 7 of [13], is that if the superconducting region is simply- connected, then the vector potential can be assumed to be constant in this region: A = C (2) -where C denotes a constant vector. The vector potential is continuous across the sur- face. We can use this condition when the source dipole is oriented transversely.

5

2. The image of a radial dipole on a superconducting sphere

The following figure 2 shows the configuration when a point magnetic dipole m1 is placed near a superconducting sphere of radius a. The sphere is centred at the origin of the coordinates O and the dipole is located at a distance d1 ( > a) from O, pointing radially outwards. Due to the Meissner effect [15], the magnetic field inside the sphere vanishes. The dipole may also be thought of as a tiny current loop with area A → 0, carrying anti-clockwise current (when viewed from the top) I → ∞, such that |m1 | = IA is finite. This dipole induces a surface current of unknown functional form on the sphere. The total magnetic field outside the sphere is the field due to the dipole plus the field due to the unknown surface current. How to find the field and the surface current? Figure 2: Magnetic dipole placed radially in front of a superconducting sphere (shaded sphere). Magnetic field B = 0 inside the sphere. In the figure, d1 = a + u. The method of images asserts that this problem could be solved by assuming that there exists a distribution of the image dipole inside the sphere, which together with the real dipole, satisfies the boundary conditions at the location of the spherical surface. We shall assume that a point image dipole m2 , pointing radially outwards, located at a distance d2 from the origin O is sufficient to address the problem. The relevant boundary condition is given by Eq.(1). The existence of a unique real solution should follow if this is true. In the scientific literature on the method of images in electrostatics, it has been illustrated previously that an analogy between the grounded conducting sphere image problem and optical image formation by the spherical mirror can be observed if the distances to the real and image dipoles are measured from the pole P , instead of from O [5]. We shall employ the same idea (object distance u and image distance v) in this problem while implementing boundary conditions. Note that d1 = u + a and d2 = a − v.

6 Let us investigate the boundary condition of a magnetic field at P , the pole, where the real and image dipoles will have only the radial components of the field. The contribution of the real dipole (m1 ) placed outside the sphere to the total field is Bs ource (P ) = µ0 m 1 4πu3 (2 cos π(−ˆz) + sin π(−ˆx)) = 2µ0 m 1 4πu3 ˆz (3) Similarly, the contribution to the field at P from the image dipole m2 is Bi mage (P ) = µ0 m 2 4πv3 (2 cos(0)(ˆz) + sin(0)(ˆx)) = 2µ0 m 2 4πv3 ˆz (4) The boundary condition at P (given in Eq.(1)) translates to: m1 u3 + m2 v3 =0 (5) Next, let us investigate the magnetic field boundary condition at a point Q that is diametrically opposite to the pole P . There, the contribution to the field from the dipole m1 is: Bs ource (Q) = µ0 m 1 4π(u + 2a)3 (2 cos π(−ˆz) + sin π(−ˆx)) = 2µ0 m 1 4π(u + 2a)3 ˆz (6) Similarly, the contribution to the field from the image dipole m2 : Bi mage (Q) = µ0 m 2 4π(2a − v)3 (2 cos π(−ˆz) + sin π(−ˆx)) = 2µ0 m 2 4π(2a − v)3 (ˆz) (7) The last two equations lead to the following boundary condition at Q: m1 (u + 2a)3 + m2 (2a − v)3 =0 (8) From Eq. 5, we express m1 and m2 in terms of u, v, a, and a non-zero constant, say, k: m1 u3 = − m2 v3 = k =⇒ m1 = u3 k, m2 = −v3 k (9) Using Eq. 8 and Eq.(9), we deduce that: u3 k (u + 2a)3 = v3 k (2a − v)3 =⇒ u u + 2a = v 2a − v =⇒ (2au − 2av) = 2uv =⇒ 1 u − 1 v = − 1 a (10) Eq. (10) is the mirror equation in this problem. Notice that Eq.(10) also means that v = au/(a+u) = au/d1 . This is the same as the corresponding mirror equation observed

