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Section 1: Main results of and

Electrostatics

Charge density

The most fundamental quantity of electrostatics is . Charge comes in two varieties, which are called “positive” and “negative”, because their effects tend to cancel. Charge is conserved: it can not be created or destroyed. Charge is quantized. The fact is that electric charges come only in discrete lumps – integers multiples of the basic unit charge e. However, the fundamental unit of charge is so tiny that in macroscopic applications this charge quantization can be ignored. Continuous distribution of charge is described in terms of the . The charge density ρ()r is the local charge per unit , so that the total charge in volume V is Qdq==∫∫ρρ()rr dV = ∫ () dr3 . (1.1) VV V V 3 d r ρ(r) r Fig.1.1 O

If charge q is localized in a point of space, say r0, this charge is called a point charge. In this case the charge density is given in terms of the delta function:

3 ρδ()rr=−q (r0 ). (1.2)

For N point charges qi located at positions ri (i = 1…N) the charge density is given by

N 3 ρδ()rr=−∑ qi (ri ). (1.3) i=1 In addition to volume charge density we will also be dealing with surface and line charge densities. The density σ ()r is given the local surface charge per unit surface, such that the charge in a small surface da at point r is then dq= σ ()r da . Similarly, the line charge density λ()r is given by the local line charge per unit dl, such that the charge in a small element of length dl at point r is then dq= λ()r dl .

Coulomb’s law

All of electrostatics originates from the quantitative statement of ’s law concerning the force acting between charged bodies at rest with respect to each other. Coulomb, in a series of experiments, showed experimentally that the force between two small charged bodies separated in air a distance large compared to their dimensions • varies directly as the magnitude of each charge, • varies inversely as the square of the distance between them, • is directed along the line joining the charges, and • is attractive for the oppositely charged bodies and repulsive if the bodies have the same type of charge.

1 Therefore, according to Coulomb’s law the force acting on charge q located at point r due to charge q1 located at point r1 is qq rr− Fr()= 11, (1.4) 4πε 3 0 rr− 1 where ε0 is the of free space. Although the thing that eventually gets measured is a force, it is useful to introduce a concept of an . The electric field is defined as the force exerted on a unit charge located at position vector r. It is a vector function of position, denoted by E. F = qE, (1.5) where F is the force, E the electric field, and q the charge. According to Coulomb the electric field at the point r due to point charges qi located at positions ri (i=1…N) is given by 1 N rr− Er()= q i , (1.6) 4πε ∑ i 3 0 i=1 rr− i which is the superposition of field produced by the individual point charge. The electric field can be visualized in terms of lines starting on positive charges and terminating on negative charges.

3 d r' r- r' E(r) ρ(r')

r' r Fig.1.2 O

If there is continuous distribution of the charge it can be described by a charge density ρ()r′ as is shown schematically in Fig.1.2. In this case the sum in Eq.(1.6) is replaced by an integral: 11rr−′ rr−′ Er()==dq ρ()r′ d3 r′ . (1.7) ∫∫33 44πε 00VVrr−−′′πε rr Similarly we can write expressions for the field produced by surface and line charge densities.

Gauss’s law There is an important integral expression, called Gauss’s law, which relates the of electric field across a closed surface and the electric charge inside this surface electric E(r). The flux of electric field through a surface S is the integral of the normal component of the electric field over the surface, ∫ En⋅ da , where n is the normal to this surface. The flux over any closed surface is a measure of the total S charge inside. Indeed for charges inside the closed surface field lines starting on positive charges should terminate on negative charges inside or pass through the surface. On the other hand, charges outside the surface will contribute noting to the total flux, since its field lines pass in one side and out the other. This is the essence of Gauss’s law. For a continuous charge density ρ(r), Gauss’s law takes the form: 1 En⋅ da=ρ() r d3 r= Q , (1.8) v∫∫S ε 0 V

2 where V is the volume enclosed by S and Q is the total charge inside the volume V. This equation is one of the basic equations of electrostatics. Note that it depends upon the inverse square law for the force between charges, the central nature of the force, and the linear superposition of the effects of different charges. Therefore, Gauss’s law is the consequence of Coulomb’s law. Differential form of Gauss’s Law Gauss’s law (1.8) can be thought of as being an integral formulation of the law of electrostatics. We can obtain a differential form by using the theorem. The states that for any well-behaved A(r) defined within a volume V surrounded by the closed surface S the relation An⋅=∇⋅da A d3 r (1.9) v∫∫S V holds between the of the divergence of A and the of the outwardly directed normal component of A. The equation in fact can be used as the definition of the divergence. The divergence theorem allows us to write Eq.(1.8) as follows

