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32 Inductance CHAPTER OUTLINE 32.1 32.2 32.3 32.4 32.5 32.6 Self-Inductance RL Circuits Energy in a Magnetic Field Mutual Inductance Oscillations in an LC Circuit The RLC Circuit ANSWERS TO QUESTIONS Q32.1 The emf induced in an inductor is opposite to the direction of the changing current. For example, in a simple RL circuit with current flowing clockwise, if the current in the circuit increases, the inductor will generate an emf to oppose the increasing current. Q32.2 The coil has an inductance regardless of the nature of the current in the circuit. Inductance depends only on the coil geometry and its construction. Since the current is constant, the self-induced emf in the coil is zero, and the coil does not affect the steady-state current. (We assume the resistance of the coil is negligible.) Q32.3 The inductance of a coil is determined by (a) the geometry of the coil and (b) the “contents” of the coil. This is similar to the parameters that determine the capacitance of a capacitor and the resistance of a resistor. With an inductor, the most important factor in the geometry is the number of turns of wire, or turns per unit length. By the “contents” we refer to the material in which the inductor establishes a magnetic field, notably the magnetic properties of the core around which the wire is wrapped. Q32.4 If the first set of turns is wrapped clockwise around a spool, wrap the second set counter-clockwise, so that the coil produces negligible magnetic field. Then the inductance of each set of turns effectively negates the inductive effects of the other set. Q32.5 After the switch is closed, the back emf will not exceed that of the battery. If this were the case, then the current in the circuit would change direction to counterclockwise. Just after the switch is opened, the back emf can be much larger than the battery emf, to temporarily maintain the clockwise current in a spark. Q32.6 The current decreases not instantaneously but over some span of time. The faster the decrease in the current, the larger will be the emf generated in the inductor. A spark can appear at the switch as it is opened because the self-induced voltage is a maximum at this instant. The voltage can therefore briefly cause dielectric breakdown of the air between the contacts. Q32.7 When it is being opened. When the switch is initially standing open, there is no current in the circuit. Just after the switch is then closed, the inductor tends to maintain the zero-current condition, and there is very little chance of sparking. When the switch is standing closed, there is current in the circuit. When the switch is then opened, the current rapidly decreases. The induced emf is created in the inductor, and this emf tends to maintain the original current. Sparking occurs as the current bridges the air gap between the contacts of the switch. 