7 in [5] in the grounded conducting sphere image problem. The focal length a denotes the distance from the pole P where the image dipole would be located when the real dipole distance u → ∞. The focal length equals the radius of the sphere, just as in electrostatics. The image distance and the focal length come with a negative sign because they are in a direction opposite to that of the real dipole distance following the trend used in geometrical optics. Only, in this case, the linear magnification is different. From Eq. 9 and Eq.(10), we see: m2 = −m1 1 u3 a3 u3 (a + u)3 = − a3 d31 m1 (11) This also shows that the image dipole must be oriented in the radially inward direction. Though the solution was found without considering the possible dependence on θ, we can check that the solution satisfies the boundary conditions even when θ ̸= 0. For example, at an arbitrary point (a, θ) on the sphere, the radial component of the total magnetic field should vanish. To verify this, we must express the magnetic fields due to the dipoles in terms of θ. First, we consider the dipole m1 . Referring to figure 2, the magnetic field due to the source dipole m1 and the image dipole are given by [2]: Bs ource (a, θ) = µ0 m 1 4πr31 (2 cos θ1 ˆr 1 + sin θ1 ˆ θ 1 ) Bi mage (a, θ) = µ0 m 2 4πr32 (2 cos θ2 ˆr 2 + sin θ2 ˆ θ 2 ) (12) where (θ1 , θ2 ) and (r1 , r2 ) are defined in figure 2. We must express these in terms of the spherical polar coordinates (a, θ, ϕ) centred at O. Note that ˆr1 × ˆθ1 points into the page, in the same direction as ˆr × ˆθ. The radial component of the field with respect to the coordinate system at O is given by: Bs ource (a, θ) · ˆr = µ0 m 1 4πr31 2 cos θ1 ( ˆr1 · ˆr) + sin θ1 ( ˆ θ 1 · ˆr) (13) We need to express cos θ1 , sin θ1 , ˆr1 · ˆr, as well as ˆθ1 · ˆr in terms of θ to make progress. We first find out ˆr1 · ˆr. We start by noting that r1 = aˆr − d1 ˆ z . So, the unit vector ˆr1 can be expressed as ˆr1 = aˆr − d1 ˆ z r1 = aˆr − d1 ( cos θˆr − sin θˆθ) r1 , (14) where we expressed ˆz in terms of the ˆr and ˆθ of the coordinates centred at O. From the above equation, we can deduce the projection of ˆr1 in the radial direction: ˆr1 · ˆr = a − d1 cos θ r1 (15)

8 Using the cosine rule, we can express: cos θ1 = − d21 + r1 2 − a2 2d1 r 1 (16) Further, expressing r1 2 in terms of θ, it is possible to simplify cos θ1 : cos θ1 = − d21 + (a2 + d21 − 2ad1 cos θ) − a2 2dr1 = − 2d21 − 2a cos θ 2d1 r 1 = − d1 − a cos θ r1 (17) Similarly, using the sine rule on the same triangle, one can deduce sin θ1 = a r1 sin θ (18) The expression of ˆθ1 is tricky, as it is not immediately obvious how to proceed. Here we use the observation that ˆr1 × ˆθ1 = ˆϕ1 = ˆϕ ≡ ˆr × ˆθ. Due to this, ˆθ1 can be expressed as ˆθ1 = ˆϕ1 × ˆr1 = ˆϕ × ˆr1 = ˆϕ × aˆr − d1 ˆ z r1 (19) Again, using the expression for ˆz, we get ˆθ1 = 1 r1 (a − d1 cos θ) ˆ θ + d1 sin θˆr (20) On account of these results, the radial component of the magnetic field due to the source dipole at a point (a, θ) is given by Bs ource (a, θ) · ˆr = − µ0 4π m1 2(d1 − a cos θ)(d1 cos θ − a) − ad1 sin2 θ (a2 + d21 − 2ad1 cos θ)5 /2 (21) In the above analysis, we did not make use of any special property of the location of the source dipole being located outside the sphere. Therefore, one can find the radial component of the magnetic field at the surface point (a, θ) due to the image dipole: Bi mage (a, θ) · ˆr = − µ0 4π m2 2(d2 − a cos θ)(d2 cos θ − a) − ad2 sin2 θ (a2 + d22 − 2ad2 cos θ)5 /2 (22) We notice that by setting d2 = a2 /d1 , the denominator of Eq.(22) reduces to: a2 + a4 d21 − 2a a2 d1 cos θ 5 /2 = a5 d51 a2 + d21 − 2ad1 cos θ 5 /2 (23) On the other hand, using Eq.(11) and d2 = a2 /d1 , the numerator of Eq.(22) becomes: − µ0 4π − a3 d31 m1 2( a2 d1 − a cos θ)( a2 d1 − a cos θ) − a a2 d1 sin2 θ = − µ0 4π − a5 d51 m1 2(a − d1 cos θ)(a − d1 cos θ) − ad1 sin2 θ (24)

