V(t) = Q(t) C, (1) and the same voltage also applies to the resistor R. Consequently, the current through the resistor is. I(t) = V(t) R = Q(t)

Transcription

1 PHY 35 K. Solutions for problem set #1. Problem 7.: (a) Suppose at some time t after the switch is closed the capacitor has charge Q(t). Then at that moment of time, the voltage on the capacitor is V(t) = Q(t) C, (1) and the same voltage also applies to the resistor. Consequently, the current through the resistor is I(t) = V(t) = Q(t) C. () The same current also flows through the capacitor itself, or rather it flows from the positive plate of the capacitor through the wires and the resistor all the way to the negative plate of the capacitor. This current makes the capacitor discharge at the rate I(t), thus its charge diminishes with time according to dq = I(t) = Q(t) C (3) Solving this first-order differential equation gives us exponentially-decreasing charge Q(t) = Q exp( t/τ) (4) where Q is the original charge at time t = when the switch was closed, and τ = C (5) is the time constant of the C circuit: After time t = τ, the charge has diminished to Q = exp( 1) Q.37Q. 1

2 The voltage on the capacitor and the current through the resistor also decrease exponentially with time, V(t) = V exp( t/τ), I(t) = I exp( tτ), (6) for the same time constant τ as the charge Q(t), but V = Q C and I = V = Q C. (7) (b) The instantaneous electric power dissipated by the resistor is P = I. For the exponentially decreasing current (6), P(t) = I (t) = I exp( t/τ), (8) so integrating this power over time we obtain the net dissipated energy as W = P(t) = I exp( t/τ) = I τ. (9) In light of eq. (5) for the time constant, this energy is W = I C = C (I ) = C V. (1) On the other hand, the energy stored in the capacitor before it began to discharge through the resistor is U = C V, (11) so we see that W = U : The net energy dissipated by the resistor is precisely the energy originally stored in the capacitor. Quod erat demonstrandum.

3 (c) Now consider the circuit on textbook figure 7.5(b). For simplicity, let s neglect the internal resistance of the battery, so the voltage V on the battery stays constant at all times. Let s also assume that the capacitor was completely discharged by the time t = when the switch was closed. Suppose at some time t > the capacitor has voltage V(t) between its plates, so its charge is Q(t) = C V(t). At the same time, the voltage between the resistor s plates is V (t) = V V(t), (1) so the current flowing through the resistor and hence through the whole circuit is I(t) = V (t) = V V(t). (13) From the capacitor s point of view, the direction of this current is from the negative plate to the positive plate, through the resistor and the battery, so this current charges the capacitor. Thus, dq = +I(t) = + V V(t) so the voltage V(t) between the capacitor s plates obeys dv = 1 C dq = V V(t) C, (14). (15) The solution to this differential equation with the initial condition V() = is for the time constant exactly as in eq. (5) in part (a). V(t) = V ( 1 exp( t/τ) ) (16) τ = C, (17) As to the time dependence of the capacitor charge Q(t) and the charging current I(t), they follow from the time-dependent voltage (16): Q(t) = CV ( 1 exp( t/τ) ), I(t) = V(t) = V exp( t/τ). (18) 3

4 (d) The instantaneous power delivered by the battery is P(t) = V I(t), so for the exponentially decreasing current (18), P(t) = V V exp( t/τ). (19) The net electrical work by the battery while the capacitor is charging is the time integral of this power, thus W net = P(t) = V exp( t/τ) = V τ. () For the time constant τ = C we have found in part (c), the bet work by the battery amounts to W net = V C = C V. (1) Some of this electric work goes towards charging the capacitor, the rest is dissipated by the resistor. The capacitor s energy is U(t) = C V (t) () where the voltage increases with time according to eq. (16). Asymptotically, the capacitor voltage approaches the battery voltage V, so its energy gets to U net = C V. (3) Note that this energy is only a half of the electric work (1) delivered by the battery, so the other half of this work is dissipated by the resistor, W dissipated = W net U net = C V. (4) Note: this dissipated energy is exactly the same as in part (b) because the current I(t) through the resistor has exactly the same form as in part (b), cf. eqs. (6) and (18). 4