⎛⎞ρ 3 ∫⎜⎟∇ ⋅−E dr =0 . (1.10) V ⎝⎠ε 0 Since this equation is valid for an arbitrary volume V, we can put the integrand equal to zero to obtain ρ ∇⋅E = , (1.11) ε 0 which is the differential form of Gauss’s law of electrostatics.

Scalar The single equation (1.11) is not enough to specify completely the three components of the electric field E(r). A vector field can be specified completely if its divergence and are given everywhere in space. Thus we look for an equation specifying curl E as a function of position. Such an equation follows directly from Coulomb’s law (1.7) because it can be written as follows: 11 Er()=− ∇∫ ρ()r′ dr3 ′ . (1.12) 4πε 0 rr− ′ Since the curl of the of any well-behaved function of position vanishes ( ∇×∇ψ =0 , for all ψ), it follows immediately from this equation that ∇ ×=Ε 0. (1.13) Eq. (1.13) is the consequence of the central nature of the force between charges and of the fact that the force is a function of relative distances only. According to Eq.(1.12) the electric field (a vector) is derived from a scalar by the gradient operation. Since one function of position is easier to deal with than three, it is worthwhile to define the scalar potential Φ(r) by the equation: E = −∇Φ . (1.14)

Then the scalar potential is given in terms of the charge density by 1()ρ r′ Φ=()r ∫ dr3 ′ . (1.15) 4πε 0 rr− ′

3 where the integration is over all the space, and Φ is arbitrary only to the extent that a constant can be added to the right-hand side of this equation. The scalar potential has a physical interpretation when we consider the work done on a test charge q in transporting it from one point A to point B in the presence of an electric field E(r), as is shown in Fig. 1.3.

Fig.1.3

The force acting on the charge at any point is F = qE, so that the work done in moving the charge from A to B is BBBB Wdqdqdqdq=−Fl ⋅ =− El ⋅ = ∇Φ⋅ l = Φ= Φ −Φ . (1.16) ∫∫∫∫(B A ) AAAA For a localized charge distribution it is convenient to place a reference point at infinity so that the potential at point r is defined as follows: r Φ()rE=−∫ ⋅dl. (1.17) ∞ This shows that qΦ can be interpreted as the of the test charge in the electrostatic field. It also follows from eq. (1.16) that v∫ El⋅ d = 0. (1.18) S This is consistent with ∇×Ε =0 because of the Stokes's theorem saying that if A(r) is a well-behaved vector field, S is an arbitrary open surface, and C is the closed curve bounding S, then v∫∫Al⋅=∇×⋅dd( An) a. (1.19) CS where dl is a line element of C, n is the normal to S, and the path C is traversed in a right-hand screw sense relative to n. For a charge density ρ()r placed in the external potential Φ()r the potential energy is U=Φ∫ ρ()rr ()d3r. (1.20) This is different from the potential energy of a charge distribution ρ()r which is defined as the energy which is required to build this distribution by bringing the charges from infinity. In this case Φ()r is the potential produced by the charge distribution ρ()r and the energy is 1 W=Φρ()rr ()d3r, (1.21) 2 ∫ Where ½ appears due to double counting. This energy can also be rewritten in terms of electric field alone ε WE= 0 ∫ 23dr. (1.22) 2 all space

4 Poisson and Laplace Equations We see that the behavior of an electrostatic field can be described by the two differential equations: ρ ∇⋅E = , (1.23) ε 0 ∇ ×=Ε 0. (1.24) the latter equation being equivalent to the statement that E is the gradient of a scalar function, the scalar potential Φ: E = −∇Φ . (1.25) Equations (1.23) and (1.25) can be combined into one partial differential equation for the single function Φ(x): ρ ∇Φ=−2 . (1.26) ε 0

This equation is called the Poisson equation. In regions of space that lack a charge density, the scalar potential satisfies the Laplace equation: ∇2Φ=0 . (1.27)