9 Dividing Eq.(24) by Eq.(23), we see that the factor (a5 /d51 ) cancels, and we can find the fraction in Eq.(22) in terms of m1 , a, d1 and θ. Adding Eq.(21) and Eq.(22), we can show that µ0 4π m1 2(d1 − a cos θ)(d1 cos θ − a) − ad1 sin2 θ (a2 + d21 − 2ad1 cos θ)5 /2 + µ0 4π m2 2(d2 − a cos θ)(d2 cos θ − a) − ad2 sin2 θ (a2 + d22 − 2ad2 cos θ)5 /2 =0 (25) Similarly, we can verify that the total vector potential on an arbitrary point (a, θ) on the sphere satisfies: m1 × ˆr1 r21 + m2 × ˆr2 r22 = m1 sin θ1 r21 (ˆz × ˆr1 ) + m2 sin θ2 r22 (ˆz × ˆr2 ) = m1 a sin θ (a2 + d21 − 2ad1 cos θ)3 /2 ˆϕ + m2 a sin θ (a2 + d22 − 2ad2 cos θ)3 /2 ˆϕ = 0 (26) when d2 = a2 d1 and m2 = − a3 d3 1 m1 . Note that this shows that Eq.(2) holds true in this case. Eq.(25) and Eq.(26) validate all the boundary conditions. We note that the image dipole configuration is not liable to make sure that the total field (or vector potential) must be zero inside the sphere.

3. Transverse dipole

Now, we consider the case in which a source dipole is oriented in the transverse direction relative to the superconductor sphere. In figure 3, we show that the dipole m1 is located on the z axis at a distance d1 from O the centre of the superconducting sphere where d1 > a, the radius of the sphere. It points to the x direction, i.e. transverse to the radial direction. The boundary condition is still given by Eq.(1) and Eq.(2). Let us try to find a possible image configuration (i.e. the location and direction of the image dipole) using Eq.(1). From the symmetry of the problem, it is clear that the image dipole must be conceived on the z axis itself. If m1 is oriented along the x axis, the image dipole (say m2 ) cannot be oriented along y or z axis for symmetry reasons. If m2 were indeed directed in the z direction, then the radial component of the field due to m1 , cannot cancel that due to m2 at all points on the sphere. Then, points would be found on the surface where the radial field component would remain nonzero. So, the image dipole may be aligned with the x axis, as shown in Fig 3. Further, we can see that there cannot be a single image point dipole in this case. This is because the radial components of the magnetic fields due to the source and image dipoles would not sum up to zero at the points under the location of the proposed image dipole. For example, consider the point S where the x axis punctures the sphere. In these regions, the field lines would be headed in similar directions. However, this does not happen as much at the points on the surface that are between the location of the image dipole and P , say at R, since

10 Figure 3: Magnetic dipole placed with transverse orientation before a superconducting sphere in these points, the field lines are oppositely directed. Therefore, we find that another distribution of the magnetic dipole is required at the bottom of the originally proposed single dipole that would be directed in the opposite direction to compensate for the non-vanishing radial component of the field at the points located on the surface.