5 Problem 7.5: The voltage on the battery V = E ri is equal to the voltage V = I on the load, E r I = I, (5) hence the current through the circuit is I = E r +. (6) The power delivered to the load is P = V I = I = (r +) E. (7) To maximize this power as a function of the load s resistance we need = r. Indeed, ( ) dp d = 1 E (r +) (r +) 3 = E r (r+) 3, (8) thus the power increases with for < r but decreases for > r, with the maximum at = r. Problem 7.7: (a) For the magnetic field B pointing into the page and the metal bar moving to the right, the direction of the vector product v B is up-the-page. Consequently, the Lorentz force F = ( e)v B pushes the electrons in the moving bar down-the-page, which makes the current flow in the opposite direction, up-the-page. That is, the current flows up through the moving bar; through the rest of the circuit it flows counter-clockwise. In particular, in the resistor at the left end of the circuit, the current flows down. Now consider the magnitude of this current. The EMF induced in the moving bar is E = l (v B) = lvb. (9) I assume the resistance of the resistor is much larger that the resistance of the moving bar or the connecting wires, so we may approximate the net resistance of the circuit as. 5

6 Consequently, the current through the circuit due to the EMF (9) induced in the moving bar is I = E net E = lvb. (3) (b) The magnetic force on the bar due to the current (3) is F = I l B. (31) For the current flowing up-the-page through the bar and the magnetic field direction intothe-page, the direction of this force is to the left, opposite to the bar s velocity, while its magnitude is F = I lb = l B v. (3) (c) By Newton s second law, m dv = F. (33) As we saw in part (b), the force on the moving bar is in the opposite direction to the bar s velocity, while the force s magnitude (3) is proportional to the speed. Therefore, dv = F m = l B m v. (34) The solution of this differential equation is an exponentially decreasing velocity, ( v(t) = v exp t ) τ for τ = m l B. (35) 6

7 (d) As the bar slows down, its kinetic energy is dissipated as the Joule heat in the resistor. Indeed, the current (3) due to EMF induced in the moving bar dissipates power P = I = l B v, (36) and this power is precisely the rate at which the moving bar looses its kinetic energy, du kin = v F = +v F = v l B = P. (37) Consequently, the net energy dissipated while the bar slows down to a complete stop should equal to the initial kinetic energy of the bar. Let s check this fact directly from the equation (36): W = P(t) = l B v (t) = l B v e t/τ = l B For the time constant (35) obtained in part (c), this formula yields W = l B ( v τ = m ) l B = v v τ. (38) m, (39) which is precisely the initial kinetic energy of the particle. Quod erat demonstrandum. Problem 7.8: (a)intheplaneofthefigure7.18, themagneticfieldabovethewireisdirected into-the-page, while its magnitude depends only on the up-the-page coordinate y, specifically B(y) = µ I πy (4) Consequently, the flux through the square loop is a Φ = dx s+a s dyb(y) = a s+a s dy µ I πy = a µ I π ln s+a s. (41) 7

8 (b) The EMF induced in a moving straight wire is E = l (v B). For the magnetic field pointing into-the-page while the loop moves up-the-page, the direction of the vector product v B is left. Consequently, there is no EMF induced in the vertical wires, while there is leftward EMF induced in the horizontal wires. elative to the loop, the EMF in the bottom wire is clockwise while the EMF in the top wire is counterclockwise. The magnitudes of these EMFs are E bottomwire = a v B(y = s) = a v µ I πs, E topwire = a v B(y = s+a) = a v µ I (4) π(s+a), with a larger EMF in the bottom wire. Consequently, the net EMF induced in the square loop is clockwise, so the current in the loop flows clockwise. As to the magnitude of the net EMF, E net = E bottomwire E topwire = avµ a π Let s compare this magnitude with the Faraday s law, ( 1 s 1 ). (43) s+a E net = dφ, (44) The magnetic flux (41) from part (a) changes with time because s changes when the loop moves up, Consequently, d ln s+a s = v d ds ds ln s+a s = v. (45) = v ( 1 s+a 1 ), (46) a and therefore dφ = aµ I π d ln s+a s = + aµ ( I 1 π s 1 ) v, (47) s+a in perfect agreement with the net EMF (43) induced in the loop. 8

