Problem Set #5: 5.7,5.9,5.13 (Due Monday, April 8th)

Transcription

1 Chapter 5 Magnetostatics Problem Set #5: 5.7,5.9,5.13 (Due Monday, April 8th) 5.1 Biot-Savart Law So far we were concerned with static configuration of charges known as electrostatics. We now switch to magnetostatics where the charges are allowed to move but in a way that the created magnetic field is static corresponding to a steady state distribution of charges, ρ =0. (5.1) t Consider a total charge Q inside some volume V.ThenthechangeofQ with time can be only due to the flow of charges through the boundary, i.e. Q t = J ˆn da, (5.2) where J is the current density measured in units of charge per unit area pre unit time. Using the divergence theorem we can rewrite (5.2) as ρ(x)d 3 x = J(x)d 3 x (5.3) t which must be satisfied for an arbitrary volume element. Thus, wegeta differential equation ρ + J =0 (5.4) t known as the continuity equation, but in magnetostatics the equation is further simplified J =0, (5.5) 55

2 CHAPTER 5. MAGNETOSTATICS 56 due to the steady state condition (5.1). It is an experimental fact that the steady states (5.1) produceanobservable magnetic phenomena which depends on the current density J. Consider awireoflength l, crosssectionalarea a, pointinginˆl and carrying an electric current I, then at a distance r from the wire there is a magnetic field which is -directlyproportional l, -directlyproportionaltoi, -inverselyproportionaltothesquareofthedistancer and -pointsinthedirectionnormaltotheplaneparalleltoˆl and ˆr, i.e. In a fixed coordinate system (and in SI units), B = k I l r 2 (ˆl ˆr ). (5.6) B(x) = µ 0 J(x ) (x x ) a l, (5.7) 4π x x 3 where the current density is J = aˆl. I (5.8) To obtain an integral expression we replace a l with a volume integral over d 3 x, B(x) = µ 0 J(x ) (x x ) d 3 x. (5.9) 4π x x 3 This is the Biot-Savart Law of magnetostatics (which is analogous to the Coulomb s Law of electrostatics) discovered first by Oersted, and elaborated by Biot and Savart and later by Ampere. Ampere s experiment showed that the force on a current element dl in a magnetic field is given by df = I (dl B). (5.10) This implies that the total force on a current density distribution is F = J(x) B(x)d 3 x (5.11) and the total torque N = x (J(x) B(x)) d 3 x. (5.12)

3 CHAPTER 5. MAGNETOSTATICS 57 For example, if the magnetic field B is generated by a closed current loop #2 then the force which a closed current loop #1 experiences can be calculated by substituting (5.9) into(5.10). Note that the volume integral is replaced by line integral since J = Iδ(x )dl. (5.13) and the resulting force is F = µ 0 4π I dl1 (dl 2 (l 1 l 2 )) 1I 2 l 1 l 2 3 = µ 0 4π I (dl1 dl 2 )(l 1 l 2 )+dl 2 (dl 1 (l 1 l 2 )) 1I 2 l 1 l 2 3 = µ 0 4π I 1I 2 (dl1 dl 2 )(l 1 l 2 ) l 1 l µ 0 4π I 1I 2 the second term vanishes for closed loops, i.e. F = µ 0 4π I 1I 2 (dl1 dl 2 )(l 1 l 2 ) l 1 l 2 3. Thus, the parallel currents attract and antiparallel current repel. ( dl 2 dl 1 1 ) (5.14) l 1 l 2 (5.15) 5.2 Ampere s Law Using ( ) J(x ) x x = J(x ) 1 x x + 1 x x J(x ) = J(x ) 1 x x = J(x ) (x x ) x x 3 (5.16) we can rewrite (5.9) as B(x) = µ 0 J(x ) (x x ) 4π x x 3 d3 x = µ ( ) 0 4π J(x ) x x d3 x. (5.17)