Dipole potential At a large distance from a localized change distribution the electrostatic potential exhibits a multipole nature. This follows from eq.(1.15) if we assume that rr ′ .The first non-vanishing term gives a monopole contribution to the potential: Q Φ=()r , (1.28) 4πε 0r where Qd= ∫ ρ()r′ 3r′ , (1.29) is the total charge. If Q = 0, the potential is given by a dipolar term 1 pr⋅ Φ=()r 3 , (1.30) 4πε 0 r where p is the dipole prr= ∫ ρ()′ ′′dr3 . (1.31) Energy in a slowly varying electric field If the external electric field varies slowly over the region where a charge distribution ρ()r is localized one expend the electrostatic energy given by (1.20) U=Φ∫ ρ()rr ()d3r (1.32) in multipoles. By expending the potential Φ()r around the origin of coordinates in a Taylor series we obtain Φ=Φ+∇Φ⋅+=Φ−(rrrE ) (0) ( ) ... (0) (0) ⋅+r ..., (1.33) r=0

5 so that

UdrdrQ≈Φ−⋅=Φ−⎡⎤⎡⎤ρρ()rr33 (0) ()rE (0) p⋅E ⎣⎦⎣⎦∫∫ . (1.34) The first term is a monopole contribution and the term is the dipole contribution to the electrostatic energy. Higher order terms are not included. If electric field is applied to a medium made up of large number of or , the charges bound in each will respond to applied field which will results in the redistribution of charges leading to a polarization of the medium. The polarization P(r') is defined as the dipole moment per unit volume. Polarization is a macroscopic quantity. The potential produced by polarization P(r') is 1 Pr()′′⋅−( r r) Φ=()r dr3 ′ . (1.35) ∫ 3 4πε 0 V rr− ′ As was shown in Phys.913, this eq. can be rewritten 1()1()∇⋅′ Pr′′Pr⋅ n Φ=−()r ∫∫dr3 ′ + v da. (1.36) 44πε 00VSrr−−′πε rr′ As follows from this expression, the polarization of the medium produces an effective charge which can be interpreted as bound charge or polarization charge. There are two contributions to the bound charge – bulk and surface. The bulk polarization charge density is given by

ρP ()rP= −∇ ⋅ ()r. (1.37) The surface polarization charge density is σ ()rPrn= ()⋅ P . (1.38) If the integration in eq.(1.36) is performed over all space the second term can formally be omitted (it is due to the abrupt change of the polarization on the surface of the ). We can, therefore, make a general statement that the presence of the polarization produces an additional polarization charge so that the total charge density becomes

ρtotal= ρρρ free+=−∇⋅ P P . (1.39) The respective electrostatic equations involve a macroscopic electric field which is the average over volume which includes a large number of atoms. The accurate procedure for the macroscopic averaging will be discussed later. Taking into account eq.(1.39) we can write the divergence of E as follows: 1 ∇ ⋅=E[ρ −∇⋅P]. (1.40) ε 0 It is convenient to define the electric displacement D,

DE= ε 0 + P, (1.41) Because this field is generated is generated by free charges only. Using the electric displacement the Gauss’s law takes the form ∇ ⋅=D ρ . (1.42) In the integral form it reads as follows: Dn⋅=daρ() r d3 r . (1.43) v∫∫S V

6 This is particularly useful way to represent Gauss’s law because it makes reference only on free charges. Connecting D and E is necessary before a solution for the electrostatic potential or fields can be obtained. For a linear response of the system the displacement D is proportional to E,

DEP= ε 0 +=εE, (1.44) where ε is the electric permittivity. If the dielectric is not only isotropic, but also uniform, then ε is independent of position. The Gauss’s law (1.42) can then be written ρ ∇ ⋅=E . (1.45) ε In this case all problems in that medium are reduced to those with no electric polarization, except that the electric fields produced by given charges are reduced by a factor ε/ε0. The reduction can be understood in terms of a polarization of the atoms that produce fields in opposition to that of the given charge. One immediate consequence is that the of a is increased by a factor of ε/ε0 if the empty space between the electrodes is filled with a dielectric with dielectric constant ε/ε0. Boundary conditions For solving electrostatics problems one needs to know boundary conditions for the electric field. Consider a boundary between different media, as is shown in Fig.1.4. The boundary region is assumed to carry idealized surface charge σ. Consider a small pillbox, half in one medium and half in the other, with the normal it to its top pointing from medium 1 into medium 2. According to the Gauss’s law D ⋅=ndaσ A , (1.46) v∫ S where the integral is taken over the surface of the pillbox and A is the area of the pillbox lid. In the limit of zero thickness the sides of the pillbox contribute nothing to the flux. The contribution from the top and bottom surfaces to the integral gives A()DDn21− ⋅ , resulting in

