PHY 35 K. Solutions for the second midterm exam. Problem 1: a The boundary conditions at the oil-air interface are air side E oil side = E and D air side oil side = D = E air side oil side = ɛ = 1+χ E. 1 For the system in question, the electric field E = k/r ˆr is spherically symmetric with respect to the metal sphere s center. Since the sphere floats in oil equator-deep, its center lies in the same plane as the oil-air interface. Consequently, at that interface, the radial electric field happens to be parallel to the surface, thus air side E oil side = E = k side ˆr, Eair r oil side = E = 0, in good agreement with the boundary conditions 1. b For the system in question, the electric field E, the displacement D, and the polarization P = D ɛ 0 E are as follows { k E = ˆr in air and in oil, r 0 inside the sphere, ɛ 0 k ˆr in air, r D = ɛɛ 0 k r ˆr in oil, 0 inside the sphere, 0 in air, ɛ 1ɛ 0 k P = r ˆr in oil, 0 inside the sphere. By inspection, the displacement D and the polarization P fields have zero divergences inside any of the three media, hence ρ f = D = 0, ρ b = P = 0, 6 which means that there are no free or bound volume charges anywhere in the system, only the surface charges. 3 4 5 1
As to the boundaries between the metal, the oil, and the air, the free and the bound surface charge densities obtain from the discontinuities of the components of the D and P fields, σ f = discd, σ b = discp. 7 In particular: At the oil-air boundary θ = π, r > R, the E, D, and P fields on both sides of the boundary are parallel to its surface or vanish altogether like P in the air, thus D = P = 0 on both sides, and the discontinuities 7 vanish. Therefore, there are no free or bound charges at the oil-air boundary. At the metal-air boundary r = R, θ < π, there is radial displaicement field in air but not in metal, while P = 0 on both sides. Consequently, there are no bound charges at this boundary, but there is free charge density σ f = D in air r @r = R = ɛ 0k R. 8 Finally, at the metal-oil boundary r = R, θ < π, there are both radial D and P fields in oil but not in metal, hence there are both free and bound charges at this boundary: σ f σ b = +D = P in oil r in oil r @r = R = + χ + 1ɛ 0k R, 9 @r = R = χɛ 0k R, 10 Note: the free charge density on the oil side of the sphere is ɛ = χ + 1 times larger than on the air side, but the net charge densities are the same on both sides: σ net [oil side] = σ f + σ b = χ + 1ɛ 0k R χɛ 0k R = ɛ 0k R = σ f [air side] = σ net [air side]. 11
c The voltage between the metal sphere and the infinity obtains from integrating the Coulomb-like electric field 3, V = R E r dr = R k r dr = k R, 1 At the same time, the charge stored in the system acting as a capacitor is the net free charge for the metal sphere, Q = πr σ f [oil side] + πr σ f [air side] = πr χ + 1ɛ 0k R + πr ɛ 0k R = πχ + ɛ 0 k. 13 Consequently, the capacitance is C = Q V = πχ + ɛ 0R = 4πɛ 0 R ɛ + 1. 14 d The net electrostatic energy of s system including dielectrics is U = 1 ρ free V d 3 Vol. 15 Note that only the free charges contribute to this integral, not the bound charges, For the system in question, all the free charges are sitting on the surface of the metal ball, where the electric potential is uniform, hence U = 1 QV, 16 where the net free charge Q and the potential V are already calculated in part c. Thus, U = 1 πχ + ɛ 0k k R = πχ + ɛ 0 k. 17 R 3