4 CHAPTER 5. MAGNETOSTATICS 58 Then it is convenient to define a vector potential A = µ 0 J(x ) 4π x x d3 x (5.18) such that B = A (5.19) and, thus, the Gauss Law for magnetic fields (in the absence of magnetic charges) must be satisfied B = ( A) =0. (5.20) As an aside, note that there is a freedom in defining A which is called the gauge freedom or gauge symmetry. Indeed, if we redefine A new (x) =A old (x)+ ψ(x), (5.21) where ψ(x) is an arbitrary function, then the magnetic field would not change, B new = A new = (A old + ψ) = A old + ψ = A old = B old. (5.22) So, we have a freedom of choosing A which would make the calculations simple without effecting physically observable quantities such as B. This is similar to the freedom we had in choosing the electric potential since redefinitions Φ new (x) =Φ old (x)+c (5.23) would not change the physically observable electric field E new = Φ new = (Φ new + C) = Φ old = E old. (5.24) The new (gauge) transformation or new (gauge) symmetry is much bigger since it involves and arbitrary function ψ(x) and not just a constant C. For example one can choose a Coulomb gauge such that A =0. (5.25) (This is still not uniquely defined since we can add to the vector potential a gradient of an arbitrary scalar function which satisfies Laplace equation.) Of course to fully appreciate the gauge symmetry of electrodynamics we need to look at the quantum electrodynamics where the vector potential appears as a more fundamental quantity. For example in the double slit experiment a shift

5 CHAPTER 5. MAGNETOSTATICS 59 in the interference pattern is observed if there is a non-vanishing magnetic flux in-between the two trajectories B ˆn da = ( A) ˆn da = A dl. (5.26) This is known as the Aharonov-Bohm effect which disagrees with classical electrodynamics but can be explained using quantum mechanics where the same electron can simultaneously travel along two different trajectories. This shows that vector potential is more fundamental then magnetic field. If we take a curl of (5.17) andmakeuseof A = ( A) 2 A (5.27) and 1 x x 2 1 x x = 1 x x (5.28) = 4πδ(x x ) (5.29) then we get B(x) = µ ( ) 0 4π J(x ) x x d3 x = µ ( ) 0 4π J(x ) x x d3 x µ ( ) 0 1 J(x ) 2 d 3 x 4π x x = µ 0 4π J(x ) 1 x x d3 x + µ 0 J(x) = µ 0 4π J(x 1 ) x x d3 x + µ 0 J(x). (5.30) In the case of magnetostatics (5.5) we get the Ampere s law in differential form, B = µ 0 J(x), (5.31) or in Coulomb gauge A = µ 0 J (5.32) ( A) 2 A = µ 0 J (5.33) 2 A = µ 0 J (5.34)

6 CHAPTER 5. MAGNETOSTATICS 60 and by integrating over some surface we obtain the Ampere s law in integral form, B n da = µ 0 J n da (5.35) B dl = µ 0 I. (5.36) Note that (5.31) impliesthatthemagneticfieldhasanon-vanishingcurling only if there is a current; and in a region with no currents the magnetic field is curl free and, thus, completely determined by the boundary conditions. 5.3 Magnetic multipole moments Similarly to the multipole of expansion of the (scalar) electric potential, Φ, one can construct the multipole expansion of the vector (magnetic) potential, A. Consider the vector potential at x due to a localized distribution of currents (near origin) parametrized by x 0, where A(x) = µ 0 4π J(x ) x x dx (5.37) B A (5.38) B = µ 0 J (5.39) If we expand then A(x) = µ 0 1 4π x 1 x x = 1 x + x x +... (5.40) x 3 J(x )dx + µ 0 1 4π x 3 (x x ) J(x )dx +... (5.41) For an arbitrary pair of function f(x) and g(x) and localized J(x) we have gj fd 3 x = f (gj)fd 3 x + fgj n da (5.42) gj fd 3 x = fg Jfd 3 x fj gd 3 x (5.43)

7 CHAPTER 5. MAGNETOSTATICS 61 or gj fd 3 x + fg Jfd 3 x + fj gd 3 x =0. (5.44) By substituting in (5.44), f(x )=1and g(x )=x i we get 0+ x i Jfd 3 x + J i d 3 x = 0, (5.45) J i d 3 x = x i Jfd 3 x (5.46) and f(x) =x i and g(x) =x j we get x i J jd 3 x + x i x j Jfd 3 x + x i J jd 3 x + x j J id 3 x = 0 (5.47) x j J id 3 x = x i x j Jfd(5.48) 3 x. However for the steady states of magnetostatics the current is divergence-less (i.e. J =0)and(5.46), (5.48) aresimplified J i d 3 x = 0 (5.49) (x ij j + x jj i )d 3 x = 0 (5.50) From (5.49) themagneticmomentofalocalizedcurrentdensity mustvanish and (5.41) becomes A(x) = µ 0 1 (x x ) J(x )dx +... (5.51) 4π x 3