3.1. Use of vector potential boundary condition

To find the image configuration, we adopt another approach. We can always construct a magnetic dipole as a tiny current loop. Let us now treat a current loop as a successive set of straight current elements. We resort to the boundary condition of A. Under the Coulomb gauge condition, the vector potential A satisfies ∇ · A = 0. In this gauge, the vector potential satisfies a vector Poisson’s equation and is given by: A = µ0 4π Idl′ |r − r′ | (27) The expression for the vector potential is akin to the expression for the electrostatic scalar potential except for the difference in the numerator where one would have an integral over a charge element instead of a current element. Otherwise, just like the scalar potential, the vector potential must be continuous across any surface. The key point to note here is that one may perhaps take the integral over the current element as a vector equivalent of the integral over a charge element. Inside the superconducting body the vector potential satisfies ∇ × A = 0. As both the divergence and the curl of the vector field are zero in the simply-connected region inside the sphere, we can choose its value to be a constant (see the discussion after Eq.(7.4) and the beginning of section 7.2 in chapter 7 of [13]). When we combine this with the continuity of the vector potential across the surface, we can say that A is a constant on the surface.

11 This resembles the condition of Φ = 0 on the surface of a conducting sphere for an electrostatic image problem. Beyond the surface, it assumes a functional form (note the kink in the vector potential in figures 7.4b, 7.7b, 7.8b of [13]). Specification of the boundary condition on A makes the solution to Laplace’s equation unique. Figure 4: Trapping of magnetic field It is possible to see that in the case of a non-simply-connected region, e.g. inside a hollow superconducting cylinder, the magnetic vector potential is not constant. In this case, magnetic flux is trapped inside the hole and gets quantized in the units of h/2e [14]. Applying Ampere’s law around a loop C lying completely within the hole, we find that: C A · dl = B · dS = n h 2e , (28) where h is Planck’s constant, e is elementary charge of an electron and n is an integer. As a result, the vector potential depends on the radius of the loop. However, for a simply-connected superconductor, there is no trapping of magnetic flux. So perhaps we can take the value of the vector potential to be zero. In fact, we have already seen that for the radial orientation of the dipole, the vector potential adds up to zero (Eq.(26)) at the surface. In fact, if we choose the constant vector A = 0 using the gauge freedom in A, then it is possible to validate the boundary conditions in B, as shown below. First, let us consider a tiny flat patch on the surface of the sphere. We construct a Gaussian pillbox of very small height on the patch. Then, the circumference of the patch acts as a boundary of the surface of the pillbox outside the sphere. Any non-zero normal component of the magnetic field would manifest as non-zero magnetic flux. If we apply Ampere’s law to this surface and the loop, then the total magnetic flux coming out of the surface can be shown to be zero. Φ⊥B = B · dS = A · dl = 0 · dl = 0 (29)

12 Because A = 0 on the surface of the sphere. In the above equation, dl represents the differential length element around the circumference and dS represents the differential area element on the upper surface of the pillbox. This validates the boundary condition B ⊥ = 0 (Eq.(1)). Similarly, we can construct another Amperian loop outside the superconductor whose one side grazes along the surface of the superconductor. In this case, the Gaussian pillbox is lying on the surface. Again, using the Ampere’s law, we find that: Φ| | B = B · dS = A · dl = surf ace 0 · dl + outside A · dl ̸= 0 (30) Where the first line integral is evaluated on the surf ace of the superconductor where we set A = 0. But the other integral gives a non-zero contribution, since A is non- trivial outside the superconducting surface. This shows that the magnetic field must have components parallel to the surface. Since the vector potential A is indeterminate up to the gradient of a gauge function χ: A → A′ = A+∇χ, therefore, it is safe to state that the A = C in a simply-connected superconductor. This justifies Eq.(2) referred to in [13]. In the following, we show the use of the boundary condition that the vector potential is a constant on the surface in the context of the problem where the source dipole is oriented transverse to the sphere. Mathematically, it implies that the differential of A (i.e. a very small change in A) can be made arbitrarily close to zero at the surface: dA → 0. Since a given current element can always be written as a linear superposition of the radial and transverse elements, we can consider the two cases separately.