Alternatively, we may treat the whole system as a capacitor whose net electrostatic energy is simply U = 1 CV. 18 Since we have already calculated the capacitance and the voltage in part c, the energy obtains as U = 1 πχ + ɛ 0R k = πχ + ɛ 0 k. 19 R R Problem : The magnetic force on the second wire stems from the magnetic field of the first wire. For an infinite straight wire along the z axis, the magnetic field is or in Cartesian coordinates B 1 = µ 0I 1 π B 1 x, y, z = µ 0I 1 π ˆφ s, 0 x ŷ y ˆx x + y. 1 The force of this field on the second wire is F = I d l B 1 nd wire where d l is the infinitesimal wire s length vector. In light of eq. E. for the wire, d l = dl sin θ ŷ + cos θ ẑ, 3 hence df = I d l B 1 = I dl sin θ ŷ + cos θ ẑ µ 0I 1 x ŷ y ˆx π x + y µ 0 I 1 I x sin θŷ ŷ = 0 + x cos θẑ ŷ = ˆx = πx + y dl. y sin θŷ ˆx = ẑ y cos θẑ ˆx = +ŷ 4 In this formula, x and y are coordinates of the appropriate point of the second wire. Ac- 4
cording to eq. E., x = a fixed, y = sin θ l, 5 hence df = µ 0 I 1 I πa + l sin θ dl a cos θ ˆx + l sin θ sin θ ẑ cos θ ŷ. 6 It remains to integrate this force over the whole length of the second wire: F net = l=+ l= df = + µ 0 I 1 I πa + l sin θ a cos θ ˆx + l sin θsin θẑ cos θŷ dl = µ 0I 1 I π + µ 0I 1 I π a cos θ ˆx + dl a + sin θ l sin θsin θẑ cos θŷ + l dl a + sin θ l 7 using integrals E.3 = µ 0I 1 I π a cos θ ˆx π a sin θ + µ 0I 1 I π sin θsin θẑ cos θŷ 0 = µ 0I 1 I tan θ ˆx. Note: when θ = 0 or θ = 180, the force between the two wires become infinite. Indeed, for θ = 0 or θ = 180, the two wires are parallel, so the force per unit length does not diminish with l ±, hence infinite net force. As to the direction of the force 7, it s ˆx for θ < 90 and +ˆx for θ > 90. In both cases, the force is along the shortest distance between the two wires; it is attractive for θ < 90 and repusive for θ > 90. 5
Problem 3: a A steady current density must have zero divergence. Let s verify that for the current density E.4: J = J x x + J y y + J z z = 0 + 0 + 0 = 0, 8 where the first two terms vanish because the current does not depend on x or y, and the third term vansih because the current does not have a z component, J z = 0. Thus, the current density E.4 indeed has zero divergence, which allows it to be steady. We cannot prove its steadiness from the eq. E.4 alone, but at least it can be steady. b First, the translational symmetries of the current E.4 in the x and y directions make the vector potential independent of the x and y coordinates, thus Ax, y, z = Az only. 9 Second, the current E.4 is symmetric under 180 rotation around the x axis, so on that axis and hence for z = 0 but any x and y the vector potential should point in the ˆx direction, Az = 0 = A 0ˆx 30 for some constant A 0. Finally, the current E.4 has helical symmetry that is, it s invariant under simultaneous translations in the z directions and rotations around the z axis, Jz + z = Rotate[angle = k z, axis = ẑ] Jz. 31 In particular, Jz = Rotate[angle = kz, axis = ẑ] J0. 3 The vector potential should have a similar symmetry, hence Az = Rotate[angle = kz, axis = ẑ] A0 = A 0ˆx = A 0 coskz ˆx + sinkz ŷ. 33 Quod erat demonstrandum. 6
c In general, the Ampere s Law for the B field translates to the equation A = A A = µ 0 J 34 for the vector potential A. But the vector potential E.5 has obviously zero divergence for the same reasons the current E.4 has zero divergence so eq. 34 becomes the Poisson equation A = µ 0 J. 35 Let s verify this Poisson equation for the current E.4 and the potential E.5. Since the potential E.5 depends only on the z coordinate, its Laplacian is simply its second derivative, Az = d Az. 36 dz Calculating this derivative components we have J x = d dz a coskz = A0 k coskz, 37 J y = d dz a sinkz = A0 k sinkz, 38 J z = 0, 39 thus in vector notations Jz = A 0 k coskz ˆx + sinkz ŷ. 40 Comparing this formula to the current density E.4, we see that indeed Az = µ 0 Jz provided A 0 k = µ 0 J 0. 41 This nails down the overall coefficient A 0 of the vector potential E.5, A 0 = µ 0J 0 k. 4 7