⊥⊥ DD21− = σ , (1.47) where D⊥ is the component of the electrical displacement perpendicular to the surface. Eq. (1.47) tells us that there is a discontinuity of the D⊥ at the interface which is determined by the surface charge.

A

E2 D2

σ

E1 D1 l

Fig. 1.4 Schematic diagram of the boundary surface between different media.

7 Now we consider a rectangular contour C such that it is partly in one medium and partly in the other and is oriented with its plane perpendicular to the surface. Since the curl of electric field is zero we have v∫ El⋅ d = 0. (1.48)

&& & For the rectangular contour C of height this integral is equal to (E21− El) , where E is the component of electric field parallel to the surface. This implies that

&& E2= E1, (1.49) i.e. the tangential component of electric field is always continuous. The electrostatic potential is continuous across the boundary.

Magnetostatics Electric currents In electrostatics electric fields which are constant in time are produced by stationary charges. In magnetostatics magnetic fields that are constant in time are produces by steady currents. If the mobile volume charge density is ρ and the is v, then J= ρv. (1.50)

Fig. 1.5 The current crossing a surface S can be written as I==⋅ JdaJn da . (1.51) ∫∫⊥ SS In particular, the total charge per unit time leaving a volume V is v∫∫Jn⋅=∇⋅da( J) d3 r . (1.52) SV Because charge in conserved, whatever flows out through the surface must come at the expense of that remaining inside: d ∂ρ ∫∫∫()∇⋅J dr33 =−ρ dr =− dr3. (1.53) VVdt V∂t The minus sign reflects the fact that an outward flow decreases the charge left in V. Since this applies to any volume, we conclude that ∂ρ ∇⋅J =− . (1.54) ∂t This is the precise mathematical statement of local . It is called the .

8 Magnetostatics deals with steady currents which are characterized by no change in the net charge density anywhere in space. Consequently in magnetostatics ∂ρ /∂=t 0 and therefore ∇ ⋅=J 0 . (1.55) We note here that a moving point charge does not constitute a steady current and therefore cannot be described by laws of magnetostatics. Biot and Savart Law Magnetic phenomena is convenient to describe in terms of a B. Biot and Savart (in 1820), first, and (in 1820-1825), in much more elaborate and thorough experiments, established the basic experimental laws relating the magnetic field B to the electric currents and established the law of force between one current and another. These experiments can be summarized in two relatively simple expressions. The magnetic field produced by a current distribution of density Jr()′ at point r is given by μ Jr()′′×−( r r) Br()= 0 dr3 ′ . (1.56) 4π ∫ rr− ′ 3 where μ0 is the permeability of free space. This is Biot and Savart law which is an analog of the Coulomb’s law in electrostatics. The magnetostatic force produced by a magnetic field Br( ) on a object carrying a volume Jr() is FJrBr=×∫ () ( )dr3 . (1.57) For a moving charged particle this equation is actually represents the fvB= q( × ) , (1.58) where q and v are a charge and velocity of the particle.

Differential Equations of Magnetostatics and Ampere's Law The magnetic field (1.56) can be written in the form μ Jr()′ Br()=∇×0 dr3 . (1.59) 4π ∫ rr− ′ It follows immediately from here that ∇ ⋅=B 0 . (1.60) This is the first differential equation of magnetostatics. Calculating the curl of B from eq. (1.59) for steady-state currents, ∇ ⋅=J 0 , leads to

∇ ×=Bμ0J. (1.61) This is the second equation of magnetostatics. This is a differential form of Ampere’s law. The integral form of the Ampere’s law can be obtained by applying Stokes’s theorem. Integrating the normal component of the vectors in the left- and right-hand side of Eq.(1.61) over open surface S shown in Fig.1.6 we obtain: ∇×Bn ⋅da =μ Jn ⋅ da . (1.62) ∫∫0 SS