3.2. Image current elements

In the following, we find the image current elements in transverse and radial directions. 3.2.1. Image for transverse current element First, we consider the case of a source element carrying current I (t) s is oriented in the transverse direction. It is passing through S and has a length ∆u, as shown in figure 5a. The superscript (t) stands for the transverse orientation of the current element. Let the image current element carrying image current I (t) i have the length ∆v and is located at the point I inverted with respect to the sphere. Let us say that the distance OS = d1 . Then, the image distance OI = d2 = a2 /d1 . Using the condition of constant vector potential on the

13 surface, we can set: dA( t) s + dA( t) i = 0 =⇒ I s (t) ∆u |− → O Q − −→ OS| + I i (t) ∆v |− → O Q − −→ OI| = 0 =⇒ Is (t) ∆u a2 + d21 − 2ad1 cos θ + Ii (t) ∆v a2 + d22 − 2ad2 cos θ =0 =⇒ Ii (t) = − a2 + (a2 /d1 ) 2 − 2a(a2 /d1 ) cos θ a2 + d21 − 2ad1 cos θ ∆u ∆v Is (t) = − a d1 a2 + d21 − 2ad1 cos θ a2 + d21 − 2ad1 cos θ ∆u ∆v Is (t) =⇒ Ii (t) = − a d1 d1 (a2 /d1 ) Is (t) = − d1 a Is (t) (31) (a) Transverse current element (b) Radial current element Figure 5: Image current elements for (a) transverse and (b) radial orientation of source current element. The Source and the Image current elements are denoted by S and I in the figure (I does not represent current). The length elements ∆u and ∆v are greatly exaggerated. 3.2.2. Image for radial current element Next, we consider the case of a source current element du carrying current Is , directed radially outwards and located at a distance u = d1 − a from the pole P , as shown in figure 5b. Comparing this configuration with the electrostatic case, we can conceive an image element dv carrying current Ii , located at v = a d1 (d1 − a). To check the direction of that current we investigate the boundary condition of A( r) (the superscript (r) stands for the radial orientation of the current

14 element) due to the source and image current elements: dA( r) s + dA( r) i = 0 =⇒ I s (r) du |− → O Q − −→ OS| + I i (r) dv |− → O Q − −→ OI| = 0 =⇒ Is (r) du a2 + d21 − 2ad1 cos θ + Ii (r) dv a2 + d22 − 2ad2 cos θ =0 =⇒ Ii (r) = − a2 + (a2 /d1 ) 2 − 2a(a2 /d1 ) cos θ a2 + d21 − 2ad1 cos θ du dv Is (r) = − a d1 du dv Is (r) Now, we know that as the source moves further, the image comes closer to the centre. In other words, u and v increase in opposite directions relative to the pole P . Hence du/dv is a negative quantity. This was also pointed out in the context of longitudinal magnification in [5]. Its magnitude can be calculated from the mirror equation Eq.(10): |du/dv| = u2 /v2 = d21 / a2 . As a result, the image radial current (along with direction) is given by: Ii (r) = − a d1 − u2 v2 Is (r) = d1 a Is (r) (32)

3.2.3. Image for a radial current element from the closure of current loop We can

3.2.3. Image for a radial current element from the closure of current loop We can
deduce the form of the image current element from the knowledge of the transverse
current element’s image and the closure of the current loops.

