PHYS 110B - HW #1 Fall 2005, Solutions by David Pace Equations referenced as Eq. # are from Griffiths Problem statements are paraphrased

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1 PHYS 110B - HW #1 Fa 2005, Soutions by David Pace Equations referenced as Eq. # are from Griffiths Probem statements are paraphrased [1.] Probem 6.8 from Griffiths A ong cyinder has radius R and a magnetization given by M = ks 2 ˆφ. Figure 6.13 in Griffiths represents this cyinder and k is a constant. For points inside and outside the cyinder find the magnetic fied due to M. Soution The magnetic fied due to M is that resuting from the bound currents that M generates. These currents are reated to the magnetization by, J b = M Eq Kb = M ˆn Eq (1) It shoud be immediatey noted that J b = 0 outside of the cyinder (i.e. for s > R) because the magnetization is zero in that region. The bound voume current inside the cyinder may be found using the expression for cur in cyindrica coordinates. A of the terms except one are zero. J b = ks 2 ˆφ (2) = 1 s [ ] s (s ks2 ) ẑ (3) = 1 s (3ks2 )ẑ (4) = 3ksẑ (5) To find the bound surface current use ˆn = ŝ. The probem describes this as a ong cyinder, which means that end effects may be negected. This surface current is ony defined at the surface where s = R. K b = kr 2 ˆφ ŝ (6) since ˆφ ŝ = ẑ. = kr 2 ẑ (7) A of the current is directed aong the axis of the cyinder so this probem exhibits symmetry. Ampere s aw may be used to find the magnetic fied. B d = µ o I enc (8) where I enc incudes both free and bound currents. The eft side of (8) wi aways be equa to B φ (2πs) because the magnetic fied must aways be in the ˆφ direction. This probem 1

2 reduces to a matter of finding the magnetic fied a certain distance, s, away from a currentcarrying wire. For the region inside the cyinder, s < R, the encosed current is ony that due to the bound voume current density. I enc = = s 0 s Using (8) and (11) the magnetic fied inside the cyinder is, 0 J b d a (9) 3ksẑ s ds dφ (10) = 3k(2π) s3 3 = 2πks3 (11) B s<r = µ o ks 2 ˆφ (12) The tota encosed current for the region outside the cyinder, s > R, considers the bound voume and bound surface current contributions. The current on the surface is given by the surface current density times the circumference of the cyinder,. I enc = R 0 J b d a + K b (13) = R 0 3ks s ds dφ + ( kr 2 )(2πR) (14) = 3k(2π) R3 3 2πkR3 (15) = 0 (16) Since the tota encosed current is zero, the magnetic fied is aso zero. [2.] Probem 6.9 from Griffiths B s>r = 0 (17) A short cyinder of ength, L, and radius, a, features a permanent magnetization of M. This magnetization is parae to the cyinder axis. Find the bound currents and sketch the magnetic fied of this cyinder for the geometries where L a, L a, and L a. Soution Being uniform and parae to the cyinder s axis means that the magnetization may be written as, M = M o ẑ (18) where M o is a constant. The bound currents may be found using (1), J b = M o ẑ = 0 (19) K b = M o ẑ ŝ = M o ˆφ (20) 2

3 where ˆn = ŝ for the surface current density because the reevant surface is the side of the cyinder. Figure 1: L a: Athough not ceary indicated by the drawing, the magnetic fied inside the cyinder is aigned with the magnetization that is shown. This fied is uniform inside the cyinder. Figure 2: L a: In this case the cyinder ooks ike a singe oop of current. The fied in this configuration is essentiay that of a physica magnetic dipoe. 3

4 Figure 3: L a: The magnetic fied ines outside the cyinder (which are not abeed) once again match those of a dipoe. Inside the cyinder the magnetization is aigned with the magnetic fied (dashed ines). [3.] Professor Carter Probem An infinitey ong cyinder of radius a has its axis aong the z-axis. Its magnetization is given in cyindrica coordinates by M = M o (s/a) 2 ˆφ, where Mo is a constant. Find the bound currents and verify that the net charge transferred by the current is zero. Find B everywhere in space using two methods: (a) Cacuate B using the cacuated bound currents, and (b) Use the auxiiary fied, H, and the modified form of Ampere s aw to find B. Soution Use (1) to sove for the bound currents. Since this is an infinitey ong cyinder, the ony surface is the one at s = a. K b,s=a = M oa 2 a 2 ˆφ ŝ (21) = M o ẑ (22) The magnetization has ony a ˆφ component, so there wi be no ˆφ component of the bound voume current. The foowing is appicabe in the region, 0 s a, the interior of the cyinder. J b = 0 ( ) Mo s 2 ŝ + 1 [ (s M )] os 2 ẑ (23) z a 2 s s a 2 = 1 [ ] M o s 3 ẑ (24) s s a 2 = 3M os a 2 ẑ (25) 4