239 240 Inductance Q32.8 A physicist’s list of constituents of the universe in 1829 might include matter, light, heat, the stuff of stars, charge, momentum, and several other entries. Our list today might include the quarks, electrons, muons, tauons, and neutrinos of matter; gravitons of gravitational fields; photons of electric and magnetic fields; W and Z particles; gluons; energy; momentum; angular momentum; charge; baryon number; three different lepton numbers; upness; downness; strangeness; charm; topness; and bottomness. Alternatively, the relativistic interconvertability of mass and energy, and of electric and magnetic fields, can be used to make the list look shorter. Some might think of the conserved quantities energy, momentum, … bottomness as properties of matter, rather than as things with their own existence. The idea of a field is not due to Henry, but rather to Faraday, to whom Henry personally demonstrated self-induction. Still the thesis stated in the question has an important germ of truth. Henry precipitated a basic change if he did not cause it. The biggest difference between the two lists is that the 1829 list does not include fields and today’s list does. Q32.9 The energy stored in the magnetic field of an inductor is proportional to the square of the current. 1 Doubling I makes U = LI 2 get four times larger. 2 Q32.10 The energy stored in a capacitor is proportional to the square of the electric field, and the energy stored in an induction coil is proportional to the square of the magnetic field. The capacitor’s energy is proportional to its capacitance, which depends on its geometry and the dielectric material inside. The coil’s energy is proportional to its inductance, which depends on its geometry and the core material. On the other hand, we can think of Henry’s discovery of self-inductance as fundamentally new. Before a certain school vacation at the Albany Academy about 1830, one could visualize the universe as consisting of only one thing, matter. All the forms of energy then known (kinetic, gravitational, elastic, internal, electrical) belonged to chunks of matter. But the energy that temporarily maintains a current in a coil after the battery is removed is not energy that belongs to any bit of matter. This energy is vastly larger than the kinetic energy of the drifting electrons in the wires. This energy belongs to the magnetic field around the coil. Beginning in 1830, Nature has forced us to admit that the universe consists of matter and also of fields, massless and invisible, known only by their effects. Q32.11 The inductance of the series combination of inductor L1 and inductor L 2 is L1 + L 2 + M 12 , where M 12 is the mutual inductance of the two coils. It can be defined as the emf induced in coil two when the current in coil one changes at one ampere per second, due to the magnetic field of coil one producing flux through coil two. The coils can be arranged to have large mutual inductance, as by winding them onto the same core. The coils can be arranged to have negligible mutual inductance, as separate toroids do. Q32.12 The mutual inductance of two loops in free space—that is, ignoring the use of cores—is a maximum if the loops are coaxial. In this way, the