9 (c) When the loop moves to the right parallel to the current-carrying wire the magnetic flux through the loop does not change, so we expect zero net EMF induced in the loop. To see how this works, consider the EMFs E = l (v B) induced in each side of the square loop. For B pointing into-the-page while the velocity vector v points right, the vector product v B points up-the-page. Consequently, there is not EMF induced in the horizontal top or bottom wires, while there is upward EMF induced in the vertical left or right wires. By symmetry, the upward EMFs in the left and the right wires are equal, but their directions relative to the loop are opposite: clockwise in the left wire but counterclockwise in the right wire. Thus, the EMFs induced in the two vertical wire precisely cancel each other, and the net EMF in the loop is zero. Problem 7.1: Inside a long solenoid, the magnetic field is uniform and points along the solenoid s axis. For a circular loop of radius a/ completely inside the solenoid and to its axis, the magnetic flux is Φ = B Area = B π(a/). (48) When the magnetic field changes with time, the flux also changes, Φ(t) = B(t) π(a/), (49) which leads to the EMF induced in the loop, E = dφ = db π(a/). (5) For the harmonically oscillating magnetic field B(t) = B cos(ωt), this EMF is E = π(a/) B dcos(ωt) = +π(a/) B ω sin(ωt). (51) Consequently, there is a harmonically oscillating current in the loop, I(t) = E(t) = π(a/) B ω sin(ωt). (5) 9

10 Problem 7.17: (a) First, let s find the magnetic flux through the loop as a function of the current I in the solenoid. The magnetic field inside the solenoid (without an iron core) is B = µ Inẑ (53) while outside the solenoid the magnetic field is negligible. Consequently, the magnetic flux through any loop surrounding the solenoid is simply the net flux through the solenoid itself, Φ = B a solenoid = µ In πa. (54) When the current through the solenoid changes, this flux changes at the rate dφ = di µ n(πa ), (55) which induces the EMF in the loop surrounding the solenoid, E = dφ = di µ n(πa ). (56) Consequently, there is a current in the loop I loop = E = di µ n(πa ). (57) The sign here reflects the Lenz rule: the current in the loop oppose the changes of the current in the solenoid. Thus, if the current in the solenoid is clockwise (as shown on figure 7.8) and increasing, then the current in the loop is counterclockwise. 1

11 (b) Now suppose instead of changing the current in the solenoid, we simply pull the loop off the solenoid and move it far away. Moving the loop leads to the motional EMF, which also obtains from the Faraday s law, E = dφ, (58) whatever the reason for changing of the magnetic flux. Consequently, the current in the loop is I loop (t) = E(t) = 1 dφ, (59) Integrating this current over time, we obtain the net charge which has flown through the loop, Q loop = I loop = 1 dφ = 1 Φ = Φ init Φ final In particular, when the loop is removed from the solenoid to a far-away place without a magnetic field, we have Φ final = and therefore (6) Q loop = Φ init. (61) If the loop initially surrounds the solenoid as in figure 7.8, then the initial flux through the loop is the flux in the solenoid as in eq. (54), hence Q loop = µ In(πa ). (6) For example, for a solenoid of radius a = 1 cm and density n = 1 loops/m carrying current I = 1 A, and the loop having net resistance = 4 Ω (which mostly comes from the resistor), the net charge flowing through the loop as it is removed is about Q = 1 7 C. 11

12 Problem 7.19: The mathematical relation between the induced electric field and the time-dependent magnetic field is very similar to the relation between the magnetics field itself and the electric current density. Indeed, compare the equations E = B t, E = assuming ρ = (63) and B = µ J, B =. (64) Similar equations call for similar solutions, hence the Biot Savart Laplace-like formula for the induced electric field: E(r,t) = 1 B(r,t) r r 4π t r r 3 d3 Vol. (65) For the toroidal coil in question, there is strong magnetic field inside the coil but no field outside the coil. For a thin toroid (compared to its radius), we may integrate over the cross-section while treating the r r as approximately constant, thus cross section B t = dφ ˆφ (66) where Φ is the net magnetic flux through the toroid, and hence E(r) = dφ 1 4π length dr r r r r 3. (67) The integral here is precisely the same integral as for calculating the magnetic field of a circular current loop, cf. the textbook example 5.6 (page 7). For generic points r, this 1

13 integral is a mess of elliptic functions, but for the r lying on the symmetry axis of the toroid, the integral simplifies to length dr r r r r 3 = πa (a +z ẑ, (68) ) 3/ exactly as in the textbook equation (5.41). For the present problem, eq. (68) means that on the toroid s axis, the electric field is E(,,z) = dφ a (a +z ẑ (69) ) 3/ Finally, the magnetic field inside the toroid is explained in the textbook example 5.1 (pages 38 39), B inside = µ IN πs ˆφ, (5.6) For a thin toroid with w a this field is approximately uniform (inside the toroid), so the flux through the toroid is simply Φ = hw B φ = µ IN hw πa. (7) While the geometry of the toroid should be time-independent, the current I may change with time, which leads to a time-dependent magnetic flux, changing at the rate dφ = µ Nhw πa di. (71) Plugging this formula into eq. (69) for the electric field, we finally arrive at E(,,z) = ( ẑ) di µ Nhwa 4π(a +z. (7) ) 3/ 13

