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2 6.2x Suppose I place a small dipole M at various locations near the end of a large solenoid. At which point is the magnitude of the force on the dipole eatest? D) Not enough information to answer E) There is no net force on a dipole r-hz) 1:. ot es*- (Acre dts.,..ts,,3 4(e (I.t.sf- ryie-((r 6.7 A small chunk of material (the "tan cube") is a placed above a solenoid. It magnetizes, weakly, as shown by small arrows inside. What kind of material must the cube be? A) Dielectric Conductor C) Diamagnetic ) Paramagnetic E) Ferromagnetic

3 6.7 Predict the results of the following b experiment: a paramagnetic bar and a diamagnetic bar are pushed inside of a solenoid. a) The paramagnet is pushed out, the diamagnet is sucked in b) The diamagnet is pushed out, the paramagnet is sucked in c) Both are sucked in, but with different force d) Both are pushed out, but with different force c e ERK6 1 Which type of magnetic material has the following properties: 1) The atoms of the material have an odd number of electrons 2) The induced atomic magnetic dipoles align in the same direction as an applied magnetic field 3) Thermal energy tends to randomize the induced dipoles A. Ferromagnetic B. Diamagnetic C. Paramagnetic c l''''`.`" mat )

4 11.8 Magnetic field caused by magnetized matter 555 J,=.111 at' a.% Figure Nonuniform magnetization is equivalent to a volume current density. current density than the one to the left of it. The current in each loop is greater than the current in the loop to the left hy am- = = v. av ' ( 11.57) At every interface in this row of blocks there is a net current in the.y direction of magnitude AL see Fig 11 20(e). To get the current per unit area flowing in the x direction we have to multiply by the number of blocks per unit area, which is IRAN. Az.). Thus = I 1 = k Ay Az ) Another ay of getting an.v-directed current is to have a y component of magnetization that varies in the direction. If you trace through

5 276 Chapter 6 Magnetic Fields in Matter Now, a rotating spherical shell, of uniform surface charge a, corresponds to a surface current density K =av = a cor sin t) It follows, therefore, that the field of a uniformly magnetized sphere is identical to the field of a spinning spherical shell, with the identification a Rto Referring back to Ex I conclude that B = ittom, (6.16) inside the sphere, while the field outside is the same as that of a perfect dipole, 4 m = 3 /rr3m. Notice that the internal field is uniform, like the electric field inside a uniformly polarized sphere (Eq. 4.14), although the actual formulas for the two cases are curiously different (i in place of )P The external fields are also analogous: pure dipole in both instances. ('-) 'F%,4 7iAlt OtaCi --F04.41 Problem 6.7 An infinitely long circular cylinder carries a uniform magnetization 11 parallel to its axis. Find the magnetic field (due to M) inside and outside the cylinder. Problem 6.8 A long circular cylinder of radius R carries a magnetization M = hn - 0. here k is a constant, c is the distance from the axis, and 0 is the usual azimuthal unit vector (Fig. 6.13). F'Dlthe magnetic field due to M. for points inside and outside the cylinder. }ICI RE 6.13 FIGURE 6.14 'it is no accident that the same factors appear in the 'contact" term for the fields of electric and magnetic dipoles (Eqs and 5.94). In fact. one good way to model a perfect dipole is to take the limit I R 0) of a polarized/magnetized sphere.

6 Problem V xm = J1, = 1 (.5 ks2) =- s (3ks2)2 = 3ksi, Kb = Mx il = ks 2(c1;x ) = --A /?2zs as So the bound current, flows up the cylinder, and returns down the surface. [Incidentally, the total current should be zero... is it? Yes, for f.4 da = (3ks)(27s ds) = 271-kW, while f Kb dl = ( k R2)(271 R) = 27rkR3.] Since these currents have cylindrical symmetry, we can get the field by Ampere's law: B 27s = Outside the cylinder it. ( = 0. so B = O. = po.4 (10 = 2ir k pos3. o B = poks2(i) = Pon

7 6.4 A solid cylinder has uniform magnetization M throughout the volume in the x direction as shown. What's the magnitude of the total magnetic dipole moment of the cylinder? R R2 L M B) 2TER L M C) 27R M D) TER 2M E) Something else/ it's complicated! zic v.(_) 6.5 A solid cylinder has uniform magnetization M throughout the volume in the x direction as shown. Where do bound currents show up? A) Top/bottom surface only B) Side (rounded) surface only C) Everywhere ), op/bottom, and parts of but not all of) side surface (but not in the volume) E) Something different/other combination!