maximum flux of the primary loop will pass through the secondary loop, generating the largest possible emf given the changing magnetic field due to the first. The mutual inductance is a minimum if the magnetic field of the first coil lies in the plane of the second coil, producing no flux through the area the second coil encloses. Q32.13 The answer depends on the orientation of the solenoids. If they are coaxial, such as two solenoids end-to-end, then there certainty will be mutual induction. If, however, they are oriented in such a way that the magnetic field of one coil does not go through turns of the second coil, then there will be no mutual induction. Consider the case of two solenoids physically arranged in a “T” formation, but still connected electrically in series. The magnetic field lines of the first coil will not produce any net flux in the second coil, and thus no mutual induction will be present. Chapter 32 241 Q32.14 When the capacitor is fully discharged, the current in the circuit is a maximum. The inductance of the coil is making the current continue to flow. At this time the magnetic field of the coil contains all the energy that was originally stored in the charged capacitor. The current has just finished discharging the capacitor and is proceeding to charge it up again with the opposite polarity. Q32.15 The oscillations would eventually decrease, but perhaps with very small damping. The original potential energy would be converted to internal energy within the wires. Such a situation constitutes an RLC circuit. Remember that a real battery generally contains an internal resistance. Q32.16 If R > Q32.17 The condition for critical damping must be investigated to design a circuit for a particular purpose. For example, in building a radio receiver, one would want to construct the receiving circuit so that it is underdamped. Then it can oscillate in resonance and detect the desired signal. Conversely, when designing a probe to measure a changing signal, such free oscillations are undesirable. An electrical vibration in the probe would constitute “ringing” of the system, where the probe would measure an additional signal—that of the probe itself! In this case, one would want to design a probe that is critically damped or overdamped, so that the only signal measured is the one under study. Critical damping represents the threshold between underdamping and overdamping. One must know the condition for it to meet the design criteria for a project. Q32.18 An object cannot exert a net force on itself. An object cannot create momentum out of nothing. A coil can induce an emf in itself. When it does so, the actual forces acting on charges in different parts of the loop add as vectors to zero. The term electromotive force does not refer to a force, but to a voltage. 