9 Fig. 1.6

Using the Stokes’s theorem it can be transformed into Bl⋅=dμ Jn ⋅da. (1.63) v∫∫0 CS Since the surface integral of the current density is the total current I passing through the closed curve C, Ampere’s law can be written in the form: Bl⋅=dμ I. (1.64) v∫ 0 C Just as Gauss’s law can be used for calculation of the electric field in highly symmetric situations, so Ampere's law can be employed in analogous circumstances. The basic differential laws of magnetostatics are ∇ ⋅=B 0 , (1.65)

∇ ×=Bμ0J. (1.66) According to eq. (1.65) the divergence of B is zero which implies that there are no sources which produce a magnetic field. There exist no magnetic analog to electric charge. According to eq. (1.66) a magnetic field curls around current. Magnetic field lines do not begin or end anywhere – to do so would require a nonzero divergence. They either form closed loops or extend out of infinity. Now the problem is how to solve differential equations (1.65) and (1.66). If the current density is zero in the region of interest, ∇×B =0 permits the expression of the magnetic field B as the gradient of a magnetic scalar potential, B =−∇ΦM . Then (1.65) reduces to the Laplace equation for ΦM, and all our techniques for handling electrostatic problems can be brought to bear. A large number of problems fall into this class, as was discussed in Phys. 913. A general method of attack is to exploit equation (1.65). If ∇ ⋅=B 0 everywhere, B must be the curl of some vector field A(r), called the vector potential: Br()= ∇× Ar (). (1.67) We have, in fact, already written B in this form (1.59). Evidently, from (1.59), the general form of A is μ Jr()′ Ar()=+0 dr3 ∇λ()r . (1.68) 4π ∫ rr− ′

The added gradient of an arbitrary scalar function Ψ shows that for a given magnetic induction B, the vector potential can be freely transformed according to Ar()→+∇ Ar ()λ () r . (1.69) This transformation is called a gauge transformation. We will discuss gauge transformations in detail later.

10 Such transformations on A are possible because (1.67) specifies only the curl of A. The freedom of gauge transformations allows us to make ∇⋅A have any convenient functional form we wish. If (1.67) is substituted into the equation (1.66), we find

∇ ×∇×A =μ0J. (1.70) or 2 ∇∇⋅( AA) −∇ =μ0J. (1.71) If we now exploit the freedom implied by (5.29), we can make the convenient choice of gauge, ∇ ⋅=A 0 . (1.72) In this case each rectangular component of the vector potential satisfies the Poisson equation,

2 ∇=−Aμ0J. (1.73) From our discussions of electrostatics it is clear that the solution for A in unbounded space is μ Jr()′ Ar()= 0 dr3 ′ . (1.74) 4π ∫ rr− ′

It is easy to see by taking directly the divergence of eq. (1.74) that indeed ∇ ⋅=A 0 .

Magnetic dipole moment At a large distance from a localized current distribution we can consider the asymptotic behavior of the vector potential. We have shown in Phys. 913 that the first non-vanishing contribution is given by the magnetic dipole term μ mr× Ar()= 0 . (1.75) 4π r3 where m is the : 1 mrJr=×[]()dr3 . (1.76) 2 ∫

Forces on a Localized Current Distribution If a localized current distribution is placed in an external magnetic field B(r), it experiences a force according to Ampere’s law. The general expression for the force is given by eq. (1.57) FJrBr=×∫ () ( )dr3 . (1.77) If the external magnetic field varies slowly over the region of current, a Taylor series expansion can be used to find the dominant term in the force. We expand the applied field around some suitably chosen origin within the current distribution Br= B0+⋅∇ r′′ Br ( ) + ... . (1.78) () ( ) ( ) r′=0 The force that the field exerts on a localized current distribution is then expanded as follows: F=− B0 × Jr ( )dr3 + Jr ( ) × r ⋅∇′′ Br ( )dr3+ ... . (1.79) ( ) ∫∫( ) r′=0 Now, the first integral in the last line vanishes for a localized steady state current distribution (there can't be any net flow of charge in any direction). The second integral after some transformations (see Phys. 913) gives