δr Figure 6: The source and the image dipoles are represented by the current loops ABCD, and a combination of the loop B ′ A′ D′ C ′ and the loop C ′ OD′ . The image loops carry the current in opposite directions. In the above figure 6, we consider a current loop ABCD carrying current i in the clockwise direction. We use the notation r to denote the radial coordinate instead of z and the notation i to denote the current to keep it separate from the discussion in 3.2.2. The images of the transverse elements of the arms AB and CD are B ′ A′ and D′ C ′ carrying current in the opposite direction. Using Eq.(31) we can state that the magnitudes of the current i′1 and i′2 in the arms B ′ A′ and D′ C ′ are given by respectively:

15 i′1 = r a i (33a) i′2 = r + δr a i (33b) Now, the images of the radial elements must close the current loops, and by symmetry, themselves be radial. Segments C ′ B′ and A′ D′ are the radial current elements that must carry current i′3 : i′3 = r a i (34) Because there are no other currents at the A′ and B ′ ). Clearly, Eq.(34) is identical to Eq.(32). The preceding analysis shows that irrespective of the radial or transverse orientation of the source current element, the image current element is scaled by the factor r a (or d1 a in the context of the figure 5b). It is in the same direction as the source current for the radial case, and opposite of the source current in the transverse case. The excess current i′4 = δr a i at the arm D′ C ′ must have come from the path OD′ at the junction D′ and must have gone towards O at the junction C ′ . Thus, the only way to explain the excess current is to pass it through the centre along the path C ′ OD′ . It appears due to our attempt to use a single image loop to satisfy the boundary conditions and the current closure condition.

3.3. Magnetic moment of the image dipole A ′ D ′ C ′ B ′

The first image dipole, as seen from the figure 6, has the same direction as that of the source dipole. In the limit where the size of the current loop for the source dipole is vanishingly small, we can treat (i) the loop to represent a point dipole, (ii) the coordinate r is almost along the z direction, and (iii) the arms AD, AB are nearly perpendicular to each other and the area of the loop will be given by the product AD·AB. Corresponding arms of the image loop will be scaled as well. It is seen that the arm A′ D′ will be scaled by a factor of dv/du = a2 /r2 . If the source dipole is located at r = d1 , we will have A′ D′ = a2 d21 AD. Then, the arm B′ A′ is scaled by a factor of (a2 /d1 ) /r = a2 /d1 2 . As a result, the total area of the loop B ′ A′ D′ C ′ is scaled down to a4 /d1 4 (AD · AB). To find the dipole moment, we must multiply this by the image current, whose magnitude is given by d1 a I. Therefore, the dipole moment for this image dipole |m2 | = a3 d1 3 |m1 | .

3.4. Handling excess current

The excess current I ′′ = δr a I will flow through the triangular loop C ′ OD′ . We can calculate the resulting dipole moment from the following figure 7a.

16 (a) (b) Figure 7: (a) Excess current I ′′ = δr a I passing through O forms a dipole in the direction opposite to that created by the current in the image loop A′ D′ C ′ B ′ . The excess current was denoted by i′4 in 3.2.3; (b) Splitting up of current loop C ′ OD′ as an infinite series of tiny current loops representing linearly diminishing magnetic dipoles. The compensating dipole moment must be equal to the area of the triangle multiplied by the excess current: µ = 1 2 C ′ D′ · OX δr a I ≈ 1 2 a2 d1 2 AB · a2 d1 δr a I = a3 d1 3 1 2 AB · δr · I = a3 d1 3 m1 2 (35) To check the consistency with the work by Lin [7], let us contemplate a linear magnetic dipole density of dipole moment per unit length λ = dµ′ dz = m1 ad1 z along the z axis between O and OX = a2 /d1 . In that case, we can verify that the total dipole moment matches the above dipole moment due to compensating current. µ′ = a2 /d1 0 m1 ad1 z dz = m1 ad1 z2 2 a 2 /d1 0 = a3 d1 3 m1 2 (36) Comparison of Eq.(36) with Eq.(35) shows the equivalence of this method with Lin’s results.

