In Coulomb gauge, the vector potential is then given by

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1 Physics 505 Fa 007 Homework Assignment #8 Soutions Textbook probems: Ch. 5: 5.13, 5.14, 5.15, A sphere of raius a carries a uniform surface-charge istribution σ. The sphere is rotate about a iameter with constant anguar veocity ω. Fin the vector potentia an magnetic-fux ensity both insie an outsie the sphere. The charge ensity for a uniformy charge sphere of raius a is simpy ρ x = σδ x a Since the sphere is rotating with constant anguar veocity ω, the veocity at any point x on the sphere is given by v = ω x. This aows us to write the current ensity as J = ρ v = σ ω x δ x a In Couomb gauge, the vector potentia is then given by A x = µ 0 4π J x x x 3 x = µ 0σa 3 4π ω ˆx x x Ω 1 where x = a. There are severa ways to perform the anguar integra. One quick metho is to reaize that the integra is a vector quantity. Then, by symmetry, once the Ω integra is performe, the ony irection it can point in is given by ˆx. This aows us to write ˆx x x Ω = frˆx where fr is a function to be etermine. In fact, by otting both sies with ˆx, we see that cos γ f = x x Ω = r < r +1 P 1 cos γp cos γω > where cos γ = ˆx ˆx an where r < = minr, a, r > = maxr, a Orthogonaity of the Legenre poynomias then seects out = 1, so that f = 4π/3r < /r> or ˆx x x Ω = 4π r < 3 r> ˆx

2 Inserting this into 1 gives A x = µ 0σa 3 3r r < r> ω x More expicity, we have µ 0 σa ω x r < a A = 3 µ 0 σa 4 ω x r > a 3r3 The magnetic inuction is now given by B in = A in = µ 0σa 3 ω x = µ 0σa ω 3 r < a an B out = A out = µ 0σa 4 ω x 3 r 3 = µ 0σa 4 3 3ˆxω ˆx ω r 3 r > a The magnetic inuction insie the sphere is uniform an parae to the axis of rotation, whie the magnetic inuction outsie is a ipoe pattern with magnetic moment m = 4π 3 σa4 ω 5.14 A ong, hoow, right circuar cyiner of inner outer raius a b, an of reative permeabiity µ r, is pace in a region of initiay uniform magnetic-fux ensity B 0 at right anges to the fie. Fin the fux ensity at a points in space, an sketch the ogarithm of the ratio of the magnitues of B on the cyiner axis to B 0 as a function of og 10 µ r for a /b = 0.5, 0.1. Negect en effects. For a ong cyiner negecting en effects we may think of this as a twoimensiona probem. Since there are no current sources, we use a magnetic scaar potentia Φ M which must be harmonic in two imensions. Since H = Φ M, we orient the uniform magnetic fie H 0 aong the +x axis an write H 0 ρ + α ρ cos φ, ρ Φ M ρ, φ = βρ + γ ρ cos φ, a δρ cos φ, > b < ρ < b ρ < a Of course, the genera harmonic expansion wou be of the form A m ρ m + B m ρ m cos mφ + C m ρ m + D m ρ m sin mφ. However here we have areay use the shortcut that a matching conitions for m 1 ea to homogeneous equations amitting ony a trivia zero soution.

3 The magnetostatic bounary conitions eman that H φ an B ρ are continuous at both ρ = a an ρ = b. This resuts in four equations for the four unknowns α, β, γ an δ. The magnetic fie an magnetic inuction components are H φ = 1 H 0 + α ρ φ Φ ρ sin φ, ρ > b M = β + γ ρ sin φ, a < ρ < b δ sin φ, ρ < a an B ρ = µ µ 0 H 0 + α ρ Φ ρ cos φ, ρ > b M = µ β + γ ρ cos φ, a < ρ < b µ 0 δ cos φ, ρ < a The resuting matching conitions at a an b are H 0 + α b = β + γ b, H 0 + α b = µ r β + γ b β + γ a = δ, β γ a = 1 µ r δ where µ r = µ/µ 0. These equations may be sove to yie α = 1 µ r µ 1 r b a H 0 β = µ 1 r H 0 γ = 1 1 µ 1 r a H 0 δ = 4 1 H 0 where a [ = 1+µ r 1+µ 1 r +1 µ r 1 µ 1 1 r = µ r + 1 µ r 1 a ] b µ r b 3 Substituting these coefficients in gives the magnetic scaar potentia ρ b a µ r µ 1 r, ρ > b ρ Φ M = H 0 cos φ 1 + µ 1 r ρ + 1 µ 1 r a, a < ρ < b ρ 4 ρ, ρ < a We see that the magnetic inuction for ρ < a is uniform, pointe aong the same irection as B 0, but reuce by a factor of 4/. The other two regions contain a ipoe fie in aition to a uniform component. Since H = Φ M = 4/ H 0ˆx for ρ < a, the ratio of B on axis ρ = 0 to B 0 is given by B = 4 B 0 = 4µ r µ r + 1 µ r 1 a/b 4