8 CHAPTER 5. MAGNETOSTATICS 62 Moreover, (5.50) canbeusedtorewrite(5.51) as A(x) = µ 0 1 dx x 4π x 3 j x jj iˆx i i,j=1,2,3 = 1 µ 0 1 ( ) dx x 2 4π x 3 j x i J j x jj i ˆx i i,j=1,2,3 = 1 µ 0 1 dx ˆx 2 4π x 3 1 (x 2 (x 1J 2 x 2J 1 ) x 3 (x 3J 1 x 1J 3 )) 1 µ 0 1 dx ˆx 2 4π x 3 2 (x 3 (x 2J 3 x 3J 2 ) x 1 (x 1J 2 x 2J 1 )) 1 µ 0 1 dx ˆx 2 4π x 3 3 (x 1 (x 3J 1 x 1J 3 ) x 2 (x 2J 3 x 3J 2 )) = 1 µ 0 1 dx x ((x 2 4π x 2J 3 3 x 3J 2 ) ˆx 1 +(x 3J 1 x 1J 3 ) ˆx 2 +(x 1J 2 x 2J 1 ) ˆx 3 ) = 1 µ π x x ( x J(x )dx ). (5.52) 3 The same calculation could have been done using the so-called Levi-Civita (or completely antisymmetric tensor) defined as +1 if (i, j, k) is (1, 2, 3), (3, 1, 2) or (2, 3, 1) ɛ ijk = 1 if (i, j, k) is (1, 3, 2), (2, 1, 3) or (3, 2, 1) (5.53) 0 if i = j, j = k, or i = k. For example, the cross product of three-vectors can be expressed as A B = i,j,k=1,2,3 ɛ ijkˆx i A j B k, (5.54) or using Einstein summation notation (i.e. always sum over repeated indices) as A B = ɛ ijkˆx i A j B k. (5.55) For instance or (A B) k = ɛ ijk A i B j (5.56) (A B) 1 = A 2 B 3 A 3 B 2 (5.57) (A B) 2 = A 3 B 1 A 1 B 3 (5.58) (A B) 3 = A 1 B 2 A 2 B 1. (5.59)

9 CHAPTER 5. MAGNETOSTATICS 63 Then A(x) = 1 µ π x 3 = 1 µ π x 3 µ 0 = 1 2 4π 1 x 3 x ( dx i,j x j ( x i J j x jj i ) ˆx i dx ɛ ijkˆx i x j (x J) k i,j x J(x )dx ). (5.60) If we now define a magnetic dipole moment as m 1 x J(x )dx (5.61) 2 then (5.52) takesthefollowingform A(x) = µ 0 m x (5.62) 4π x 3 which is similar to the electric potential produced by the electric dipole moment (4.10), Φ(x) = 1 p x 4πɛ 0 x. (5.63) 3 The magnetic dipole can also be expressed through the magnetic dipole moment density m = M(x )dx. (5.64) defined as M(x) 1 2 x J(x ). (5.65) Then the magnetic field is given by B = A = µ 0 4π m x x 3 = µ 0 4π 3ˆr (ˆr m) m r 3 (5.66) to be compared with the electric field produced due to the electric dipole moment (4.14), E = 1 3ˆr (ˆr p) p. (5.67) 4πɛ 0 r 3

10 CHAPTER 5. MAGNETOSTATICS 64 For example, if the current I flows in an arbitrary closed loop of area A confined to the z =0plane then the magnetic moment point along z axis with magnitude determined from (5.61), i.e. m = I x dl 2 = IA. (5.68) 5.4 External magnetic field Consider a localized current density from the previous section in the presence of external magnetic field, B(x). AccordingtoAmpere slawthedistribution would experience a force (5.11), F = J(x) B(x)d 3 x (5.69) and the resulting torque (5.12), N = x (J(x) B(x)) d 3 x. (5.70) If the external magnetic field is a slowly varying function (near appropriately chosen origin), then one can expand it (around the origin) to the linear order, B(x) B(0) + (x ) B(0). (5.71) Then the force (5.69) isgivenby F = d 3 xj(x) (B(0) + (x ) B(0)) ( ) = ɛ ijkˆx i d 3 x J j (x ) B k (0) + ɛ ijkˆx i d 3 x J j (x )x B k (0) (5.72) For a localized distribution of currents the first (or zeroth order) term vanishes and the leading contribution comes from the second (or linear order term) F i = ɛ ijk ( = 1 2 ɛ ijk = 1 2 ɛ ijk ) d 3 x J j (x )x l B k x l d 3 x ( J j (x )x l J ) l(x )x j B k (0) x l d 3 x (J(x ) x ) k B k (0) + 1 x i 2 ɛ ijk d 3 x (J(x ) x ) i x k B k (0) = ɛ ijk m k x i B k (0) = ɛ ijk (m ) j B k (0) (5.73)

