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Introduction to Electrodynamics - ch06 - Chapter 6...

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Chapter 6 Magnetostatic Fields in Matter Problem 6.1 N B B J.Lo 1 [3( A ) A ] A A A A B J.Lo ml A = m2 X 1; 1 = _ 4 "3 mi. r r - ml ; r = y; ml = mlZ; m2 = m2Y' 1 = -- 4 3z, 1rr 1rr N J.Lomlm2 ( A A ) J.Lomlm2A H 2 1 b 2 I S IN J.Lo(abI)2A I F . I . . =-- 4 ~ yxz = -- 4 ~x. ereml = 1ra , m2 = . a = -- 4 ~x. ma onentatIOn: 7rr 7rr r 'downward I (-z). Problem 6.2 elF = I dl X B; tIN = r X elF = Ir X (dI X B). Now (Prob. 1.6): r X (dI X B) + dl X (B X r) + B X (r X ill) = O. But d[r X (r X B)] = dr X (r X B) + r X (dr X B) (since B is constant), and dr = dI, so dl X (B X r) = r X (dI X B) - d[r X (r X B)]. Hence 2r X (dI X B) = d[r X (r X B)] - B X (r X dI). dN=!I {d[r X (r X B)] '- B X (r X dI)}. :. N = !I {§d[r X (r X B)] - B X §(r X dI)}. But the first term is zero (§d(... ) = 0), and the secondintegral is 2a (Eq. 1.107). So N = -I(B X a) = m X B. qed Problem 6.3 (a) I~ Accordingto Eq. 6.2, F = 27rIRBcos(). But B = l!Q. [3(ml.r~r-mtl and B cas () = B. Y A so B cas () = 4". r ' , ~~ [3(ml .i)(i. y) - (mi' y)]. But ml . Y = 0 and i . Y = sinq" while ml . i = ml cos(). :. Bcos() = ~~3ml sin q, cas q,. F = 27rIR~~3ml sin q,cos q,. Now sinq, = ~, cosq, = yr2 - R2/r, so F = 3~mlIR2~. But IR27r = m2, so F = :!.!!!!. 2 mlm2 ~, whilefor a dipole, r, so F = 3 2 J.LO m1 4 m2. ". r 1r r (b)F = V(m2. B) = (m2. V)B = (m2:z) [~z\(~(ml. z)z - ml)] = ~mlm2z d~ (z\), ~ ~ 2ml -3:\ z or,since z = r: I F = - 3J.Loml m2 A I 21r r4 z. 113
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114 CHAPTER 6. MAGNETOSTATIC FIELDS IN MATTER Problem 6.4 dF = J {(dyy) X B(O,y,O) + (dzz) X B(O,t,z) - (dyy) X B(O,y,t) - (dzz) X B(O,O,z)} = J{-(dY y) X JB(O, y, t) - B(O, y, On +(dz z) X JB(O, t, z) - B(O, 0, zn} ~ y ~ t8B ~ t8B 8z 8y :::}Jt2 { zx8 8 B_YX8 8 B } . [ NotethatJdy~Blo O~t~Bl o ooandJdz~B I ~t~B I . ] Y z z ,y, z, , y O,O,z y 0,0,0 { X Y z x y Z } F = mOO 1 - 0 1 0 = m y 8Bx - x8By- x 8Bz - z8Bx 8Bz ~!!..!i... 8Bz ~!!..!i... {8y 8y 8z 8z } 8y 8y 8y 8z 8z 8z [ ~ 8Bx ~ 8Bx ~ 8Bx ] ( . . 8By 8Bz 8Bx ) =m x-+y-+z- usmgV.B=Otownte -+- = -- . 8x 8y 8z 8y 8z 8x But m. B = mBx (since m = mx, here), so V(m. B) = mV(Bx) = m (8fxz x + ~y + 8fzz z). Therefore F = V(m. B). Qed Problem 6.5 z (a) B = 110Joxy (Prob. 5.14). m. B = 0, so Eq. 6.3 says I F = 0.1 (b) m. B = mWoJox, so IF = mol1oJox.! y (c) Use product rule #4: V(p. E) = p x (V x E) + E x (V x p) + (p . V)E + (E . V)p. But p does not depend on (x, y, z), so the second and fourth terms vanish, and V x E = 0, so the first term is zero. Hence V (p . E) = (p . V)E. Qed This argument does not apply to the magnetic analog, since V x B =I 0.' In fact, V(m. B) = (m. V)B + 110(mx J). (m. V)Ba = mofx(B) = mol1oJoY,(m. V)Bb = mo-/y(l1oJoxy) = O. Problem 6.6 Aluminum, copper, copper chloride, and sodium all have an odd number of electrons, so we expect them to be paramagnetic. The rest (having an even number) should be diamagnetic. Problem 6.7 ~fi Jb = VxM = 0; Kb = M X ii = M(p. The field is that of a surface current Kb = M (P, but that's just a solenoid, so the field !outside is zero, I and inside B = 110Kb = 110M. Moreover, it points upward (in the drawing), so I B = 110M.!
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115 Problem 6.8 VxM = Jb = ~! (sks2)z = ~(3ks2)Z = 3ksz, s vS s Kb = M X Ii = ks2(J> X s) = -kR2z. Sothe bound current flows up the cylinder, and returns down the surface. [Incidentally, the total current should be zero ... is it? Yes, for JJbda = JoR(3ks)(27rsds) = 27rkR3, while JKbdl = (-kR2)(27rR) = -27rkR3.] Sincethese currents have cylindrical symmetry, we can get the field by Ampere's law: B .
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