5 The next step is to show that these bound currents are not transferring charge. Notice that the bound surface current is directed opposite to the bound voume current (regardess of the sign of M o ). Surface charge density is given in units of Amperes per meter. The tota current fowing aong the cyinder s surface is given by mutipying the magnitude of K b by the ength across which it fows, the circumference of the cyinder. I surf = K 2πa (26) = 2πM o a (27) where I have not incuded the negative sign because it has been estabished that this current is in the direction opposite the bound voume current. The tota bound voume current is, I vo = J b d A (28) where d A is a differentia surface area (the circuar cross-section of the cyinder). I vo = a 2π 0 0 = 3M o a (2π) 2 = 6πM o a 2 [ s 3 3M o s ẑ s ds dφ ẑ (29) a 2 3 a 0 ] a 0 s 2 ds (30) (31) = 2πM o a 2 a 3 (32) = 2πM o a (33) The currents in (27) and (33) are equa. Since they are in opposite directions this means that there is no net transfer of charge. (a) Cacuate B using the bound currents. For s > a we can immediatey say that B = 0. Whie determining that the bound currents resut in no net charge transfer we have aso shown that there is no current encosed by an Amperian oop drawn around the cyinder. For s a, draw an Amperian oop inside the cyinder. This oop is centered on the cyinder s axis and is directed aong ˆφ. Use (8) to sove for B, 5

6 B(2πs) = µ o J b d A (34) 2π s 3M o s = µ o ẑ s ds dφ ẑ (35) 0 0 a 2 ( ) 3M o s 3 = µ o a (2π) (36) 2 3 B = µ om o s 2 a 2 ˆφ (37) where the direction of B is known because the Amperian oop is in the same direction. (b) Cacuate B using H and the modified form of Ampere s aw. We wi be using the foowing equations to sove for the magnetic fied by way of the auxiiary ( H) fied, H 1 µ o B M Eq (38) H = J f Eq (39) H d = I f,enc Eq. 6.20, integra form of 6.19 (40) The probem statement does not mention the existence of any free currents so I f,enc = 0. Free currents are put into a system by us (e.g. pugging a circuit into a power suppy or connecting a battery across a resistor), they do not spontaneousy appear. Using (40) and repacing H by way of (38) we may appy Ampere s aw. In this case the cosed oop is centered on the cyinder s axis and is a circe of radius s. The symmetry of the probem indicates that any magnetic fied wi be directed aong ˆφ (noting that d = s dφ ˆφ). [ ] 1 B M d µ = 0 (41) o 1 B d = M d µ (42) o B(2πs) = µ o M(2πs) (43) where M is the magnitude of M. B = µ o M (44) The expression in (44) is genera, but the vaue of M depends on which region we are examining. Outside of the cyinder, s > a, there is no magnetization and the magnetic fied is given by, B = 0 (45) 6

7 which agrees with the soution in (a). In part (a) we soved for the magnetic fied using conceptua arguments and our understanding of such fieds. Using the modified form of Ampere s aw did not provide a significant benefit in this case. Inside the cyinder, s a, the magnetization is non-zero, ( ) Mo s 2 B = µ o a 2 B = µ om o s 2 (46) a 2 ˆφ (47) which agrees with the soution in part (a). This time, using the modified form of Ampere s aw was a consideraby simper method. [4.] Probem 6.15 from Griffiths Find the fied inside the uniformy magnetized sphere of exampe 6.1 (p. 264). Take the foowing as a hint: There is no free current in this sphere, therefore, the H fied may be written as, H = W where W is a scaar potentia. By way of equation 6.23 it is aso known that, 2 W = M. Finay, M = 0 everywhere except at the surface of the sphere. Soution This is a probem concerning Lapace s equation, 2 W = 0. It is known that the soutions to Lapace s equation in spherica geometry with azimutha symmetry (i.e. no φ dependence) are of the form, ( Ξ = A r + B ) P r +1 (cos θ) (48) where Ξ is any function satisfying 2 Ξ = 0, the A s and B s are arbitrary constants, and the P terms are the Legendre poynomias. Inside the sphere there cannot be a 1/r term because that woud represent an infinite W at r = 0. Outside the sphere there can be no r term because that woud represent an infinite W at r. Therefore the expressions for W inside and outside the sphere may be partiay simpified and written as, W in = A r P (cos θ) (49) W out = B r +1 P (cos θ) (50) This scaar potentia is a continuous function (treat it ike eectric potentia, V, in the boundary vaue probems of eectrostatics), therefore we have our first boundary condition at the surface of the sphere, W in (r = R) = W out (r = R) (51) A R P (cos θ) = B R +1 P (cos θ) (52) 7