11 CHAPTER 5. MAGNETOSTATICS 65 and thus, F = (m ) B(0) = (m B(0)) m ( B(0)) = (m B(0)) (5.74) since the magnetic field is divergence-less (i.e. B =0). Expression for torque is found by inserting the zeroth order term of (5.71) into (5.70), N = x (J(x ) B(0)) d 3 x = (x B(0)) J(x )d 3 x (x J(x )) B(0)d 3 x. (5.75) The second integral vanishes for a localized distribution of currents and the first integral is the same as the second integral in (5.72)with B k (0) replaced with B(0). Thefinalexpressionis N = m B(0). (5.76) If we define the potential energy, U, asaquantitywhosenegativegradient is force F = U (5.77) then according to (5.74) thepotentialenergy(uptoanadditiveconstant)is given by U = m B. (5.78) Clearly the energy is minimized when the magnetic moment m is in the direction of the magnetic field B. 5.5 Magnetostatics in Ponderable media When electric currents a placed in a material the material might respond (or magnetize) to the generated magnetic fields analogous to how the materials respond to electric fields. The magnetization of the material can be induced or can be permanent as in permanent magnets. To obtain a macroscopic description of such systems one must average (or coarse-grain) over macroscopically small, but microscopically large regions. The averaging is particularly simple for steady state currents of magnetostatics B micro =0 B macro =0. (5.79)

12 CHAPTER 5. MAGNETOSTATICS 66 This means that we can use a concept of coarse-grained vector potential defined as before A macro B macro. (5.80) The molecules in ponderable media can be considered as small loops of current with non-vanishing magnetic moment which do not move on macroscopic distances. However, according to (5.78) they should all rotate in the direction of the external magnetic field to minimize the potential energy. The overall effect of the induced magnetic dipole moments is defined macroscopically as a sum over averaged dipole moment density, M(x) = i N i m i, (5.81) where m i is the magnetic dipole moment and N i is the number density of the ith type of molecule. Consider a small volume V of the ponderable media. Then the vector potential at some distant point x can be expanded into multipole moments as A = µ 0 4π [ J(x ) x x V + M(x ) (x x ) x x 3 V If we treat V as an infinitesimal d 3 x then 1 A(x) = 4πɛ 0 1 = 4πɛ 0 1 = 4πɛ 0 [ J(x d 3 x ) x x + M(x ) (x x ) x x 3 d 3 x [ J(x ) x x + M(x ) ] ] ( 1 x x )] (5.82) d 3 x 1 x x [J(x )+ M(x )]. (5.83) It is convenient to separate the macroscopic magnetic field into two parts B(x) µ 0 = H(x)+M(x). (5.84) with contributions from the excess (or applied) currents and the magnetization (or induced) currents H = J (5.85) M = J m. (5.86)

13 CHAPTER 5. MAGNETOSTATICS 67 Then the combined effect is given by B µ 0 = H + M = J + J m. (5.87) Note a similar role played by P, D and E in electrostatic described by to the role played by M, H and B in magnetostatics P = ρ p (5.88) D = ρ (5.89) E = 0 (5.90) M = J M (5.91) H = J (5.92) B = 0. (5.93) The simplest types of materials are diamagnetic (i.e. µ<µ 0 )andparamagnetic (i.e. µ>µ 0 )whoseresponsetotheappliedfieldislinearand isotropic, i.e. B = µh (5.94) where µ is the magnetic permeability typically very close to µ 0.Ifthematerial is homogeneous (uniform) then all of the solutions obtained in the vacua are valid with a trivial replacement µ µ 0. If the medium, for example, consists of two regions with different magnetic properties, then we can derive boundary conditions that H and B must satisfy. An integral over a closed contour enclosing the boundary gives H dl = ( H) ˆn da = J nda (5.95) and or ˆn (H 2 H 1 )=K (5.96) M dl = ( M) ˆn da = J M nda (5.97) and ˆn (M 2 M 1 )=K m (5.98) where K and K m are the surface current densities of the applied and induced magnetization currents respectively. Moreover, a volume integral enclosing