14 Problem 7.51: First, let s find the magnetic field of the moving wire. Since the wire is moving at a constant velocity v that is much less than the speed of light c, we may use the quasi-static approximation: The magnetic field B(r,t) is approximately the same as the field of the stationary wire located wherever the wire happens to be at time t. In particular, at t = when the wire runs along the z axis, we have B(t =,r) = µ I π ˆφ s = µ I π xŷ yˆx x + y. (73) For future convenience, let s express this field in terms of the vector potential, B = A for A(t =,r) = µ I π ln(s)ẑ = µ I 4π ln(x +y )ẑ. (74) At other times t, we have similar formulae for the magnetic field and the vector potential in terms of the relative coordinates x X wire (t) = x and y Y wire (t) = y vt, thus B(t,r) = µ I π xŷ (y vt)ˆx x + (y vt), (75) A(t,r) = µ I 4π ln( x +(y vt) ) ẑ. (76) Note: in a coordinate frame moving with the wire, the magnetic field (73) and the vector potential (74) are time-independent. But in the lab frame, the magnetic field at any fixed point (x,y,z) changes with time because the distance to the moving wire changes with time, thus ( ) B(t;x,y,z). (77) t fixed lab frame(x,y,z) Consequently, in the lab frame there is electric field induced by the time-dependent magnetic field. 14

15 The induced electric field E(t, r) obtains from solving the Induction Law equation as well as the Gauss Law for ρ, E(t,r) = B(t,r) t, E(t,r) =. (78) The simplest way to solve these is in terms of the vector potential A(t,r) and the scalar potential V(t, r), E(t,r) = A(t,r) t V(t,r). (79) The vector potential here should be exactly as in eq. (76), so that A(r,t) is the timedependent magnetic field; this will make the electric field (77) obey the Induction Law for any V(t,r). As to the scalar potential, it should take care of the Gauss Law by obeying the Poisson equation V(t,r) = t( A(t,r) ). (8) Fortunately, the vector potential (76) at hand has zero divergence A = A z z = (81) hence the scalar potential should obey V =, and the simplest solution to this Laplace equations is simply V(x,y,x,t). Consequently, E induced (t,r) = t A(t,r) = ( µ I t 4π ln( x +(y vt) ) ) ẑ = µ I ln ( x +(y vt) ) ẑ 4π t = + µ I v(y vt) 4π x +(y vt) ẑ, (8) 15

16 In particular, at time t = when the wire runs along the z axis, E(t = ;x,y,z) = µ Iv π yẑ x +y, (83) or in cylindrical coordinates E(t = ;s,φ,z) = µ Iv π sinφ s ẑ. (84) Problem 7.: (a) Let s assume the little loop is much smaller than the big loop or the distance between the loops, a b,z, so the magnetic field of the big loop is approximately uniform over the little loop, B big (r littleloop) const. (85) Evaluating this approximately uniform field at the little loop s center which happens to lie on the big loop s axis we find B big (r littleloop) B big (,,z) = µ I big b (b +z ẑ, (86) ) 3/ where the second equality here is the textbook equation (5.41) for the magnetic field of a circular loop. Consequently, the magnetic flux (of the big loop field s) through the little loop is Φ little a little B big (,,z) = πa B z big (,,z) = µ I big πa b (b +z. (87) ) 3/ 16

17 (b) Now consider the magnetic flux of the little loop through the big loop. Again, we assume that the little loop is very small compared to the distance to the big loop, and this allows us to use the dipole approximation, where B little (r ) = µ 4π 3(m ˆr )ˆr m r 3 (88) m = (πa )I littleˆẑ (89) is the magnetic moment of the little loop, and the radius vector r is taken relative to the little loop s center. In cylindrical coordinates, the magnetic field (88) becomes B little (s,φ,z) = µ (πa ( )I little z s 4π (z +s ) 5/ ẑ + 3z ) s (z +s ) 5/ ŝ where z = z z littleloop. Now let s calculate the flux of this magnetic field through the big loop. Spanning the big loop with a flat disk of radius b in the xy plane hence at constant z = while z = z littleloop, we have (9) Φ big = disk B z little (s,φ)d A = = µ (πa )I little b b πsds µ (πa )I little 4π (z s )sds (z +s ) 5/ changing variables from s to ν = s +z = µ (πa )I little = µ (πa )I little = µ (πa )I little = µ (πa )I little b +z z (3z ν) 1 dν ν 5/ b +z d ( z ν 3/ + 1 ) ν 1/ z (( 1 b +z z ) (b +z ) 3/ b (b +z ) 3/ z s (z +s ) 5/ ( )) 1 z z z 3 (91) 17