3.5. Validation of the result for radial dipole

In this case, every point on the current loop representing the source dipole m1 is equidistant from the centre. Not only that, all elements are transverse to the

17 superconducting sphere. As a result, the image current should be given by Eq.(31). Suppose the current loop for the source dipole is modelled as a tiny current loop located at a distance d1 from the centre of the sphere on the z axis. In that case, the length of the image dipole’s current elements will be scaled down by a factor a2 d1 2 . Assuming that the source current loop carries a current I, the magnetic dipole moment of the source dipole would be m1 = AI ˆz, where A represents the area of the loop, A → 0. On the other hand, the magnetic dipole moment for the image loop will be m′2 = − a4 d1 4 A · d1 a I ˆz = − a3 d1 3 (AI ˆz) = − a3 d1 3 m1 (37) This validates the fact that the model based on current loops is effective in explaining both the radial and transverse orientation of the magnetic dipole placed in front of a superconducting sphere.

4. Summary

In this work, we solved the problem of a magnetic dipole placed in front of a superconducting sphere using the method of images in a way accessible to undergraduate physics courses. The case of the radial dipole was solved using the boundary condition that B⊥ = 0. The case of the transverse orientation of the source dipole has been studied by Lin [7] in the past. We approached the problem through an intuitive pathway of representing magnetic dipoles by current loops and finding images of the elements of the current loop. Invoking the similarity between the vector potential and the electrostatic potential under the Coulomb gauge, and the boundary condition on vector potential, we determined the image of the transverse current element. Further, we deduced the image of the radial current element using two different methods: (a) boundary condition of A and (b) loop closure method. Using these results, we determined the image configuration of a current loop that represents a source magnetic dipole. This result was shown to be consistent with Lin’s work and also with the case of a radially oriented source dipole. So, the novelty of the present work is not in discovering the current element method, but in demonstrating that the method can be used to solve the problem where the dipole is oriented transversely relative to the sphere. This task is not trivially accomplished, since it requires conceiving a continuous distribution of image magnetic dipoles down to the centre of the sphere. But the treatment in section 2 is also useful due to its elegance and the practice of vector algebra. Additionally, it presents an alternative way of solving the problem. Overall, this example of the method of images involving superconductors is interesting from a pedagogical perspective because the source is a vector quantity, unlike the grounded conducting sphere image problem. Moreover, it presents us with the unique feature of a continuous image dipole distribution. This does not happen when an electric dipole is placed near a conducting sphere [4]. For completeness, let us mention that if we place the magnetic dipole in front of a sphere made of normal conductor (like Aluminium or Copper) instead of a superconductor, then the boundary

18
conditions Eq.(1) and Eq.(2) cannot be used. The relevant boundary conditions should
be:
(i) electric field should vanish inside the conductor
(ii) Normal component of the magnetic field and the tangential component of the
electric field should be continuous across the surface.
The method of images can still be used if charges change value or position slowly relative
to the relaxation time of the conductor [24]. See the qualitative discussion related to
Fig. 4.2.3 in section 4.2 of [24]. For superconductors, we can solve a problem as shown
in this paper. We expect that this work will help the teaching-learning process and will
gain the appreciation of the students and teachers.

5. Importance in physics pedagogy

The teachers who teach intermediate and advanced electricity and magnetism courses
for undergraduate or master’s level students will benefit from this work. One limiting
factor in this work is that superconductivity is not a common undergraduate topic.
However, with some basic introduction to electromagnetism of superconductors, as
shown in Matsushita [13], this work builds up an analogue to an electrostatics problem.
It provides a version of the solution that is more undergraduate-ready than the cited
literature, adding pedagogical appeal to the work.

6. Conflict of interests

The authors declare no conflict of interests, financial or otherwise, with any known
party.