8 The orthogonaity of the Legendre poynomias means that we can equate each term separatey. A R = B (53) R +1 The probem ony asks for the fied inside the sphere, but the fied inside must sti adhere to boundary conditions that reate to the fied outside. In (53) we reate the A terms (which we want to sove) to the B terms that we do not care about. After finding another reation between these variabes we can negect the B terms and finish the probem. There is another condition at the sphere s surface that wi provide the second equation we need to sove for the A terms. H above H beow = (M above M beow) Eq (54) where the above and beow subscripts are equivaent to outside and inside respectivey. In this spherica geometry the perpendicuar components at r = R are those aong the ˆr direction. Beginning with the right side of (54), M above = M out = 0 since there is no magnetization outside of the sphere. This means that the perpendicuar component of M must aso be zero. From exampe 6.1 we may write the magnetization of the sphere as M = M o ẑ, where M o is a constant. Using ẑ = cos θ ˆr sin θ ˆθ, the ˆr component of M in is, and this is the right side of (54). M in = M o cos θ (55) For the eft side of (54) we keep ony the ˆr component of the H = W expression. Taking the expression for the gradient in spherica coordinates we get (recaing that there is aready a negative sign in the expression), H out = r W out = ( + 1)B r +2 P (cos θ) = ( + 1)B R +2 P (cos θ) (56) Hin = r W in = A r 1 P (cos θ) = A R 1 P (cos θ) (57) where the ast step in each ine incorporated the fact that r = R. Taking a of the information just cacuated aows us to rewrite (54) as, ( ) ( + 1)B P R +2 (cos θ) A R 1 P (cos θ) = M o cos θ (58) The right side incudes a cos θ term, which is equivaent to P 1 (cos θ). As such, the eft side of the above expression must have ony = 1 terms. A other vaues of A and B are zero. Rewriting (58), (2)B 1 R 3 P 1 (cos θ) + A 1 R 0 P 1 (cos θ) = M o P 1 (cos θ) (59) 8 2B 1 R 3 + A 1 = M o (60)

9 From (53) we have, and this may be put into (60) as foows, B 1 = R 3 A 1 (61) 2 R 3 R3 A 1 + A 1 = M o (62) A 1 = M o 3 (63) The scaar potentia inside the sphere is, W in = A 1 r cos θ = M o r cos θ (64) 3 Now we can sove for H inside the sphere, H in = W in = [ M o r 3 r cos θ ˆr + 1 r θ ] M o r cos θ ˆθ 3 (65) = M o 3 (cos θ ˆr sin θ ˆθ) (66) = M o 3 ẑ (67) The magnetic fied, B, may be found using (38) and the fact that M in = M o ẑ, B in = µ o ( H in + M in ) (68) ( = µ o M ) o 3 ẑ + M oẑ (69) = 2 3 µ om o ẑ (70) = 2 3 µ o M (71) where (71) is the soution given as Griffiths equation The method used here is just another way to sove the probem. [5.] Professor Carter Probem A coaxia cabe consists of a conducting wire of radius a surrounded by a conducting cyindrica tube of radius c. A current I fows down the wire and returns aong the outer conductor (see figure 6.24), in both cases the current is distributed uniformy. Two different magnetic materias exist in the space between the two conductors of the cabe: the first, with susceptibiity χ m,1, fis the space from a < r < b and the second, with susceptibiity χ m,2, fis the space from b < r < c. (a) Find the magnetic fied in the region between the conductors. 9