18 Altogether, Φ big = µ I little b (πa ) (b +zlittle. (9) )3/ Alternative solution: In the dipole approximation to the little loop s magnetic field, the vector potential is A little (r ) = µ 4π m ˆr r, (93) or in cylindrical coordinates A little (r ) = µ (m = πa I little ) 4π s (s +z ) 3/ ˆφ. (94) In terms of this vector potential, the magnetic flux through the big loop is Φ big = big loop A little (r ) dr. (95) Geometrically, the big loop is located at fixed s = b and fixed z = z = z littleloop (relative to the little loop), while φ span the full circle from to π. Consequently, A little (r ) dr = µ (πa )I little 4π b b +zl bdφ, (96) )3/ so the magnetic flux (95) is simply Φ big = µ (πa )I little 4π b b +zl πb = µ I little (πa ) )3/ b (b +zlittle. (9) )3/ 18

19 (c) In terms of mutual inductances, eq. (87) from part(a) becomes Similarly, eq. (9) from part (b) becomes Φ little = M l,b I big (97) where M l,b = µ b πa (b + z. (98) ) 3/ Φ big = M b,l I little (99) where M b.l = µ b πa (b + z. (1) ) 3/ By inspection of eqs (98) and (1), the two mutual inductances are equal, M l,b = M b,l. (11) Quod erat demonstrandum. Problem 7.3: By symmetries of the coaxial cable, the magnetic field has form B(s,φ,z) = B(s only) ˆφ (1) where B(s) obtains from the Ampere s law: πs B(s) = µ I[through disk of radius s] = I (s /a ) inside the inner wire, I between the wires, outside the whole cable, (13) In the limit of the outer wire and the insulation being very thin compared to the inner wire s radius a, this gives us B(s) = µ Is πa for s < a, for s > a. (14) There are two ways to calculate the self-induction of a wire coil, loop, or a circuit: (a) as 19

20 a ratio of the magnetic field s flux to the current, or (b) from the magnetic energy formula U = 1 µ whole space B d 3 Vol = LI. (15) In the flux/current method, we should calculate the magnetic flux through the same loop L that carries the current. For the coaxial cable, the loop should surround the longitudinal section of the insulator as well as the adjacent parts of the two wires. But it is far from clear how much of the thick inner wire should be surrounded by this loop: All the way to the inner wire s axis? As little as possible? Half-way through the inner wire s radius? Something else? Let me illustrate all these choices by green lines (dashed or solid) on the following diagram of the cable s cross-section: outer wire inner wire insulation Since there is non-zero magnetic field inside the inner wire, different loops on this diagram would contain different magnetic fluxes, and it s not at all clear which of these fluxes would give us the right answer for the cable s self-inductance. Indeed, In my own first attempt at solving this problem I had assumed the right loop should extend all the way to the center of the inner wire, and I ve ended up with self inductance (per unit of cable s length) L length = µ 4π = H/m. (16) Alas, thisanswer iswrong by afactor of, aswe shall see inamoment. (But forthe purposes of grading, this particular wrong answer would get you full credit as I would not penalize a mistake I have made myself.)

21 Since it s not clear over which loop we should integrate the magnetic flux, let s use the other method for calculating the cable s self-inductance, namely the magnetic energy formula (15). Given the magnetic field (14), its net energy per unit of cable s length l obtains as U = 1 µ whole = l ( ) µ I µ πa B d 3 Vol = 1 µ l a ds(πs) dsπs s = li µ 8π a 4 πa4 4 ( B(s) = µ ) Is πa Θ(a s) = li µ 8π. (17) Identifying this magnetic energy as inductor energy U = 1 LI, we obtain the cable s self inductance per unit length as L l = U li = µ 8π = H/m. (18) Problem 7.9: The geometry of the coil is shown on textbook figure 7.34 (page 37): rectangular crosssection (b a) h, where a is the inner radius of the toroid and b is the outer radius. Assuming the toroid is uniformly and densely wound, the magnetic field vanishes anywhere outside the coil, while inside the coil B inside = µ NI πs ˆφ. (19) 1

