107 Chapter Electrostatic Fields in Matter Problem 4.1 −30 E = V / x = 500 /10−3 = ×105 Table 4.1: α / 4πε = 0.66 ×10 , so α = 4π (8.85 × 10−12 )(0.66 ×10 −30 ) = 7.34 ×10 −41 p = α E = ed ⇒ d = α E / e = (7.34 × 10−41 )(5 ×105 ) /(1.6 ×10 −19 ) = 2.29 × 10−16 m d / R = (2.29 ×10−16 ) /(0.5 ×10 −10 ) = 4.6 ×10−6 To ionize, say d = R Then R= α E / e = αV / ex ⇒ V = R ex / α = (0.5x 10−10 )(1.6x10 −19 )( 10−3 )/(7.34x10 −41 ) = 108V Problem 4.2 First find the field, at radius r, using Gauss’ law: ∫ E.da = ε Qenc , or 1 E= Qenc 4πε r 4π q r −2 r / a 4q  a −2 r / a  a2  Qenc = ∫ ρ dr = e r dr =  − e  r + ar + ÷ π a ∫0 a  2    2q  −2 r / a a2  r r2  −2 r / a − e ( r + ar + ) − e (1 + + ) = = q a   a a   r r [Note: Qenc (r → ∞) = q ] So the field of the electron could is Ee = q  r r2  −2 r / a − e (1 + + ) The proton will be shifted from r = to the 4πε r  a a  point d where E e = E (the external field): q  d d  −2 d / a  E= 1 − e 1 + + 2 ÷ 4πε d  a a   108 Expanding in powers of (d/a): e 3 d d d2   2d   d   d  d  4d  −2 d / a  = 1−  + − + = − + − + − e + +  ÷ ÷  ÷  ÷  ÷  ÷ a a a2   a   a  3!  a   a  3 a     d d d2  d  4d  − 1 − +  ÷ −  ÷ + ÷1 + + 2 ÷  ÷ a a a   a  3 a    d d2 d d2 d3 d2 d3 d3 γ −γ − − 2 + + + − 2 − + + a a a a a a a a3 4d   ÷ + higher order terms 3 a  −2 d / a = = = E= q  d3  = (qd ) = p α = 3πε a  ÷ 4πε d  a  4πε 3a 3πε a [Not so different from the uniform sphere model of Ex.(see Eq 4.2) Note that 3 this result predicts 4πε α = a = (0.5 ×10 ) = 0.09 ×10 m , compared with an experimental value (table 4.1) of 0.66 x 10 −30 m Ironically the “classical” formula (Eq 4.2) is slightly closer to the empirical value.] −10 30 Problem 4.3 ρ ( r ) = Ar Electric field (by Gauss’s Law): Ñ ∫ E.da = E (4π ) = 1 Qenc = ε0 ε0 ∫ r r r2 r 4π A r Ar Ar 4π r d r , or E= = This “internal” 4π r ε 4ε field balances the external field E when nucleus is “off-center” an amount d : ad 4ε = E ⇒ d = 4ε E A So the induced dipole moment is p = ed = 2e ε A E Evidently p is proportional to E1 For Eq 4.1 to hold in the weak – field limit , E must be proportional to r , for small r , which mean that must go to a constant (not zero) at the origin : ρ (0) ≠ (nor infinite) 109 Problem 4.4 αq q Field of q : 4π ∈ r rˆ Induced dipole moment of atom : p = α E = 4πε r rˆ 0 Field of this dipole , at location of q ( θ = π , in Eq.3.103 ) : E = 1  2α q   ÷ 4πε r  4πε r  ( to the right ) Force on q due to this field : ( attractive ) Problem 4.5 p Field of p1 at p2 ( θ = π in Eq 3.103 ) : E1 = 4πε r θˆ ( points down ) pp o Torque on p2 : N = p2 × E1 = p2 E1 sin 90 = p2 E1 = 4πε r ( points into the page) 2p p Torque on p1 : N1 = p1 × E2 = 4πε r (points into the page ) Problem 4.6 Use image dipole as shown in Fig (a) Redraw , placing pi at the origin , Fig (b) Ei = p 4πε ( z ) ( cos θ rˆ + sin θθˆ ) ; p = p cos θ rˆ + p sin θθˆ ( out of the page ) But sin