10 (b) Cacuate the magnetization and find the bound currents; make sure these are consistent with the answer to part (a). Soution Begin by noting that the probem writes the radia coordinate in terms of r, whie Griffiths uses s. This soution wi use s so that the method better matches that given in the text and used on other probems. (a) Let the current in the center wire be aong the ẑ direction. The current in the outer she must then be aong the ẑ direction, but this wi not matter since we ony need to sove for the fieds inside the coaxia cabe. The free currents are given in the probem and they exhibit cyindrica symmetry. We can use (40) to sove for H and then find B. H φ (2πs) = I f (72) H in = I 2πs ˆφ (73) where H in is the fied in the region between the conductors. This does not depend on the materias inside the coaxia cabe because H is determined entirey from free currents. There is a reation between B and H fieds in inear media (it is safe to assume that a magnetic materias studied in this course wi be inear), B = µ o (1 + χ m ) H Eqs and 6.32 (74) This shows how the materias affect the fied. There is a vaue of B for the region occupied by each materia. Caing these regions 1 and 2 for containing materias with χ m,1 and χ m,2 respectivey, (b) In inear media the magnetization is given by, B 1 = µ o (1 + χ m,1 ) I 2πs ˆφ (75) B 2 = µ o (1 + χ m,2 ) I 2πs ˆφ (76) M = χ m H Eq (77) The magnetization wi therefore vary according to which materia we are considering. This is simiar to the manner in which the poarization in a region of space depends on the dieectric properties of any materias there. The magnetization in each region is, M 1 = χ m,1 I 2πs ˆφ (78) M 2 = χ m,2 I 2πs ˆφ (79) where H from part (a) has been used (it s sti vaid because nothing in the probem has changed, we are just appying a different method to sove for the magnetic fied). 10

11 Once again, the bound current densities are given by (1). Beginning with the voume bound current densities, J b,1 = χ m,1i 2πs ˆφ = 1 [ ( s χm,1i )] ẑ = 0 (80) s s 2πs J b,2 = χ m,2i 2πs ˆφ = 1 [ s s There is no bound voume current. ( s χm,2i 2πs )] ẑ = 0 (81) There are three surfaces at which to determine bound surface current density, s = a, s = b, and s = c. This is where the surfaces of the magnetic materias may be found. For s = a ony the magnetization of materia 1 matters, and the norma vector to the surface is directed aong ŝ, K b = χ m,1i 2πs ˆφ ( ŝ) = χ m,1i ẑ at s = a (82) 2πa At s = b there are two separate bound surface current densities to consider. Each materia makes a contribution and the tota density on that surface is their sum. The direction of the norma vector is reversed for each region. The tota surface current density at s = b is, K b,1 = χ m,1i 2πb ˆφ ŝ = χ m,1i 2πb ẑ (83) K b,2 = χ m,2i 2πb ˆφ ( ŝ) = χ m,2i 2πb ẑ (84) K b = I 2πb (χ m,2 χ m,1 )ẑ at s = b (85) At s = c the same method is empoyed, K b = χ m,2i 2πc ˆφ ŝ = χ m,2i ẑ at s = c (86) 2πc A of the bound currents are directed aong the ẑ axis (positive or negative). Ampere s aw in the form of (8) can be used to find B and make sure it agrees with the vaue given in part (a). Cyindrica symmetry is maintained, so we have, B φ (2πs) = µ o I enc (87) where I enc is the tota encosed current (bound and free). The free current is a simpe constant that is given. The bound current is found from the bound surface current density. This density is constant, so the tota bound current on any surface is found by mutipying the current density by the ength across the surface. This ength is the circumference in our cyindricay symmetric system. 11

12 In region 1 our Amperian oop encoses the free current and the current on the surface at s = a, I enc = I + K b,a L = I + χ m,1i 2πa (2πa) = I + χ m,1i = (1 + χ m,1 )I (88) The magnetic fied in region 1 is, which agrees with the soution in part (a). B 1 = µ o (1 + χ m,1 ) I 2πs ˆφ (89) Region 2 has a different vaue for the tota encosed current, [ I enc = I f + I s=a + I s=b = I + χ m,1 I + (χ m,2 χ m,1 ) I ] 2πb (90) 2πb = I + χ m,1 I + χ m,2 I χ m,1 I = (1 + χ m,2 )I (91) The magnetic fied in region 2 is, which agrees with the soution in part (a). µ o (1 + χ m,2 ) I 2πs ˆφ (92) In this probem the magnetic materias can increase or decrease the fied. In inear materias they do not change the direction of the fied (e.g. a fied originay directed aong ˆφ wi not be redirected aong ẑ in the presence of a inear magnetic materia). Taking this into account it is sensibe that the magnetic fied inside the cabe retains the 1/s dependence that we normay see in the fied due to a current aong a wire. [6.] Professor Carter Probem A conducting sab, parae to the xy pane and extending from z = a to z = a, carries a uniform free current density j = j oˆx. The conductor (with χ m = 0) is surrounded by a materia with susceptibiity χ m = 0 (you can imagine that this materia covers a of space). Find the magnetic fied (everywhere in space) and the ocation of a bound currents. (Do the bound currents transfer net charge in this case?) Soution Figure 4 shows the set up for this probem. This is a modified form of probem 5.14 in Griffiths (see figure 5.41 in Griffiths for another reference). The sab is a conductor, therefore the magnetic fied inside of it may be determined using the unmodified version of Ampere s aw given in (8). Figure 5 shows the Amperian oop used to determine this fied (we wi extend this oop to sove for the fied outside of the conductor). The magnetic fied inside the conductor has no z component. Each individua current eement generates a fied whose z component cances out the z component of the fied generated by the neighboring current eement. The resuting fied is directed entirey aong 12