d Given the vector potential Ar, the magnetic field obtains as its curl B = A. For the potential E.5 which depends only on the z coordinate, the curl simplifies to B x = da y = d A 0 sinkz = A 0 k coskz, 43 dz dz B y = + da x = + d A 0 coskz = A 0 k sinkz, 44 dz dz B z = 0, 45 or in vector notations Bz = A 0 k coskz ˆx + sinkz ŷ = µ 0J 0 k coskz ˆx + sinkz ŷ. 46 Problem 4: For the current loops whose diameter is much smaller than the distance ebtween them, we may approximate each loop as a pure magnetic dipole of magnetic moment m = I πa pointing to the loop which for the loops at hand means vertically up. Thus, m 1 = m = πa I ẑ. 47 In the dipole approximation, the torque on the second dipole relative to its own center follows from the magnetic field of the first dipole the second dipole s location, τ = m B 1 r for B 1 r = µ 0 4π where r = r 1 = r r 1 = Hˆx + lẑ, hence 3ˆr m 1 ˆr m 1 r 3, 48 τ = µ 0 4πr 3 3m 1 ˆrm ˆr m m 1. 49 For the two equal and parallel dipoles, m 1 = m = m, this formula simplifies to τ = 3µ 0 4πr 3 m ˆrm ˆr = 3µ 0 m rm r 50 4πr5 8
where for the geometry at hand m r = mẑ Lˆx + Hẑ = +mlŷ, m r = mẑ Lˆx + Hẑ = mh, 51 r = L + H. Consequently, the torque on the second dipole relative to its own center is τ = 3µ 0mHmL ŷ 4πL + H 5/ = 3µ 0m 4π LH L + H 5/ ŷ = 3πµ 0a 4 I 4 LH L + H ŷ. 5 5/ The +ŷ direction of this torque means clockwise in the x, z plane: the left side of the wire circle is pushed up up while the right side is pulled down. Now consider the torque on the first dipole relative to its own center. By symmetry between the two wire loops, we may reuse eq. 49 for this torque without any change except reversing the direction of thee unit vector ˆr. But the formula 49 is manifestly even in this unit vector, so reversing its direction makes no overall difference. Thus, τ 1 = τ = 3πµ 0a 4 I 4 LH L + H ŷ, 53 5/ the torques on eqach dipole relative to its own center are equal in both magnitude and direction. Note that τ 1 + τ 0 does not violate the angular momentum conservation because the two torques are taken relative to different relative points. Consequently, the net torque relative to the same pivot point is not τ 1 + τ but τ net = τ 1 + τ + r 1 F 1, 54 and it is this net torque which must vanish for the sake of the angular momentum conservation. In a perfect world I should have asked you to calculate the force 1 between the two magnetic dipoles and then verify that the net torque 53 indeed vanishes. But the exam time is too short, and the algebra for calculating the force is quite messy, so I have limited this problems to calculating just the τ 1 and the τ. 9
Extra material for problem 4: For the students who are curious how the net torque 54 cancels for the two dipoles in this problem, here is the explanation. The magnetic force form the second dipole on the first dipole is F 1 = m B 1 r, 55 where for the two equal and parallel magnetic moments m 1 = m = m, m B 1 r = µ 0 4πr1 3 3m ˆr 1 m = µ 0 4πr1 5 3m r 1 m r1. 56 For the geometry at hand, this means m B 1 r = µ 0 4π 3mH m L + H L + H 5/ = µ 0m 4π H L L + H. 57 5/ Taking the gradient of this expression with respect to the second dipole s position r = L, 0, H, we have where for fl, H = f L ˆx + f H ẑ 58 fl, H = H L L + H, 59 5/ f L = = L L + H 5/ 5LH L L + H 7/ L L + H 7/ L H 10H + 5L = 3L 1H, 60 f H = +4H L + H 5/ 5HH L L + H 7/ H = L + H 7/ 4L + 4H 10H + 5L = 9L 6H, 61 10
and hence fl, H = L3L 1H ˆx + H9L 6H ẑ L + H 7/. 6 In terms of the gadient of the magnetic energy of the second dipole, this translates to the force F 1 = m B 1 r = µ 0m 4π fl, H = µ 0m 4π L3L 1H ˆx + H9L 6H ẑ L + H 7/ 63 = 3πµ 0a 4 I 4 LL 4H ˆx + H3L H ẑ L + H 7/. Thus, It remains to cross this force vector by the vector distance between the two dipoles, r 1 = r r 1 = Lˆx + Hẑ. 64 r 1 F 1 = 3πµ 0a 4 I [ L + H 7/ Lˆx + Hẑ LL 4H ˆx + H3L H ] 65 where [ ] = LH3L H ˆx ẑ = ŷ + HLL 4H ẑ ˆx = +ŷ = ŷlh 3l H L + +4H = L + H = LHL + H ŷ, 66 and therefore r 1 F 1 = 3πµ 0a 4 I Comparing this formula to eq. 5 we see that and therefore Quod erat demonstrandum. 4 LH L + H ŷ. 67 5/ r 1 F 1 = τ = τ 1 68 τ net = τ 1 + τ + r 1 F 1 = 0. 69 11