θ cos θ = ( ) sin 2θ , so N = p sin 2θ (out of thae page ) 4πε ( 16 z ) For < θ < π 2, N tends to rotate p counterclockwise ; for π < θ < π , N rotates p clockwise Thus the stable orientation is perpendicular to the surface –either ↑ or ↓ 110 Problem 4.7 Say the field is uniform and points in the y direction First slide p in from infinity along the x axis-this takes no work , since F is ⊥ dl (If E is not uniform , slide p in along a trajectory ⊥ the field.) Now rotate (counterclockwise ) into final position The torque exerted by E is N = p × E = pE sin θ zˆ The torque we exert is N = pE sin θ clockwise , and dθ is counterclockwise , so the net work done by us is negative : U= θ ∫ π pE sin θ dθ = pE ( − cos θ ) θ π π  = − pE  cos θ − cos ÷ = − pE cos θ = − p.E 2  qed Problem 4.8 U = − p1.E2 , but E2 = 1 1 3( p2 rˆ)rˆ − p2 ] So U =  p1 p2 − ( p1rˆ ) ( p2 rˆ )  [ 4πε r 4πε r  Problem 4.9 ( a ) F = ( p.∇ ) E ( Eq.4.5) ; E = qed q q xxˆ + yyˆ + zzˆ rˆ = 4πε r 4πε ( x + y + z )  ∂ ∂ ∂ q x Fz =  px + py + pz ÷ ∂x ∂y ∂z  4πε0 ( x + y + z )  =    2x    + py p − x  4πε0  ( x + y + z ) 2 ( x + y + z )     = q  pz x  − p x + p y + p z = ( ) x y z  4πε0  r r  4πε0 q q   2y − x  + pz  ( x2 + y + z )     p 3r ( p.r )   3−  r r  x     111 F= q  p − ( p.rˆ ) rˆ  4πε r  ( b) E= { } 1 1  p ( −rˆ )  ( − rˆ ) − p = 3 ( p.rˆ ) rˆ − p  (This is from 4πε r 4πε r  Eq 3.104; the minus signs are beause r points toward p , in this problem ) F = qE = q 3 ( p.rˆ ) − p  4πε r  [Note that the forces are equal and opposite , as you would expect from Newton’s third law ] Problem 4.10 (a) σ b = P.nˆ = kR; ρb = −∇.P = − ∂ r kr ) = − 3kr = −3k ( r ∂r r (b) For r < R , E= 3ε ρ rrˆ (Prob 2.12 ) , so E = − ( k ε ) r For r > R, same as if all charge at center ; but 4  Qtot = ( kR ) ( 4π R ) + ( −3k )  π R ÷ = , so E = 3  Problem 4.11 ρb = 0;σ = P.nˆ = ± P (plus sign at one end-the one P points toward ; minus sign at the other-the one P points away from ) 112 (i) L >> a Then the ends look like point charges , and the whole thing is like a physical dipole , of length L and charge Pπ a See Fig (a) (ii) L R)   R3 R P cos θ ˆ P r = ,  ( r > R ) , 2 ε r ε r  0   So V ( r , θ ) =  Pr cos θ  ( r < R ) ,  ,  3ε P.r = 3ε    ( r < R )  Problem 4.13 Think of it as two cylinders of opposite uniform charge density ± ρ Inside the field 113 At a distance s from the axis of a uniformly charge cylinder is given by Gauss’s law : E 2π sl = ε ρπ s l ⇒ E = ( ρ 2ε ) s For two such cylinders , one plus and one minus The net field (inside) is E = E+ + E− = ( ρ 2ε ) ( s+ − s− ) But s+ − s− = −d , so E = ρ d ( 2ε ) , where d is the Vector from the negative axis to positive axis In this case the total dipole 2 moment of a chunk of length l is P ( π a l ) = ( ρπ a l ) d So ρ d = P, and E = − P ( 2ε ) , for s < a Outside , Gauss s law give E 2π sl = ’ ρ a sˆ ρπ a 2l ⇒ E = , For one