4L 4L , then the oscillator is overdamped—it will not oscillate. If R < , then the oscillator is C C underdamped and can go through several cycles of oscillation before the radiated signal falls below background noise. SOLUTIONS TO PROBLEMS Section 32.1 P32.1 P32.2 Self-Inductance ε =L jFGH IJ K ∆I 1.50 A − 0.200 A = 3.00 × 10 −3 H = 1.95 × 10 −2 V = 19.5 mV ∆t 0.200 s e Treating the telephone cord as a solenoid, we have: ja f e e j 2 −7 −3 µ N 2 A 4π × 10 T ⋅ m A 70.0 π 6.50 × 10 m = L= 0 A 0.600 m fFGH a 0 − 0.500 A ∆I = −2.00 H ∆t 0.010 0 s P32.3 ε = −L P32.4 L= P32.5 ε back = −ε = L I FG 1 V ⋅ s IJ = JK H 1 H ⋅ A K NΦ B LI → ΦB = = 240 nT ⋅ m 2 I N a b 2 = 1.36 µH . 100 V through each turn g ja fa f dI d =L I max sin ω t = LωI max cos ω t = 10.0 × 10 −3 120π 5.00 cos ω t dt dt f b g a e f a f ε back = 6.00π cos 120π t = 18.8 V cos 377 t 242 P32.6 Inductance From ε = L From L = FG ∆I IJ , we have H ∆t K L= NΦ B , we have I ΦB a fe P32.8 N ε =L e j e 500 f= 19.2 µT ⋅ m 2 . j (a) At t = 1.00 s , ε = 360 mV (b) At t = 4.00 s , ε = 180 mV (c) ε = 90.0 × 10 −3 2t − 6 = 0 ja e f t = 3.00 s . FG 450 IJ b0.040 0 Ag = H 0.120 K (a) B = µ 0 nI = µ 0 (b) Φ B = BA = 3.33 × 10 −8 T ⋅ m 2 (c) L= 188 µT NΦ B = 0.375 mH I B and Φ B are proportional to current; L is independent of current a f e 2 *P32.11 ja H 4.00 A dI d 2 = 90.0 × 10 −3 t − 6t V dt dt (d) P32.10 −3 −4 µ N 2 A µ 0 420 3.00 × 10 L= 0 = = 4.16 × 10 −4 H 0.160 A dI dI −ε −175 × 10 −6 V ε = −L → = = = −0.421 A s dt dt L 4.16 × 10 −4 H when P32.9 g LI e 2. 40 × 10 = = j 2 P32.7 b ε 24.0 × 10 −3 V = = 2.40 × 10 −3 H . 10.0 A s ∆I ∆t (a) −3 µ N 2 A µ 0 120 π 5.00 × 10 = L= 0 0.090 0 A (b) Φ ′B = j 2 = 15.8 µH µm µ N2A ΦB → L = m = 800 1.58 × 10 −5 H = 12.6 mH µ0 A e j We can directly find the self inductance of the solenoid: ε = −L dI dt +0.08 V = − L 0 − 1.8 A 0.12 s L = 5.33 × 10 −3 Vs A = e j µ µ N π F 200 m I = = G J A H 2π N K µ0N2A . A Here A = π r 2 , 200 m = N 2π r , and A = N 10 −3 m . Eliminating extra unknowns step by step, we have 5.33 × 10 A= −3 4 × 10 −3 µ N 2π r 2 Vs A = 0 A WbmA 5.33 × 10 −3 AVs = 0.750 m 0 2 2 0 40 000 4π A m2 = e j 10 −7 40 000 m 2 Tm A A Chapter 32 P32.12 L= 243 NΦ B NBA NA µ 0 NI µ0N 2 A = ≈ ⋅ = I I I 2π R 2π R FIG. P32.12 P32.13 ε = ε 0 e − kt = − L dI = − ε 0 − kt e dt L dI dt ε 0 − kt dq e = kL dt ε Q = 20 . k L I= If we require I → 0 as t → ∞ , the solution is z z Q = Idt = Section 32.2 P32.14 I= ∞ ε 0 − kt ε e dt = − 20 kL k L 0 RL Circuits ε R e1 − e j : − Rt L 0.900 ε R = ε R 1 − e − R a3.00 s f 2.50 H FG Ra3.00 sf IJ = 0.100 H 2.50 H K exp − R= 2.50 H ln 10.0 = 1.92 Ω 3.00 s a f ε e1 −Re j I (A) −t τ P32.15 (a) At time t, It = where L τ = = 0.200 s . R After a long time, I max = e ε 1 − e −∞ R 1 0 R a0.500f Rε = ε e1 − e R so 0.500 = 1 − e − t 0. 200 s . − t 0. 200 s j a e Isolating the constants on the right, ln e − t 0. 200 s = ln 0.500 (b) 0.5 j=ε . At I t = 0.500 I max af j t (s) 0 f t = −0.693 0.200 s and solving for t, − or t = 0.139 s . Similarly, to reach 90% of I max , 0.900 = 1 − e − t τ and t = −τ ln 1 − 0.900 . Thus, t = − 0.200 s ln 0.100 = 0.461 s . a a f a I max f f 0.2 0.4 FIG. P32.15 0.6 244 P32.16 Inductance Taking τ = IR + L L , R dI = 0 will be true if dt Because τ = P32.17 P32.18 0 −t τ L = 2.00 × 10 −3 s = 2.00 ms R τ= (b) I = I max 1 − e − t τ = (c) I max = I= I 0 Re − t τ L , we have agreement with 0 = 0 . R (a) .00 V I J e1 − e j FGH 64.00 ΩK e −0. 