4 This may be potte as foows B/ B a / b = a / b = 0.1 og 10 µ r 5.15 Consier two ong, straight wires, parae to the z axis, space a istance apart an carrying currents I in opposite irections. Describe the magnetic fie H in terms of a magnetic scaar potentia Φ M, with H = Φ M. a If the wires are parae to the z axis with positions, x = ±/, y = 0, show that in the imit of sma spacing, the potentia is approximatey that of a twoimensiona ipoe I sin φ Φ M + O /ρ πρ where ρ an φ are the usua poar coorinates. We start with the magnetic inuction for a singe wire ocate at x = y = 0 an carrying current in the +ẑ irection. By eementary appication of Ampère s aw, we have B = µ 0I πρ ˆφ H I = πρ ˆφ Assuming H = Φ M, an using = ˆρ ρ + ˆφ 1 ρ φ in poar coorinates, we see that the above magnetic fie may be obtaine from a magnetic scaar potentia Φ M = I π φ 5 Note that this expression is mutipe vaue. The reason for this is that the current source at the origin ie the wire carrying current I vioates the cur-free conition H = 0 at the origin. Athough the magnetic scaar potentia is mutipe vaue when we go aroun the origin, the physica magnetic fie H is we efine everywhere except at ρ = 0. For two parae wires with currents ±I in the ẑ irection an positions x, y = ±/, 0, we superpose 5 to obtain tan 1 Φ M = I π φ 1 φ = I π y x / tan 1 y x + /

5 where φ 1 an φ are as inicate x,y Using gives ϕ ϕ 1 / / Φ M = I π tan 1 tan 1 α tan 1 β = tan 1 α β 1 + αβ y x + y /4 = I π tan 1 /ρ sin φ /ρ where we have use ρ = x + y an y = ρ sin φ. In the imit of sma spacing, the arctan may be expane in powers of /ρ. The eaing term gives the esire resut Φ M I π ρ sin φ + O /ρ 3 7 Note that ony o powers of /ρ appear in the Tayor expansion. In fact, the fu expansion of 6 yies [ 3 sin 3φ + 1 Φ M = I π = I π ρ sin φ n + 1 n=0 ρ ρ 80 n+1 sin[n + 1φ] 5 sin 5φ + ] ρ This form of Φ M = A m /ρ m sinmφ satisfies Lapace s equation in twoimensiona poar coorinates, as it must. The series converges for ρ > /, so it may be consiere an outsie soution for this probem. b The cosey space wires are now centere in a hoow right circuar cyiner of stee, of inner outer raius a b an magnetic permeabiity µ = µ r µ 0. Determine the magnetic scaar potentia in the three regions, 0 < ρ < a, a < ρ < b, an ρ > b. Show that the fie outsie the stee cyiner is a two-imensiona ipoe fie, as in part a, but with a strength reuce by the factor F = Reate your resut to Probem µ r b µ r + 1 b µ r 1 a

6 Using the approximation 7, we write the magnetic scaar potentia in a three regions as α ρ sin φ, ρ > b Φ M = βρ + γ ρ sin φ, a < ρ < b 8 I π ρ + δρ sin φ, ρ < a where α, β, γ an δ are constants to be etermine. Note that we aow the ipoe potentia 7 in the interior region to be moifie by the aition of δρ sin φ because of the presence of the stee cyiner. This expression for Φ M is simiar to that of corresponing to a cyiner in a region of initiay uniform magnetic inuction B 0. However, here the source term I/π/ρ sin φ is a ipoe source for ρ < a as oppose to a uniform source H 0 ρ cos φ for ρ > b. In any case, we match H φ an B ρ at both ρ = a an ρ = b. The components H φ an B ρ are H φ = 1 ρ φ Φ M = α ρ cos φ, β + γ ρ cos φ, I π ρ > b a < ρ < b ρ δ cos φ, ρ < a an α B ρ = µ µ 0 ρ sin φ, ρ Φ M = µ β + γ ρ sin φ, µ 0 I π ρ > b a < ρ < b ρ + δ sin φ, ρ < a The resuting matching conitions at a an b are α b = β + γ b β + γ a = I π a + δ α b = µ r β + γ b β γ a = 1 µ r I π a + δ This simutaneous system of four equations for four unknowns may be sove to yie α = I where π β = I π 1 µ 1 r b γ = I 1 + µ 1 r π δ = I π a b µ r µ 1 r a b = 1 [ µ r + 1 µ r 1 a ] µ r b