cylinder For ε0 2ε s ρ a  sˆ+ sˆ−  the combination , E = E+ + E− =  − ÷, where 2ε  s+ s−  d s± = s m ; −1 −1  s±  d  d 1 d  s.d  1 d  s.d  =  s m ÷ s + ms.d ÷ ≅  s m ÷1 m ÷ ≅  s m ÷ ± ÷ s±   s   s  s   s   = ( s.d ) d  1 s±s m ÷  s  s 2 ( keeping only 1st order terms in d )   sˆ+ sˆ−   ( s.d ) d   s    s ( s.d ) ( s d ) − d ÷  − ÷ =  s + s − ÷−  s − s s + ÷ =  2 s 2    s  s  s+ s−  s   E ( s) = a2  ( P.sˆ ) sˆ − P  , for s > a 2ε s  Problem 4.14 114 Total charge on the dielectric is Qtot = Ñ ∫ s σ b da + ∫ν ρ b dτ = Ñ ∫ s P.da − ∫ν ∇.Pdτ But the divergence theorem says Ñ ∫ P.da = ∫ ∇.Pdτ , s ν so Qenc = qed Problem 4.15 ∂  2k k  r ÷= − ; r ∂r  r  r + P.rˆ = k b r = b,  σ b = P.nˆ =   − P.rˆ = − k a r = a. (a) ρb = −∇.P = − Q enc Gauss’s law ⇒ E = 4πε r rˆ For r < a , Qenc = , so E = For r > b , Qenc =0 (Prob 4.14) , so E = r  −k  −k  2 ÷( 4π a ) + ∫a  ÷4π r dr = −4π ka − 4π k ( r − a ) = −4π kr ; so a r      For a < r< b , Qenc =  E = − ( k ε r ) rˆ (b) Ñ ∫ D.da = Q f enc = ⇒ D = everywhere D = ε E + P = ⇒ E = ( −1 ε ) P, so E = (for r < a and r > b ); E = ( k ε r ) rˆ (for a < r < b) Problem 4.16 (a) Same as E0 minus the field at the center of sphere with uniform polarization P The latter (Eq 4.14) is − P 3ε So E = E0 + 3ε P 1 D = ε E = ε E0 + P = D0 − p + p, So D = D0 − P 3 115 (b) Same as E0 minus the field of ± charges at the two ends of the “needle” – but these are small , and far away , so E=E0 D = ε0 E = ε0 E0 = D0 − P, so D = D0 − P (c) Same as E0 minus the field of a parallel-plate capacitor with upper plate at σ = P The later is E = E0 + P − ( ε ) P, s D = ε E = ε E0 + P, so D = D0 o ε0 Problem 4.18 (a) Apply ∫ D.da = Q f to the Gaussian surface shown DA = σ A ⇒ D = σ (Note : D =0 inside the metal plate ) This is true in both slabs ; D points down (b) D = ε E ⇒ E = σ ε1 in slab , E = σ ε in slab But ε = ε 0ε r , so enc ε1 = 2ε ; ε = ε E1 = σ 2ε , E2 = 2σ 3ε −1 (c) p = ε χ e E , so P = ε χ e d ( ε ε r ) = ( χ e ε r ) ; χ e = ε r − ⇒ P = ( − ε r ) σ P1 = σ 2, P2 = σ (d) V = E1a + E2 a = ( σ a 6ε ) ( + ) = 7σ a 6ε (e) ρb = 0; σ b = + P1 at bottom of slab ( 1) = σ 2, ( 2) = σ ( ) = −σ σ b = + P2 at bottom of slab 3, σ b = − P1 at top of slab ( 1) = −σ 2, σ b = + P2 at top of slab 3, (f) In slab : total surface charge above : σ − ( σ ) = σ total surface charge below : ( σ ) − ( σ 3) + ( σ 3) − σ = −σ 2, In slab : total surface charge above : σ − ( σ ) + ( σ ) − ( σ 3) = 2σ 3, total surface charge below : ( σ 3) − σ = −2σ 3, 116 problem 4.19 With no dielectric , C0 = Aε d (Eq 2.54) In configuration (a) , with +σ on upper plate , −σ on lower , D = σ between the plates E = σ ε (in air ) and E = σ ε (in dielectric) So V = Ca = Q ε0 A  =  V d  1+1 εr σ d σ d Qd  ε  + = 1+ ε ε 2ε A  ε ÷   Ca 2ε = ÷=  C0 + ε r In configuration (b) , with potential difference V : E = V d , so σ = ε E = ε 0V d ( in air ) P = ε χ eV d (in