7. Data availability statement

No new data was generated or analysed in this study.

8. Acknowledgements

We would like to thank the National Initiative on Undergraduate Science (NIUS)
Program, run by the Homi Bhabha Centre for Science Education for supporting this
research. No funding was received to carry out this work.

9. References

[1] William Thomson and Norman Macleod Ferrers. The Cambridge and Dublin Mathematical
Journal, volume I. Macmillan, 1846.
[2] David J Griffiths. Introduction to electrodynamics, 2005.
[3] Walter Greiner. Classical electrodynamics. Springer Science & Business Media, 2012.
19
[4] FC Santos and AC Tort. The electrostatic field of a point charge and an electrical dipole in the
presence of a conducting sphere. European journal of physics, 25(6):859, 2004.
[5] Kolahal Bhattacharya. Analogy of the grounded conducting sphere image problem with mirror
optics. European journal of physics, 32(5):1163, 2011.
[6] WT Norris. Charge images in a dielectric sphere. IEE Proceedings-Science, Measurement and
Technology, 142(2):142–150, 1995.
[7] Qiong-Gui Lin. Theoretical development of the image method for a general magnetic source in
the presence of a superconducting sphere or a long superconducting cylinder. Physical Review
B, 74(2):024510, 2006.
[8] Dragan Redzic, Mohsan Eldakli, and Milan Redzic. An extension of the kelvin image theory to
the conducting heaviside ellipsoid. Progress In Electromagnetics Research, 18:233–246, 2011.
[9] V Arkadiev. A floating magnet. Nature, 160(4062):330–330, 1947.
[10] William B Smythe. Static and dynamic electricity. McGraw-Hill Book Company, 2nd edition,
1950.
[11] WM Saslow. How a superconductor supports a magnet, how magnetically “soft”iron attracts a
magnet, and eddy currents for the uninitiated. American Journal of Physics, 59(1):16–25, 1991.
[12] A Bad´ıa-Maj´os. Understanding stable levitation of superconductors from intermediate
electromagnetics. American journal of physics, 74(12):1136–1142, 2006.

[13] Teruo Matsushita. Electricity and Magnetism: New Formulation by Introduction of

Superconductivity. Springer Nature, 2021.
[14] Michael Tinkham. Introduction to superconductivity. Courier Corporation, 2004.
[15] Walther Meissner and Robert Ochsenfeld. Ein neuer effekt bei eintritt der supraleitf¨ahigkeit.
Naturwissenschaften, 21(44):787–788, 1933.
[16] Ismo V Lindell. Image theory for electrostatic and magnetostatic problems involving a material
sphere. American journal of physics, 61(1):39–44, 1993.
[17] Apo Sezginer and Weng C Chew. Image of static current loop over a superconducting sphere.
IEEE transactions on magnetics, 26(3):1137–1138, 1990.
[18] DV Redˇzi´c and SS Redˇzi´c. Image current inclusion in the permeable sphere. European journal of
physics, 27(4):787, 2006.
[19] Dragan V Redˇzi´c. An extension of the magnetostatic image theory for a permeable sphere. Journal
of Physics D: Applied Physics, 39(19):4136, 2006.
[20] ZJ Yang. Interaction between a magnetic dipole and a superconducting sphere. Solid state
communications, 107(12):745–749, 1998.
[21] Mark W Coffey. Levitation force between a point magnetic dipole and superconducting sphere.
Journal of superconductivity, 13(3):381–388, 2000.
[22] Mark W Coffey. London model for the levitation force between a horizontally oriented point
magnetic dipole and superconducting sphere. Physical Review B, 65(21):214524, 2002.
[23] Mark W Coffey. Levitation force between a horizontally oriented point magnetic dipole and a
superconducting sphere. Journal of superconductivity, 15(4):257–262, 2002.
[24] David H Staelin. Electromagnetics and applications. Massachusetts Institute of Technology
Cambridge, MA, USA, 2011.