250 2.00 ε 6.00 V = = 1.50 A R 4.00 Ω 0.800 = 1 − e (d) FG IJ H K F 1I + Le I e jG − J = 0 . H τK dI 1 = I0 e−t τ − τ dt I = I 0 e −t τ : − t 2.00 ms a f a j= 0.176 A FIG. P32.17 f → t = − 2.00 ms ln 0.200 = 3.22 ms 120 ε 1 − e −t τ = 1 − e −1.80 7.00 = 3.02 A R 9.00 e ∆VR j e j = IR = a3.02fa9.00f = 27.2 V ∆VL = ε − ∆VR = 120 − 27.2 = 92.8 V P32.19 Note: It may not be correct to call the voltage or emf across a coil a “potential difference.” Electric potential can only be defined for a conservative electric field, and not for the electric field around an inductor. (a) ∆VR = IR = 8.00 Ω 2.00 A = 16.0 V a fa f and ∆VL = ε − ∆VR = 36.0 V − 16.0 V = 20.0 V . Therefore, ∆VR 16.0 V = = 0.800 . ∆VL 20.0 V a fa FIG. P32.19 f ∆VR = IR = 4.50 A 8.00 Ω = 36.0 V (b) ∆VL = ε − ∆VR = 0 P32.20 a f After a long time, 12.0 V = 0. 200 A R . Thus, R = 60.0 Ω . Now, τ = jb e g L gives R L = τ R = 5.00 × 10 −4 s 60.0 V A = 30.0 mH . P32.21 e e jFGH − τ1 IJK dI = − I max e − t τ dt j I = I max 1 − e − t τ : τ= L 15.0 H = = 0.500 s : R 30.0 Ω ε dI R = I max e − t τ and I max = dt L R dI R ε 100 V = I max e 0 = = = 6.67 A s dt L L 15.0 H (a) t=0: (b) t = 1.50 s : dI ε − t τ = e = 6.67 A s e −1.50 a0.500 f = 6.67 A s e −3.00 = 0.332 A s dt L b g b g Chapter 32 P32.22 e j I = I max 1 − e − t τ : 0.980 = 1 − e −3.00 ×10 −3 −3 τ 0.020 0 = e −3.00 ×10 τ =− L , so R τ= P32.23 τ 3.00 × 10 −3 = 7.67 × 10 −4 s ln 0.020 0 b g FIG. P32.22 ja f e L = τ R = 7.67 × 10 −4 10.0 = 7.67 mH Name the currents as shown. By Kirchhoff’s laws: I1 = I 2 + I 3 (1) +10.0 V − 4.00 I 1 − 4.00 I 2 = 0 (2) a f dIdt = 0 3 +10.0 V − 4.00 I 1 − 8.00 I 3 − 1.00 (3) From (1) and (2), +10.0 − 4.00 I 1 − 4.00 I 1 + 4.00 I 3 = 0 and I 1 = 0.500 I 3 + 1.25 A . Then (3) becomes 10.0 V − 4.00 0.500 I 3 + 1.25 A − 8.00 I 3 − 1.00 FIG. P32.23 a f dIdt = 0 b g a1.00 HfFGH dIdt IJK + a10.0 ΩfI = 5.00 V . 3 3 3 We solve the differential equation using Equations 32.6 and 32.7: .00 V I a f FGH 510.0 J 1− e a ΩK I 1 = 1.25 + 0.500 I 3 = P32.24 P32.25 a0.500 Af 1 − e 1.50 A − a0.250 A fe f − 10.0 Ω t 1.00 H I3 t = = −10 t s L L = , we get R = R C 3.00 H = 1.00 × 10 3 Ω = 1.00 kΩ . 3.00 × 10 −6 F (a) Using τ = RC = (b) τ = RC = 1.00 × 10 3 Ω 3.00 × 10 −6 F = 3.00 × 10 −3 s = 3.00 ms e je −10 t s j For t ≤ 0 , the current in the inductor is zero . At t = 0 , it starts to grow from zero toward 10.0 A with time constant τ= a f 10.0 mH L = = 1.00 × 10 −4 s . R 100 Ω a f e j a10.0 Afe1 − e At t = 200 µs , I = a10.00 A fe1 − e j = 8.65 A . For 0 ≤ t ≤ 200 µs , I = I max 1 − e − t τ = −10 000 t s j . −2.00 Thereafter, it decays exponentially as I = I 0 e a f a f I = 8.65 A e −10 000b t − 200 µsg s = 8.65 A e −10 000 t − t′ τ FIG. P32.25 , so for t ≥ 200 µs , s + 2.00 e j = 8.65 e 2.00 A e −10 000 t s = a63.9 Afe −10 000 t s . 245 246 P32.26 Inductance ε I= (b) Initial current is 1.00 A: ∆V12 = 1.00 A 12.00 Ω = 12.0 V R = 12.0 V = 1.00 A 12.0 Ω (a) a fa a f fb g ∆V1 200 = 1.00 A 1 200 Ω = 1.20 kV ∆VL = 1.21 kV . (c) dI R = − I max e − Rt L dt L dI − L = ∆VL = I max R e − Rt L . dt I = I max e − Rt L : and b 12.0 V = 1 212 V e −1 212 t so 9.90 × 10 −3 = e −606 t . 