dielectric) , so σ b = − ε χ eV d (at top surface of dielectric ) σ tot = ε 0V d = σ f + σ b = σ f − ε χ eV d , so σ f = ε 0V ( + χ e ) d = ε 0ε rV d (on top plate above dielectric) ⇒ Cb = Q 1 A A A  V V  Aε  + ε r = σ + σ f ÷=  ε0 + ε ε r ÷=  V V 2  2V  d d  d   Cb + ε r = ÷  C0 [Which is greater ? + ε r ) − 4ε r + 2ε r + 4ε r2 − 4ε r ( − ε r ) Cb Ca + ε r ( 2ε r − = − = = = > So Cb > Ca ] C0 C0 1+ εr 2(1+ εr ) 2( 1+ εr ) 2( 1+ εr ) 2 If the x axis points down : Problem 4.20 ⇒ D 4π r = ρ π r ⇒ D = ρ r ⇒ E = ( ρ r 3ε ) rˆ , for 3 r < R; D 4π r = ρ π R ⇒ D = ρ R 3r ⇒ E = ( ρ R 3ε r ) rˆ, for r >R ∫ D.da = Q f enc V = − ∫ E.dl = ∞ ρ R3 3ε r R ∞ − ρ 3ε ∫ R rdr = ρ R2 ρ R2 ρ R2   + = 1 + ÷ 3ε 3ε 3ε  2ε r  W= [1 + 21n(5/il)] z Problem 5.54 Apply the divergence theorem to the function [U x (V x V)], noting (from the product rule) that V • [U x (V x V)] = (V x V ) • (V x U) - U • [V x (V x V)]: / v [ U x ( V x V ) ] d r = J {(V x V ) • (V x U) - U • [V x (V x V)]} dr = flf [U x (V x V)] • da As always, suppose we have two solutions, B* (and Ai) and B (and A2) Define B3 = B2 - Bj (and A = A - A i ) , so that V x A3 = B3 and V x B3 = V x Bj - V x B2 = /ioJ - fM)J = Set U = V = A3 in the above identity: CHAPTER J MAGNETOSTATJCS {(V x A3) • (V x As) - As • [V x (V x As))} dr m j {(B3) | (B3) - A3 • (V x B3]} dr = j(B3fdr = |[A3X(VX A3)] • da = x B3) • da But either A is specified (in which case A3 = 0), or else B is specified (in which case B = 0), at the surface In either ease /(A x Bg) • da = So J(B 3)2dr = 0, and hence Bi = Ba qed Problem 5.55 From Eq 5.86, Btot = ~(2cos0* + sm00) There47TT fore B ■ r = B„(Z r) - ^2cos* = (B0 - cos* nr8 This is zero, for all 0, when r = R, given by £0 Evidently no field lines cross this sphere Problem 5.56 (a) J = = 9^.- a = 7rH2; m = ^-tcR2 z = %u>R2 z L = RMv - MuR2; L = Mu>R2 z (27r/w) 2?r 108 MOTttQ 2?xR? , or R= fMomo\1/s n _ Q wR* Q L Mwi?2 M " 2ir — I&H and the gyromagnetic ratio is 9= 2M (b) Because g is independent of R, the same ratio applies to all "donuts", and hence to the entire sphere (or any other figure of revolution): y v e n _ eh _ (1.60 x 1Q-19)(1.Q5 x IQ-34) Q M' 4.61 x 10~24 A m2 J(V x A) /{/H Lm 2m 4m 4(9.11 x 10~31) 4»r/*a 4?r (4HS J ^ X {j? * £*r'' Note t*)at J depends on the source point r', not on the field point r To the surface integral, choose the (x, y, z) coordinates so that r' lies on the z axis (see diagram) Then * = y/R? + (*')2 - 2Rz'cos0, while da = R2 sin 0d$dr By symmetry, the x and y components must Integrate to zero; since the z component of t is cos0, we have Problem 5.57 (a) Bavc = tw+im'U«f S-jsl _ -Mr)* -7Rx'u*6 J0 y/Jp + (z'y -2Rz'co*0 Ustu «odii» -gin0d0 tat fl , » f, v//P + (*')* - 2/k'u 2 2[2(rt + (z') ) + 2i2z'd —— 3(2hVP V / P + W-W- J = -g^p {[iZ2 + + Hz*] v/H2 + (z')2 -2tfz' - f/22 + -H^] ft] {[H2 + (*>)2 + Rz"\ \R - - [H2 + - fbt\ (R + ' Air , „ 4tt r—z z = —r , 4tril3 R2 , z*)} m n dm d0 For now we want r7 < fl,soBftve = J{3xt')dr' m (Eq 5.91), so BaVe = qed (b) This time r» > R, so Bw, = "M^f^ / ('1 * p?) ** = S/^^ J(Jxr')^ Now m JJ(rx J) B = poks24> JO Outside the cylinder Jenc = 0, so|B = Problem 6.9 = p.oM mf 115 ¥ Kt = M X n = M4> (Essentially a long solenoid) (Essentially a physical dipole) (Intermediate case) [The external fields are the same as in the electrical case; the interned fields (inside the bar) are completely different—in fact, opposite ih direction.] Problem 6.10 I fa - Mt so the field inside a complete ring would be poM The field of a square loop, at the center, is Prob 5.8: B,q = y/2 pqI/tcR Here I = Mw, and R = a/2, so y/2poMw 2y/2PQMW _ _ , = = — n e t fi e l d m g a p : 2y/2t B = poM CHAPTER MAGNETOSTATIC FIELDS IN MATM Problem 6.11 ' ;J| givenby As in Sec 4.2.3, we want the average of B = B out + Bln, where Bout is due to molecules outside a sm$ spher around point P, and B, n is due to molecules inside the sphere The average of B out is same as field at center (Pro 5.57b), and for this it is OK to use Eq 6.10, since the center is "far" from all the molecules k question: Mxi _ Afo f 47X J dr outside *out The average of Bin is g Eq 5.89—where = ±irR3M Thus the average Bin is 2^qM/3 But what is left out of the integral Aout is the contribution of a uniformly magnetized sphere, to wit: 2^M/3 (Eq 6.16), and this is precisely what Bjn puts back in So we'll get the correct macroscopic field using Eq 6.10 qed Problem 6.12 (a) M = ksz\ Jb - V x M = Kb = M X n = kR H = H0 fi0 - - -Kb This time the bound currents are small, and far away from the center, so B = Bp, | while H = j~Bo = H0 + M | H = H0 + M [Comment: In the wafer, B is the field in the medium; in the needle, H is the H in the medium; in Dm sphere (intermediate case) both B and H are modified.] B = Bo - ^MoM, H = Hn + ^M (b) (c) - M M \ -/ioM Problem 6.15 Potentials Win(r,0) = ]T Air1 Pt(cos6), (r < R ) ; Wout(rr$) = Boundary Conditions: (r>R) J (i) W i a { R , ) = Wollt(R,9), 0 -^r\Jt + ^\R = M^ = M z r = M c o s e {The continuity of W follows from the gradient theorem: W(b) - PF(a) = VW - dl = - H • dl; if the two points are infinitesimally separated, this last integral 0.) V AtRl — = R2l+lAh V Z { l + l)1lfaPl(casO) + ZlAtR?-*Pi(cose) = Mcos0 ^Combining these: + l)Rl~lAtPtica&O) - M cos 0, so At = (I ? I), and 3AX = M =» Ai = M „ Af , , _ _ M „ _ _ Thus WjnM) = y rc o s fl = —a n d hence Hln = -VWin = -y = —M, so B = /io(H + M) = /io f- ^M + M^ = CHAPTER MAGNETOSTATIC FIELDS IN MATTER 118 Problem 6.16 /H-dl = //.