8 6.21 A solid cylinder has uniform magnetization M throughout the volume in the (i) direction as shown. In which direction does the bound surface current flow on the (curved) sides? A. There is no bound surface current. B. The current flows in the -±- q) direction. The current flows in the -±s direction. ZI_DjThe current flows in thedirection. E. The direction is more complicated than the answers B, C, or D. 6.6 A sphere has uniform magnetization M in the z direction. Which formula is correct for this surface current? A) M sine 0 B) M sine cl\p C) M cos() 0 D) Mcos0 c^pi E) None of these!

9 -).4 Magnetic Vector Potential 245 For line and surface currents, Po f I,, Pol f 1, ito f K, = al A = a d. (5.66) 47T 47r 4 47r 4 (If the current does not go to zero at infinity, we have to find other ways to get A; some of these are explored in Ex and in the problems at the end of the section.) It must be said that A is not as useful as V. For one thing, its still a vector, and although Eqs and 5.66 are somewhat easier to work with than the Biot- Savart law, you still have to fuss with components. It would be nice if we could get away with a scalar potential B = VU, (5.67) hut this is incompatible with Ampere's law, since the curl of a gradient is always zero. (A magnetostatic scalar potential can be used, if you stick scrupulously to simply-connected, current-free regions. hut as a theoretical tool. it is of limited interest. See Prob ) Moreover, since magnetic forces do no work, A does not admit a simple physical interpretation in terms of potential energy per unit charge. (In some contexts it can be interpreted as momentum per unit charge.2 ) Nevertheless, the vector potential has substantial theoretical importance, as we shall see in Chapter 10. Example A spherical shell of radius R, carrying a uniform surface charge a. is set spinning at angular velocity co. Find the vector potential it produces at point r (Fig. 5.45). Solution It might seem natural to set the polar axis along w, but in fact the integration is easier if we let r lie on the z axis, so that co is tilted at an angle 1t. We may as well orient the x axis so that to lies in the z plane, as shown in Fig According to Eq. 5.66, FIGURE 5.45 FIGURE M. D. Semon and J. R. Taylor. Am. J. Phys. 64, 1361 (1996).

10 246 Chapter 5 Magnetostatics K(r' ), Air= da where K = a v, =,/R 2 + r 2 2Rr cos O', and da' =R 1 sin fr ci(r (10'. Now the velocity of a point r in a rotating rigid body is given by co x r': in this case, v = x = 0) sin 0 w cos V/ R N' COS R sin Nr sin 4;.' R cos N' = R(0 (cos 1,1/ sin O r sin (/)') + (cos sin Hi cos sin IP cos N') + (sin lir sin O' sin Notice that each of these terms, save one, involves either sin cpfi or cos 0". Since f02,72.7 sin 0' d4)1 = f cos d(/). = 0, such terms contribute nothing. There remains it0r 3aw sin ;II cosoi sin O' Mr) = I (hi) 0 /R 2 + r 2 2Rr cos f-i' Letting u cos 0', the integral becomes (R 2 +r 2 + Rru) du = R 2 + r 2 2Rru J, vir,± r 2 2Rru 3R2r 2 1 = t( +, + RrR ri 3R2r2 ( R 2 + Ri. )( R r )1. If the point r lies inside the sphere, then R > r, and this expression reduces to (2r/3R2 ); if r lies outside the sphere, so that R < r, it reduces to (2R 13r 2). Noting that (co x = wr sin V/SP', we have, finally, A(r) = 1 /10 Ra (a) x r). 3 ttor4a 3r3 (CO x r). for points inside the sphere. for points outside the sphere. (5.68) Having evaluated the integral, I revert to the "natural" coordinates of Fig. 5.45, in which w coincides with the =I z axis and the point r is at (r. 0): 14()Rwr7 r (1). (r < R. Au-, o, 0 3 (5.69) itor4wr7 sin, (r R). 3 r-. c)261--& =cr 1C 59 3 /4, cr etc) fir.d-gc-5 (7.-fo)