14 CHAPTER 5. MAGNETOSTATICS 68 the boundary leads to Bd 3 x = 0 (5.99) B ˆn da = 0 (5.100) ˆn (B 2 B 1 ) = 0. (5.101) 5.6 Linear media For linear (i.e. B = µh) andhomogeneous(i.e. µ(x) =const ) media we can rewrite (5.92) intermsofthevectorpotentialas, ( ) 1 µ A = J (5.102) ( A) 2 A = µj. (5.103) Then it is convenient to work in Coulomb gauge (i.e. A =0)sothat (5.103) reducestotheformof(5.34) withµ 0 replaced with µ, 2 A = µj. (5.104) When the current density vanishes (i.e. J =0)the solution can be obtained using a magnetic scalar potential (similarly to the electric potential), and (5.93) becomes H = Φ M. (5.105) (µ Φ M ) = 0 (5.106) 2 Φ M = 0 (5.107) in each homogeneous region (i.e. µ(x) =const ). Alternatively, one can choose a scalar potential Ψ M = µφ M.

15 CHAPTER 5. MAGNETOSTATICS 69 For example, let us calculate the effect of an unmagnitized sphere of radius a made of linear magnetic material placed in an initially uniform magnetic field B = B 0 ẑ. Since there are no applied currents we can solve a Laplace equation for the scalar potential to determine 2 Ψ M =0 (5.108) B = Ψ M. (5.109) Because of the azimuthal symmetry we can expand the solution in Legendre polynomials Ψ M (r, θ, φ) = l ( Cl r l + D l r l 1) P l (cos θ). (5.110) Outside of the sphere and at very large r the magnetic field should be B = B 0 ẑ or Ψ out M B 0r cos θ (5.111) and thus, or Ψ out M 0 for i =0 C i = B 0 for i =1 0 for i 2 = B 0r cos θ + l (5.112) D l r l 1 P l (cos θ). (5.113) Inside the sphere a finiteness of the Magnetic field implies that Ψ in M = l C l r l P l (cos θ). (5.114) To determine the constants D l s and C l s we must match the solutions at the boundary using (5.101) ( Ψ out ˆr M ΨM) in = 0 B 0 cos θ l=0 [ r Ψout M ] r=a = (l +1)D l a l 2 P l (cos θ) = l=1 [ r Ψin M ] r=a lc l a l 1 P l (cos θ)(5.115)

16 CHAPTER 5. MAGNETOSTATICS 70 and (5.96) 1 µ 0 ( ( 1 ˆr Ψ out M µ 0 1 µ 0 1 µ Ψin M [ ] θ Ψout M B 0 a d dθ P 1(cos θ)+ l=1 D l a l 1 d dθ P l(cos θ) By equating the respective coefficients we get D 0 = 0 ) ) r=a = 0 = 1 [ ] µ θ Ψin M r=a = 1 C l a l d µ dθ P l(cos θ) l=1 and which are satisfied only when B 0 2D 1 a 3 = C 1 (l +1)D l a l 2 = lc l a l 1 for l>1 1 ( B0 a + D 1 a 2) µ 0 = 1 µ C 1a 1 D l a l 1 µ 0 = 1 µ C la l for l>1. D l = C l =0for l>1 C 1 = 3µ B 0 µ +2µ 0 D 1 = µ µ 0 µ +2µ 0 a 3 B 0. By putting everything together we obtain our final solution Ψ out M = B 0r cos θ + µ µ 0 µ +2µ 0 a 3 B 0 r 2 cos θ Ψ in M = 3µ µ +2µ 0 B 0 r cos θ which should be compared with (4.68) and(4.69) obtainedbysolvingacom- pletely analogous electrostatic problem. The final expression for the magnetic field is then ( ) B out = B in = ˆr r Ψout M + ˆθ 1 r ( ˆr r Ψin M + ˆθ 1 r θ Ψout M θ Ψin M ) µ µ ( 0 a ) 3 = B 0 ẑ + B 0 (3 cos θˆr ẑ ) µ +2µ 0 r 3µ = B 0 ẑ. µ + µ 0