7 is efine ienticay with 3. Substituting these coefficients into 8 gives Φ M = I π sin φ 4 ρ, ρ > b 1 µ 1 r 1 + µ 1 r b ρ + ρ 1 ρ b a µ r µ 1 r a b ρ, ρ < a, a < ρ < b Since the ipoe fie for ρ < a is generate by the 1/ρ term in the thir ine above, we see that the externa ρ > b fie is aso a ipoe, but with strength reuce by the factor F = 4 = 4µ r µ r + 1 µ r 1 a/b 9 Note that this shieing factor is ientica to 4, which was obtaine in the soution to Probem This emonstrates that the shieing factor from insie to outsie is ientica to that from outsie to insie. c Assuming that µ r 1, an b = a + t, where the thickness t b, write own an approximate expression for F an etermine its numerica vaue for µ r = 00 typica of stee at 0 G, b = 1.5 cm, t = 3 mm. The shieing effect is reevant for reuction of stray fies in resientia an commercia 60 Hz, 110 or 0 V wiring. [The figure iustrates the shieing effect for a/b = 0.9, µ r = 100.] Taking a = b t, we may express the ratio a/b = 1 t/b. Substituting this into 9 gives 4µ r F = 4µ r + t/b t/bµ r 1 = 1 + µ 1 rt b Substituting in the above vaues gives [ t 1 + µ r 1 t b b ] 1 1 µ 1 r F 1 5 = 0.04 In other wors, the externa ipoe fie is reuce by a factor of 5 compare with the fie insie the stee shie A circuar oop of wire of raius a an negigibe thickness carries a current I. The oop is centere in a spherica cavity of raius b > a in a arge bock of soft iron. Assume that the reative permeabiity of the iron is effectivey infinite an that of the meium in the cavity, unity.

8 a In the approximation of b a, show that the magnetic fie at the center of the oop is augmente by a factor 1 + a 3 /b 3 by the presence of the iron. Since this probem invoves a current oop, we take a vector potentia approach. For a circuar oop of raius a in the x-y pane in free space, we have foun the expression for the vector potentia A oop φ = µ 0Ia r< r> +1 P 1 0P 1 cos θ where r < = minr, a an r > = maxr, a. Of course, since the current oop is centere in a spherica cavity, this cannot be the compete answer. In orer to fin the appropriate soution, we note that, by inear superposition, we may a an arbitrary soution to Lapace s equation, A = 0, to the above. Taking avantage of cyinrica symmetry, the genera soution to the homogeneous equation may be written as A 0 φ = A r + B r +1 P 1 cos θ where A an B are coefficients to be etermine. This was essentiay emonstrate in Probem 5.8. Taking an insie soution requires us to set B = 0, so that the vector potentia is we behave as r 0. Superposing A oop an A 0 then gives A φ = µ 0Ia 1 r < + 1 r> +1 + α r P 1 0P 1 cos θ 10 where we have chosen to write the homogeneous soution using a new set of expansion coefficients α for convenience. Athough the homogeneous soution cou in principe invove even terms, it is easy to see that such terms must vanish as they wou not be source by the current oop which ony generates o terms. Assuming the reative permeabiity of the iron is effectivey infinite, the bounary conitions inicate that the magnetic fie must en up being perpenicuar to the surface of the cavity. In other wors, the parae component of B must vanish at r = b. Because of azimutha symmetry, the ony parae component of interest is B θ r = b = 1 r r ra φ = µ 0Ia = µ 0Ia r=b 1 a + 1 r + + 1α r 1 b 1 a + 1 b α P 1 0P 1 cos θ P 1 0P 1 cos θ r=b

9 where we have use r < = a an r > = r when computing B θ at the surface of the cavity. Setting this parae component to zero gives in which case 10 becomes A φ = µ 0Ia α = 1 r < + 1 r> +1 a + 1 b +1 Note that for r < a this expression may be rewritten as A φ = µ 0I + ar + 1 b +1 P 1 0P 1 cos θ 1 r 1 + a +1 P 1 0P 1 cos θ + 1 a + 1 b Using gives an B r = 1 r sin θ = µ 0I a x [ 1 x P 1 x] = + 1P x θ sin θa φ r a a b +1 P 1 0P cos θ B θ = 1 r r ra φ = µ 0I 1 a r a a +1 P 1 0P 1 cos θ + 1 b This emonstrates that the -th component of the magnetic inuction is enhance by a factor F = 1 + a b compare to a current oop in free space. The fie at the center of the oop is given by setting r = 0, in which case ony the = 1 term contributes. The enhancement factor is then F 1 = a b 3 1 Note that this resut is vai, even if b is not much greater than a so ong as we assume the iron has infinite reative permeabiity.

10 b What is the raius of the image current oop carrying the same current that simuates the effect of the iron for r < b? As ong as we work in the imit b a, the = 1 enhancement factor F 1 ominates over the others in the sense that a/b +1 gets much smaer as increases. Because of this, a we nee out of an approximate image oop is the correct enhancement of the magnetic ipoe. In the ipoe approximation, an isoate current oop of raius R has magnetic inuction B z = µ 0I R at its center. For the enhancement factor F 1 given in 1, the magnetic inuction is expicity B z = µ 0I a 3 = µ 0I a b a + µ 0 I b 3 /a The first term on the right is interprete as the magnetic inuction from the rea oop, whie the secon is that from the image oop. This inicates that the raius of the image oop is R image = b3 a which is greater than b as appropriate for an image. Note that no exact singe image oop soution is possibe, as the enhancement factor 11 is epenent, an a epenent raius simpy oes not make sense. The reason this approximation makes sense for b a is that in this case the > 1 image moments are insignificant, an may be ignore.