22 According to the textbook equation (7.35), the net energy stored in this magnetic field is U = 1 µ whole = 1 µ space toroid B d 3 Vol = 1 µ µ N I 4π = µ N I 8π ( ) µ NI d 3 Vol πs h h πln b a b dz a πsds s (11) = I µ N hln(b/a). π On the other hand, treating the toroid as an inductor with self-inductance L = µ N hln(b/a) π (7.8) (see the textbook example 7.11 (page 35) for the calculation), we expect the net magnetic energy stored in this inductor to be U = I L = I µ N hln(b/a). (111) π By inspection, this is precisely the same energy as in eq. (11). Problem 7.33: (a) When the cylinder rotates with angular velocity ω, the charges on its surface move at linear velocity v = ωˆφ, which creates a surface current K = σv = σω ˆφ (11) This current is similar to what we have in a solenoid, so the magnetic field it creates is

23 similar to the solenoid s field: B = { µ σωẑ inside the cylinder, outside the cylinder. (113) Strictly speaking, this formula applies only when the current around the cylinder is steady, which would require ω = const. However, if the rotation speed ω changes slowly, we may use the quasi-static approximation B(r) B Ampere [currents at time t], thus B(r,t) = { µ σω(t)ẑ for r inside the cylinder, for r outside the cylinder. (114) The electric field induced by this time-dependent solenoid-like magnetic field is explained in detail in my notes on magnetic induction(pages 11 1): By symmetry, the induced electric field points in the circular direction ˆφ while its magnitude depends only on the cylindrical radius s, E induced = E(s) ˆφ. (115) Furthermore, the magnitude E(s) follows from the Ampere-like law E d l = πs E(s) = d Φ[through loop of radius s]. (116) Specifically, for the solenoid-like cylinder at hand, { } { } πs Φ(t) = B inside for s <, πs for s <, (t) = µ σ ω(t) π for s >, π for s >, (117) hence E induced (s) = µ σ dω In particular, at the cylinder s surface itself s s for s <, for s >. (118) E induced (@s = ) = µ σ dω ˆφ. (119) Besides the induced electric field, there is also a static electric field due to charges on 3

24 the cylinder s surface. egardless of the rotation, E static = inside the cylinder, σ ŝ ǫ s outside the cylinder. (1) The net electric field is E net (r,t) = E static (r) + E induced (r,t); (11) in particular, at the surface of the cylinder, E net (@s = ) = µ σ dω ˆφ + 1 σ ŝ. (1) ǫ This net electric field exerts forces on the charges at the cylinder s surface. While the net force on all the charges vanishes by the rotational symmetry, there is however a non-zero net torque. Indeed, a small patch of the charged surface of charge dq feels force df = (dq)e net = (dq)e induced (s = ) ˆφ + (dq)e static (s = )ŝ (13) and while the radial force of the static field has zero level arm, the tangential force of the induced field has lever arm, hence torque dτ = (dq) E induced (s = ). (14) The torques on different patches of the surface all have the same direction clockwise or counterclockwise, depending on the sign of E induced, hence τ net = Q net E induced (s = ) = πlσ ( µ σ ) dω = πµ l 4 σ dω (15) where l is the cylinder s length. 4

25 When we spin up the cylinder (starting with ω = ), this torque performs negative mechanical work W = t end t τ dφ(t) = πµ l 4 σ t end This work depends only on the final rotation frequency; indeed t fin t dω ω = hence for ω =, t fin t dω ω(t) (16) t ω(t) dω(t) = 1 ω (t fin ) 1 ω (t ) = 1 ω fin 1 ω, (17) W = πµ l 4 σ 1 ω fin. (18) This negative work is done by the magnetic forces, so whoever spins up the cylinder must perform the opposite, positive work +W = +πµ l 4 σ 1 ω fin. (19) (b) The work (19) performed by the outside forces while spinning up the cylinder is stored in the magnetic field inside the rotating cylinder. Indeed, the net energy of the magnetic field (114) is U = 1 B d 3 Vol µ whole space = 1 µ cylinder ( µ σω ) d 3 Vol = 1 µ ( µ σω ) π l = πµ l 4 σ 1 ω. (13) By inspection, this is precisely the work (19) needed to spin-up the cylinder and create the magnetic field. Quod erat demonstrandum. 5