2.00 t = 7.62 ms . L 0.140 = = 28.6 ms R 4.90 ε 6.00 V = 1.22 A I max = = R 4.90 Ω τ= (a) e I = I max 1 − e − t τ e P32.28 g Solving Thus, P32.27 FIG. P32.26 −t τ j so a f j t = −τ ln 0.820 = 5.66 ms = 0.820 : e e 0. 220 = 1.22 1 − e − t τ j a fe j (b) I = I max 1 − e −10.0 0.028 6 = 1.22 A 1 − e −350 = 1.22 A (c) I = I max e − t τ (a) and 0.160 = 1.22 e − t τ so t = −τ ln 0.131 = 58.1 ms . a (c) f For a series connection, both inductors carry equal currents at every instant, so same for both. The voltage across the pair is dI dI dI = L1 + L 2 Leq so dt dt dt (b) FIG. P32.27 dI dI dI = L1 1 = L 2 2 = ∆VL dt dt dt ∆VL ∆VL ∆VL = + Thus, Leq L1 L2 Leq where and dI is the dt Leq = L1 + L 2 . I = I 1 + I 2 and dI dI 1 dI 2 = + . dt dt dt 1 1 1 = + . Leq L1 L 2 dI dI dI + Req I = L1 + IR1 + L 2 + IR 2 dt dt dt dI are separate quantities under our control, so functional equality requires Now I and dt both Leq = L1 + L 2 and Req = R1 + R 2 . Leq continued on next page Chapter 32 ∆V = Leq (d) dI dI dI dI dI 1 dI 2 = + + Req I = L1 1 + R1 I 1 = L 2 2 + R 2 I 2 where I = I 1 + I 2 and . dt dt dt dt dt dt We may choose to keep the currents constant in time. Then, 1 1 1 = + . Req R1 R 2 We may choose to make the current swing through 0. Then, 1 1 1 = + . Leq L1 L 2 This equivalent coil with resistance will be equivalent to the pair of real inductors for all other currents as well. Section 32.3 P32.29 L= P32.30 (a) Energy in a Magnetic Field e j −4 NΦ B 200 3.70 × 10 1 1 = = 42.3 mH so U = LI 2 = 0.423 H 1.75 A I 1.75 2 2 a f e The magnetic energy stored in the field equals u times the volume of the solenoid (the volume in which B is non-zero). j a0.260 mfπ b0.031 0 mg a f LMN e 2 = 6.32 kJ j OPQ 2 68.0 π 0.600 × 10 −2 N2A L = µ0 = µ0 = 8.21 µH A 0.080 0 1 1 2 U = LI 2 = 8. 21 × 10 −6 H 0.770 A = 2.44 µJ 2 2 ja e (a) U= (b) I= FG IJ H K ε 1 2 1 LI = L 2 2 2R FG ε IJ 1 − e b H RK g − RL t R t = ln 2 L u =∈0 z ∞ *P32.34 = 0.064 8 J . j 2 P32.33 2 2 e P32.32 f 4.50 T B2 = = 8.06 × 10 6 J m3 . 2 µ 0 2 1.26 × 10 −6 T ⋅ m A U = uV = 8.06 × 10 6 J m3 P32.31 fa The magnetic energy density is given by µ= (b) a 0 2 f = Lε 2 8R2 so so E2 = 44.2 nJ m3 2 e −2 Rt L dt = − z a0.800fa500f = 27.8 J 8a30.0f ε FεI = G J 1− e b g 2R H R K 2 = 2 − RL t t= u= FG H IJ K →e b g − RL t = 1 2 L 0.800 ln 2 = ln 2 = 18.5 ms 30.0 R B2 = 995 µJ m3 2µ 0 L ∞ −2 Rt L −2 Rdt L −2 Rt L e e =− L 2R 0 2R ∞ 0 =− a f L −∞ L L e − e0 = 0−1 = 2R 2R 2R e j 247 248 P32.35 Inductance a fa 1 2 1 LI = 4.00 H 0.500 A 2 2 (a) U= (b) When the current is 1.00 A, Kirchhoff’s loop rule reads f 2 U = 0.500 J a fa f +22.0 V − 1.00 A 5.00 Ω − ∆VL = 0 . Then ∆VL = 17.0 V . The power being stored in the inductor is I∆VL = 1.00 A 17.0 V = 17.0 W . a a P32.36 fa P = I∆V = 0.500 A 22.0 V (c) f fa P = 11.0 W I= From Equation 32.7, ε e1 − e j . − Rt L R ε = 2.00 A . R At that time, the inductor is fully energized and P = I ∆V = 2.00 A 10.0 V = 20.0 W . (a) The maximum current, after a long time t , is I= a f a P32.37 a f a5.00 Ωf = 2 (b) Plost = I 2 R = 2.00 A (c) Pinductor = I ∆Vdrop = 0 (d) U= e j a10.0 Hfa2.00 Af = LI 2 2 u =∈0 Therefore ∈0 E2 2 E2 B2 = 2 2µ 0 6.80 × 10 5 V m 3.00 × 10 8 m s fa f 20.0 W 2 2 We have B = E ∈0 µ 0 = P32.38 FIG. P32.35 f = 20.0 J B2 . 