„„ =/, s o H = Jfc = VxL — _ , « K , B= (l + Xm)H = Al0 ds \ 2l(8 J Total enclosed current, for an amperian loop between the cylinders: 1+ 2^27r° + Xm)I > 80 f8 " S = | Ml + pi f B = / == = Problem 6.17 ' Prom Eq 6.20: $H • dl = H ( n s ) = I* = S1^^ ^ < a ) ; 11 (s > a) M t ' iiit itsi! Jt - X m J f (Eq 6.33), and J j - so Kfc = M X n = XmH x n Xml ira M = XmH = Xml 2m?' =Mxn= f^fz, ats = [-^fz, at r = a; ani I 2m' (s>o) J = Xm/ ?ra2 (same direction as I), Kfc = (opposite direction to I) Ib = Jb(na?) + Kb( n a ) — Xml ~ ym/ = |o| (as it should be, of course) = Problem 6.18 ==—==——= ^' By the method of Prob 6.15: For large r, we want B(r,0) Bo = B0z, so H = j^Boi, and hence W -> -j^Boz -jfcBorcosO a Potentials ■Wn(r,0) « £v4,r'P,(cos0), (r < R ) ; BWout(r,^) = — /fc-Bor cos + J2 pqfr Pj (cos 0), (r > R ) Boundary Conditions: J O ) Wi„(R,#) = Wmit(fl,»), i ( « ) - ^ M k + ^ L- o (The latter follows from Eq 6.26.) ^Hocos0 + |gg + 1) Pi (cos # j^ M/^^cosfi) = For i ^ 1, (i) Bi = so + 1) + /iiJA/i?'"1 = 0, and hence At = For / = 1, (i) => A i R = —f c B o R + B J R , and (ii) =» B0 + 2miBi/R* + nA x = 0, so A x * * -3B0/(2w»t^ (ii) =S> Ho WW*'* = "(2^) B = /iH = 3B0 3Br ZBqz (2po + p)' 3/iBo Hln = -Wm = ■z= rcosf = (2po + M) (2/IQ + #}' i+Xm \ R ITwaJ®0(2/UO + M) By the method of Prob 4.23: ■ Step 1: B0 magnetizes the sphere- M f sphere given by Eq 6.16: up a field within Bl & 3 as we did in deriving the Clausius-Mossotti equation in Prob 4.38.] Let's say V = §7rr3 Then Xm = ru use A= 10~10 m for r Then Xm = — (10~7) (4(93^f0-9'i)(T(j-»)) = ~2 x 10"M which is not ^d~Table 6.1 says X m = - l x However, I used only one electron per atom (copper has 29) and a very crude value for r Since the orbital radius is smaller for the inner electrons, they count for less (Am ~ r 2) I have also neglected competing paramagnetic effects But never mind this is in the right ball park Problem 6.20 > Place the object in a region of zero magnetic field, and heat it above the Curie point—or simply drop it on a hard surface If it's delicate (a watch, say), place it between the poles of an electromagnet, and magnetize it back and forth many times; each time you reverse the direction, reduce the field slightly Problem 6.21 V Identical to Prob 4.7, only starting with Eqs 6.1 and 6.3 instead of Eqs 4.4 and 4.5 V Identical to Prob 4.8, but starting with Eq 5.87 instead of 3.104 V U = -g^r[3cos0i cos#2 - cos(02 - 0i)]mim3 Or, using cos(02 - i ) = cos0i cos02 - sin0i sin02, U = — ff1*^ (sin sin 02 — 2cos0i cos02) - - - - -lyj d&i — g02 — w w _ My (cosi9l sin02 + 2sin^i cos02) =0=> 2sin0i cos0a = - cos0i sin02; fg = ^ff* (sin 9\ cos + c o s s i n ) = sin 0a cos02 = -4cos0i sin h the qovm* v»jr uij u,xg; Stable position occurs at minimum energy: = = Ik® { Either sin0i = sin0 = : —>~—► or or cos#i p cos 02 — : ft or t Which of these is the stable minimum? Certainly not© or®—for these m is not parallel to Bj, whereas m know m2 will line up along B j It remains to compare© (with Q\ = 02 = 0) and® (with 6\ — tt/2, 02 = -jr/2): Ux - M^jJra(~2); Ui 1) Ui is the lower energy, hence the more stable configuration Conclusion: They line up parallel, along the line joining them: (d) They'd line up the same way: Problem 6.22 F = /^ca x B = * B a + I £ d L x [ ( v V0)B0] ~ I (J> dI) x Kr° * vo)B0] = I j d\ x [(r • V0)B0J (because fdl - 0) Now (di x Bo)* = €ijkdlj(B0)k, and (r • V0) = j> ti dlj [(V0)i(Bb)jfe] ^ Lemma 1: j> n dlj = X) (proof below), so j jj| = I £ijJ#iS(Vo)/(Bo)k |Lemma 2: ]T eijkttjm - Suhm - hm&ki (proof below) = / J2 - MVo)i(£o)fc = / Y, MV0)i(Boh - *i{VoMBo)k} k,l,m k = J[(Vo)i(a ■ Bo) — a,(Vo • Bo)] But V0 • B0 - (Eq 5.48), and m = /a (Eq 5.84), so F = V0(m • B0) (the subscript just reminds us to take the derivatives at the point where m is located), qed Proof of Lemma 1: Eq 1.108 says /(c • r) dl - a x c =-c x a The jth component is f W dh = ~ EP,m cjpmCpOm- P»ck Cp — 5pi (i.e for the Zth component, zero for the others) Then / « dlj - ~ €jim«m = Em £'imR) ® ZpQXm** ( ~ - r-sr current 3 \P / 122 tan Everything is consistent, therefore, provided A = §MoXm x „ = -1, so A (l - I + |f) = or A (l + = I But or A (l - ^g*XmJ A= and hence I* + The exterior field is that of the central dipole plus that of the surface current, which, according to Prob 5.36, is also a perfect dipole field, of dipole moment w _ msurface ~ current _43/3 ^ _ 2irRz p 2(/ip - fi)m _ /ifcp - fi)m \auJK) — -TTtt I -—^surface J ~ -"T~ r>a/n i \ —To — \2/Zq current/ fiQ 4lT «J(2/io + /*) A»o(2/4) + fl) So the total dipole moment is: ™ - ** ~ M _ (Po - M) 3/um mtot — —m H m — ui T 111 v —' /-A • -■ ••■■ \ , Pq /Xo (2/io + W + At) and hence the field (for r > R) is ^ 4tr Problem 6.28 The problem is that the field inside a cavity is not the same as the field in the material itself V Ampire type The field deep inside the magnet is that of a long solenoid, Bo « /ioM From Prob 6.13: ( Sphere: B = B0 - f^oM s |/*oM; < Needle : B = B0 - fi0M = 0; [ Wafer : B = /10M (b)Gilbert type This is analogous to the electric case The field at the center is approximately that midway between two distant point charges, B0 « From Prob 4.16 (with E B, 1/eo H o, P M): ( Sphere : B = B0 + fM m < Needle: B = B0 = 0; ( Wafer : B = B0 + MoM = FIOM h»the cavities, then, the fields are the same for the two models, and this will be no test at all Yes Fund it with tl M from the Office of Alternative Medicine ... shown in Fig (a) Redraw , placing pi at the origin , Fig (b) Ei = p 4πε ( z ) ( cos θ rˆ + sin θθˆ ) ; p = p cos θ rˆ + p sin θθˆ ( out of the page ) But sin θ cos θ = ( ) sin 2θ , so N = p sin... aiming into yhe page , for points above it , and out of the page , for points below ) Above and below both plates the two fields cancel ; between the plates they add up to µ0 K , pointing in. .. the field is uniform and points in the y direction First slide p in from infinity along the x axis-this takes no work , since F is ⊥ dl (If E is not uniform , slide p in along a trajectory ⊥ the