17 CHAPTER 5. MAGNETOSTATICS Hard ferromagnets Another interesting example involves the so-called hard ferromagnets whose magnetization is nearly independent of the applied magneticfield. Inthecase of vanishing currents (5.93) iswrittenusing(5.107) as where B = 0 (5.116) µ 0 (H + M) = 0 (5.117) 2 Φ M = M (5.118) 2 Φ M = ρ M (5.119) ρ M M (5.120) is called the effective magnetic charge density. Clearly, the solutionforthe magnetic scalar potential (just like for the electric scalar potential)mustbe given by Φ M (x) = 1 4π M(x ) d 3 x. (5.121) x x For a localized distribution of M(x ) one can integrate by parts and ignore the boundary term to get Φ M (x) = 1 M(x ) 1 4π x x d3 x = 1 M(x ) 1 4π x x d3 x = 1 M(x 4π ) x x d3 x. (5.122) Note that far from the magnetization region Φ M (x) 1 ( ) 1 4π M(x )d 3 x = m x x 4πx. (5.123) 3 If the hard ferromagnets have a finite size with a boundary beyond which the magnetization vanishes, then the divergence theorem can be used to calculate the effective magnetic charge density and (5.121) generalizesto Φ M (x) = 1 4π σ M = n M (5.124) M(x ) d 3 x + 1 x x 4π ˆn M(x ) da. (5.125) x x

18 CHAPTER 5. MAGNETOSTATICS 72 Note that for a homogeneous magnetization the first term vanishes. For the problem of hard ferromagnets with currents it is convenient to work with vector potential in Coulomb gauge (i.e. A =0) H = 0 (5.126) ( ) B M = 0 (5.127) µ 0 2 A = µ 0 M (5.128) 2 A = µ 0 J M (5.129) where J M is the magnetization current (5.86). In the absence of boundary the solution was already derived in (5.83), A(x) = µ 0 M(x ) d 3 x (5.130) 4π x x and a more general solution A(x) = µ 0 4π M(x ) d 3 x 1 x x 4π ˆn M(x ) da. (5.131) x x can be obtained by including the boundary term in the differentiation by parts in (5.83). 5.8 Faraday s Law of Induction The first observation of the effect of the time-dependent magnetic fields were made by Faraday who observed that the change in time of the total magnetic flux through some surface creates an electromotive force (and therefore electric current) on the loop bounding the surface, i.e. E E dl = k d B ˆn da (5.132) dt where E is the electric filed at dl in the coordinate system where dl is at rest. To determine k we can make an assumption of Galilean invariance (i.e. physical laws are invariant under Galilean transformation (x,t) (x = x vt, t = t)). Consider a loop moving with some velocity v (whose spatial derivatives

19 CHAPTER 5. MAGNETOSTATICS 73 vanish), then d B ˆn da = dt = = B ˆn da + (v ) Bda t B ˆn da + (B v) da + v ( B) da t B ˆn da + (B v) dl (5.133) t By inserting (5.133) into(5.132) weobtain B (E k (v B)) dl = k ˆn da. (5.134) t Alternatively one can consider the same loop but in a coordinate system moving with the loop and then v =0and B E dl = k ˆn da (5.135) t where E is the electric field in the new frame. The Galilean invariance implies that E = E k (v B) (5.136) or Atestchargeatrestexperiencesanelectricforce E = E + k (v B). (5.137) F = qe (5.138) but in a moving reference frame the same charge represents a current J = qvδ(x x 0 ) and experiences an electric as well as magnetic forces, F = qe + J(x) B(x)d 3 x = q (E + v B). (5.139) Therefore the Galleon invariance implies that k =1and the Faraday s Law takes the following form E dl = d B ˆn da. (5.140) dt If we choose the reference frame where the current is held fixed E dl = B ˆn da (5.141) t

20 CHAPTER 5. MAGNETOSTATICS 74 then ( E + t B ) ˆn da =0 (5.142) must be satisfied for an arbitrary integration volume and we obtain a differential form of the Faraday s law E + B =0 (5.143) t which is a time-dependent generalization of the electrostatic equation E =0. Achangingmagneticfluxcreatesanelectromotiveforceandthus does work on a current per unit time (or power) dw dt = IE = I d dt B ˆn da. (5.144) For linear magnetic material one can show that W = 1 H Bd 3 x (5.145) 2 and the magnetic potential energy can be defined as w = 1 2µ B2. (5.146)