13 Figure 4: Geometry of the probem. The sab is infinite in the xy pane and has a thickness z = 2a. The pacement of the coordinate axes is arbitrary, so ong as they correcty represent an orthogona tripet (i.e. ˆx ŷ = ẑ) Figure 5: A view of the conducting sab in the yz pane. The horizonta axis is the y direction, and the vertica axis is the z direction. The current is coming out of the pane of the page, as shown by the vector dot. The magnetic fied due to one sma eement of the current is shown. A of the individua current eements add together to generate a B that is directed entirey aong the ±ŷ direction. ±ŷ. In the foowing steps I am going to ignore direction since we have soved for that based on the geometry of the probem. Using the dashed ine in figure 5 as our Amperian oop eads to, B(2y) = µ o (2yz)j o (93) B = µ o j o z (94) B = µ o j o z ŷ (95) 13

14 where in (95) I have incuded the vector properties of the magnetic fied. In this case, the negative sign is incuded because we know that in the +z ( z) pane the magnetic fied is directed aong ŷ (+ŷ). To sove for the magnetic fied outside of the conductor we can stretch the previousy used Amperian oop aong the z direction. The modified form of Ampere s aw, (40), can now be used. Equivaent to the previous case, the H fied wi have no z component because of the way contributions to the fied from neighboring current eements cance out. This simpification stems from this conductor being of infinite extent in the xy pane. One difference in this case is that the tota free current encosed by our oop wi have no z dependence because we are ony concerned with the region z > a. H(2y) = (2ay)j o (96) H = aj o (97) H = { ajo ŷ for z > a aj o ŷ for z < a (98) where in (98) I have assigned the directions of the fied. This H fied has no z dependence so it must be written in a form ike that above. Use (74) to sove for the magnetic fied outside of the conductor by way of our H fied, where H is given in (98). B = µ o (1 + χ m ) H (99) The bound currents may be found once we know the magnetization, M. Magnetization is reated to the H fied according to (77), M = χm H. The expressions for the bound currents are given in (1). There are two surfaces of interest: the xy panes at z = a (ˆn = ẑ) and at z = a (ˆn = +ẑ). Reca that the direction of the surface norma is determined by pointing away from the materia being investigated. K b,z=a = χ m ( aj o ) ŷ ( ẑ) (100) = χ m aj o ˆx (101) K b,z= a = χ m (aj o ) ŷ ẑ (102) = χ m aj o ˆx (103) Since H is uniform, that means H = 0 and there are no bound voume currents, J b = 0. This soution impies that there is a net transfer of charge due to bound currents. The appearance of a charge transfer is due to the fact that this is not a rea, physica system. The surface bound currents are going in the same direction, but they must be turning around at infinity and eventuay cosing in on themseves. We cannot sove for these cosed current paths (one above the conductor and one beow) because the tota surface current is infinite. Put another way, ask yoursef how much current is fowing aong the z = +a surface? This 14