11 FmB= ∇⋅( ) , (1.80) where the gradient is to be evaluated at the center of the current distribution and m is the magnetic moment (1.76). Notice in particular that there is no force if the applied magnetic induction is uniform. More generally, the force is in the direction of the gradient of the component of B in the direction of m. The potential energy of a permanent magnetic moment in an external magnetic field can be obtained from the expression for the force (1.80). If we interpret the force as the negative gradient of the potential energy U, we find U = −⋅mB. (1.81) This is well-known result which shows that the dipole tends to orient itself parallel to the field in the position of lowest potential energy. Magnetic Fields in Matter In the presence of matter, atomic give rise to effective atomic currents (or bound currents) which produce the orbital magnetic moment. In addition, electrons have their intrinsic magnetic moments due to ’s spin. Thus matter has the magnetic dipole moment. The magnetic dipole moment per unit volume is known as the M. The magnetization produces a magnetic field which can be described through the vector potential. In the magnetized object, each volume element dr3 ′ carries a dipole moment Mr()′dr3 ′ (Fig. 1.7), so the total vector potential is given by μ Mr()(′ ×− r r′ ) Ar()= 0 dr3 ′ . (1.82) 4π ∫ rr− ′ 3 A(r)

r - r'

3 d r' r

r' Fig.1.7

O It was shown in Phys.913 course that eq. (1.82) can be transformed to

μ ⎪⎧ ∇×′′Mr() Mr () ′′× n ⎪⎫ Ar()=+0 ⎨∫∫dr3 ′ v da′⎬ . (1.83) 4π ⎪⎩⎭rr−−′′ rr ⎪ The first term looks like the potential of a volume current density,

JM = ∇×M, (1.84) while the second term looks like the potential of a surface current density

KMM = × n, (1.85) where n is the normal unit vector. With these definitions

12 μ ⎪⎧ Jr()′′ Kr () ⎪⎫ Ar()=+0 ⎨ MMdr3 ′ da′⎬ . (1.86) ∫∫′v ′ 4π ⎩⎭⎪V rr−− rr ⎪ What this means is that the potential (and hence also the field) of a magnetized object is the same as would be produced by a volume current JM = ∇×M throughout the material, plus a surface current

KMM =×n, on the boundary. Instead of integrating the contributions of all the infinitesimal dipoles, as in Eq.(1.82), we first determine these bound or magnetization currents, and then find the field they produce, in the same way we would calculate the field of any other volume and surface currents. Notice the striking parallel with the electrostatics: there the field of a polarized object was the same as that of a polarization volume charge ρP =−∇⋅P plus a polarization surface charge σ P = Pn⋅ .

Incidentally, like any other steady current, JM obeys the :

∇ ⋅=J M 0 , (1.87) because the divergence of a curl is always zero. If the integration in eq. (1.86) is performed over all space the second term can be omitted (it appears in this integration due to the abrupt change of the magnetization on the surface). That implies that in general the effective current density in a magnetic medium represents a sum of free currents and magnetization currents, J+∇×M. Consequently a new macroscopic equation for the magnetic field averaged over large number of atoms in medium is

∇×BJM =μ0 ( +∇× ) . (1.88) The term ∇×M can be combined with B to define a new macroscopic field H, 1 HB= − M. (1.89) μ0 Therefore, the macroscopic equations become ∇ ×=HJ, (1.90) ∇ ⋅=B 0 . (1.91) Eq.(1.90) is the Ampere’s law for magnetostatics with magnetized materials. It represents a convenient way to find magnetic field H using only free currents. In the integral form the Ampere’s law reads v∫ Hl⋅ dI= , (1.92) where I is the electric free current passing through the loop. To complete the description of macroscopic magnetostatics, there must be a constitutive relation between H and B. For diamagnetic and paramagnetic materials and not too strong fields the relation is linear and is given by

BHM=+=μ0 ( ) μH. (1.93) where μ is the permeability of the material. Boundary Conditions Just as the electric field suffers a discontinuity at a surface charge, so the magnetic field is discontinuous at a surface current. Only this time it is the tangential component that changes. Indeed, if we apply ∇⋅B =0 in the integral form v∫ Β ⋅nda = 0. (1.94) S

13 to a thin gaussian pillbox straddling the surface (Fig. 1.8), we obtain

⊥⊥ BB21− = 0. (1.95) where B⊥ is the component of the magnetic field B perpendicular to the surface. Eq. (1.95) tells us that B⊥ is continuous at the interface. The perpendicular component of H is however discontinuous if the magnetization of the two media a different:

⊥ ⊥⊥⊥ HH21−=−−( MM 2 1) . (1.96) As for the tangential components, from Ampere’s law ∇ ×=HJ an amperian loop running perpendicular to the current (Fig. 1.8) yields Hl⋅ dHHlK=−&& =l. (1.97) v∫ ( 21) or && HHK21− = . (1.98) Thus the component of H that is parallel to the surface but perpendicular to the current is discontinuous in the amount K . A similar amperian loop running parallel to the current reveals that the parallel component is continuous. These results can be summarized in a single formula:

&& HH21− =×( Kn) . (1.99) where n is a unit vector perpendicular to the surface, pointing “upward”.

A

B2 H2

K

B1 H1 l Fig. 1.8

Magnetic scalar potential If the current density vanishes in some finite region in space, i.e. J = 0 , eq.(1.90) becomes ∇ ×=H 0 . (1.100)

This implies that we can introduce a magnetic scalar potential ΦM such that

H = −∇Φ M , (1.101) just as E =−∇Φin the electrostatics. Assuming that the medium is linear and uniform (i.e. the magnetic permeability is constant in space) eq.(1.101) together with eqs. (1.93) and (1.91) lead to the Laplace equation for the magnetic scalar potential:

2 ∇ Φ=M 0. (1.102) Therefore, one can use methods of solving differential equations to find the magnetic scalar potential and therefore the magnetic fields H and B.

14 In hard ferromagnets the situation becomes simpler. In this case the magnetization M is largely independent on the magnetic field and therefore we can assume that M is a given function of coordinates.

We can exploit eqs. BHM=+μ0 () and ∇ ⋅=B 0 to obtain

∇ ⋅=∇⋅+BHMμ0 ( ) =0 , (1.103) and hence ∇ ⋅=−∇⋅HM. (1.104) Now using the magnetic scalar potential (1.101) we obtain a magnetostatic Poisson equation:

2 ∇ Φ=−M ρM , (1.105) where the effective magnetic charge density is given by

ρM = −∇ ⋅M . (1.106)

The solution for the potential ΦM it there are no boundary surfaces is 11ρ ()r′ ∇⋅′ Mr(′) Φ=()r M dr33′ =− dr′ . (1.107) M 44ππ∫∫rr−−′rr′

In solving magnetostatics problems with a given magnetization distribution which changes abruptly at the boundaries of the specimen it is convenient to introduce magnetic surface charge density. If as hard ferromagnet has volume V and surface S we specify M(r) inside V and assume that it falls suddenly to zero at the surface S. Application of the divergence theorem to ρM (1.106) in a Gaussian pillbox straddling the surface shows that the effective magnetic surface change density is given by

σ M = nM⋅ , (1.108) where n is the outwardly directed normal. Then instead of (1.107) the potential is represented as follows 1()1()∇⋅′ Mr′′n⋅ Mr′ Φ=−()r dr3 ′ + da′ . (1.109) M ∫∫v 44ππVSrr−−′rr′ An important special case it that of uniform magnetization throughout the volume V. Then the first term vanishes; only the surface integral over σ M contributes. As an example consider a slab of magnetic material which has a uniform magnetization M oriented ether parallel (Fig. 1.9a) or perpendicular (Fig.1.9b) to the surfaces of the slab. The slab is infinite in the plane. We need to calculate the magnetic fields H and B everywhere in space.

(a) (b) + + + + H M M − − − − Fig. 1.9

Since J = 0 , ∇×H =0 and ∇ ⋅=−∇⋅HM. This implies that ρM =−∇⋅M plays a role of magnetic charge density, and H can be found like electric field E in electrostatics. In case (a), since

M=const, ρM =−∇⋅M =0 , and therefore H = 0 everywhere in space. Therefore B = 0 outside the slab and B= μ0M inside the slab. In case (b) magnetization creates positive surface charge, σ M =+M , on the top surface and negative surface charge, σ M = −M , on the bottom surface. These charges generate magnetic field, HM=− , opposite to the magnetization within the slab and no field outside, H = 0 . This makes field B zero everywhere in space.

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