15 is given by, I = K, where is the ength of the ine across which the surface current fows. The surface extends to infinity, so I =. This is a non-physica resut. As such, our previous requirements for bound currents are modified and we see an apparent charge transfer. Rest assured, in any rea, physica system this wi not be the case. [7.] Probem 6.18 from Griffiths A sphere is paced in a uniform background magnetic fied. The sphere is made of inear magnetic materia and the background fied is B o. Find the magnetic fied inside the sphere. (Prof. Carter Hint: use the method of probem 6.15, referring to exampe 4.7) Soution Professor Carter soved this probem using an iterative method in ecture (October 5, 2005). Here we wi foow the hint and sove this as a separation of variabes probem. There is no free current in the sphere; we know this because free current woud ony exist in the sphere if it was purposey put there, and then the probem statement woud describe it. Without free current the cur of H is zero and have (just as probem 6.15 describes), H = W (104) where W is again a scaar potentia. 2 W = 0 Inside and outside the sphere (105) The pan is to sove for W inside the sphere, use that vaue to sove for H in, and then use H in to sove for B in. Inside the sphere we have, W in = A r P (cos θ) (106) which is the given soution for Lapace s equation in sphericay symmetric systems. Before we can write the expression for W outside the sphere we have to determine the vaue of W as r. The background fied is B o = B o ẑ. The scaar potentia is in terms of H and eads to (see exampe 4.7 in Griffiths to prove this to yoursef), W (r ) = H o r cos θ = B o µ o r cos θ (107) where this makes use of B = µ o H outside the sphere. The expression for the potentia outside of the sphere contains this term and the genera sum, W out = B o C r cos θ + µ o r P (cos θ) (108) +1 where C is an unknown constant. At the surface of the sphere, W in (r = R) = W out (r = R). A R P (cos θ) = B o µ o R cos θ + C R +1 P (cos θ) (109) 15

16 Noting that cos θ = P 1 (cos θ), (109) gives two separate expressions, A 1 R = B o µ o R + C 1 R 2 for = 1 (110) A R = C R +1 for 1 (111) In the equations above there are (essentiay) two unknowns and one equation. We can use another boundary condition at the surface of the sphere to get another equation. In probem 6.15 we used (54), but in this probem we don t know anything about the magnetization inside the sphere. We do, however, have information about the magnetic fied across this boundary and we may consider, Soving for Babove, which is evauated at r = R, B above B beow = 0 Eq (112) Babove = µ o Hout = µ o r W out (113) [ ] = µ o (114) B o µ o cos θ + = B o P 1 (cos θ) + ( + 1)C R +2 P (cos θ) µ o ( + 1)C R +2 P (cos θ) (115) where the perpendicuar components are the ˆr components of the gradient. To find B beow we foow a simiar method, but in this case B = µh, where µ = µ o(1 + χ m ). Since the sphere is made of inear magnetic materia we know that its susceptibiity may be written as χ m. B beow = µh in = µ r W in (116) = µa R 1 P (cos θ) (117) The expressions (117) and (115) are equa according to the boundary condition given in (112). B o 2µ oc 1 R 3 = µ o (1 + χ m )A 1 for = 1 (118) ( + 1)C 1 R +2 = A R 1 for 1 (119) 16

17 Soving for the 1 terms first, insert (111) into (119), ( + 1) ( A R 2+1) = A R +2 R 1 (120) ( + 1)R 2+1 = R 1 R +2 (121) ( + 1)R 2+1 = R 2+1 (122) 1 = (123) = 1 2 (124) so the 1 terms ony exist for the = 1/2 vaue. Our sum for the potentia requires that the s be positive whoe numbers, therefore, the ony terms that exist in the fina soution are the = 1 terms. Sove for the = 1 terms by inserting (118) into (110), A 1 R = B o R + 1 ] [ R3 (µ µ o R 2 o (1 + χ m )A 1 + B o ) 2µ o ( A µ ) o(1 + χ m ) 2µ o (125) A 1 = B o 1 (µ o (1 + χ m )A 1 + B o ) (126) µ o 2µ o = 3B o 2µ o (127) A 1 = 3B o 2µ o A 1 = 3B o µ o ( 2µ o 2µ o + µ o (1 + χ m ) ) ( 1 ) 3 + χ m (128) (129) For the sake of eventuay matching the soution presented in ecture, I wi rewrite A 1 as, A 1 = B ( ) o 1 (130) µ o 1 + χm 3 Now that we have A 1 we know the vaue of W inside the sphere, W in = B ( ) o 1 r cos θ (131) µ o 1 + χm 3 17

18 The vaue of H in is, [ H in = W in = r W inˆr + 1 ] r θ W inˆθ [ = B ( o 1 µ o 1 + χm 3 = B ( o 1 µ o 1 + χm 3 )] ( cos θˆr + 1 (r)( sin θ)ˆθ r ) (132) (133) ) ẑ (134) Finay, the magnetic fied inside the sphere is, B in = µ H in = µ o (1 + χ m ) Bo µ o ( χm 3 ) ẑ (135) which agrees with the iterative method of ecture. = 1 + χ m B 1 + χm o ẑ (136) 3 18