2µ 0 and u= so B 2 =∈0 µ 0 E 2 = 2.27 × 10 −3 T . The total magnetic energy is the volume integral of the energy density, u = Because B changes with position, u is not constant. For B = B0 FG R IJ HrK 2 , u= B2 . 2µ 0 FB GH 2 µ 2 0 0 I FG R IJ JK H r K 4 . Next, we set up an expression for the magnetic energy in a spherical shell of radius r and thickness dr. Such a shell has a volume 4π r 2 dr , so the energy stored in it is j FGH 2π µB R IJK rdr . e 2 0 dU = u 4π r 2 dr = 0 4 2 We integrate this expression for r = R to r = ∞ to obtain the total magnetic energy outside the sphere. This gives U= 2π B02 R 3 µ0 = e j e6.00 × 10 mj e1.26 × 10 T ⋅ m Aj 2π 5.00 × 10 −5 T 2 −6 6 3 = 2.70 × 10 18 J . Chapter 32 Section 32.4 P32.39 249 Mutual Inductance af I1 t = I max e −αt sin ω t with I max = 5.00 A , α = 0.025 0 s −1 , and ω = 377 rad s dI 1 = I max e −α t −α sin ω t + ω cos ω t . dt b g a fh a fh dI 1 = 5.00 A s e −0.020 0 − 0.025 0 sin 0.800 377 + 377 cos 0.800 377 dt b At t = 0.800 s , g b g c c dI 1 = 1.85 × 10 3 A s . dt Thus, ε 2 = −M P32.40 ε 2 = −M bε g 2 max dI 1 : dt M= −ε 2 +3.20 V = = 1.73 mH . dI 1 dt 1.85 × 10 3 A s dI 1 = − 1.00 × 10 −4 H 1.00 × 10 4 A s cos 1 000t dt e j b je g = 1.00 V ε2 M= P32.42 Assume the long wire carries current I. Then the magnitude of the magnetic field it generates at µ I distance x from the wire is B = 0 , and this field passes perpendicularly through the plane of the 2π x loop. The flux through the loop is dI 1 dt z = 96.0 mV = 80.0 mH 1.20 A s P32.41 z za f Φ B = B ⋅ dA = BdA = B Adx = FG H z IJ K µ 0 IA 1.70 mm dx µ 0 IA 1.70 = ln . 2π 0. 400 mm x 2π 0.400 The mutual inductance between the wire and the loop is then M= FG H a f e IJ K M = 7.81 × 10 −10 H = 781 pH P32.43 je ja f 1 4π × 10 −7 T ⋅ m A 2.70 × 10 −3 m N 2 Φ 12 N 2 µ 0 IA N µ A 1.70 = = 2 0 1.45 = ln 1.45 I1 2π I 0. 400 2π 2π e j (a) −6 N B Φ BA 700 90.0 × 10 = = 18.0 mH M= 3.50 IA (b) −6 Φ A 400 300 × 10 = = 34.3 mH LA = IA 3.50 (c) ε B = −M e j a fb dI A = − 18.0 mH 0.500 A s = −9.00 mV dt g 250 *P32.44 Inductance The large coil produces this field at the center of the small coil: N 1 µ 0 I 1 R12 e 2 x 2 + R12 j 32 . The field is normal to the area of the small coil and nearly uniform over this area, so it produces flux N 1 µ 0 I 1 R12 Φ 12 = e 2 2x + j 3 2 R12 π R 22 through the face area of the small coil. When current I 1 varies, this is the emf induced in the small coil: ε 2 = −N2 P32.45 2 2 N 1 N 2 πµ 0 R12 R 22 dI 1 dI 1 N 1 N 2 πµ 0 R12 R 22 d N 1 µ 0 R1 π R 2 I = − = − M = so M . 1 3 2 32 dt 2 x 2 + R 2 3 2 dt dt 2 x 2 + R12 2 x 2 + R12 1 e j e e j With I = I 1 + I 2 , the voltage across the pair is: ∆V = − L1 dI 1 dI dI dI dI − M 2 = − L 2 2 − M 1 = − Leq . dt dt dt dt dt dI 1 ∆V M dI 2 = + dt L1 L1 dt − So, −L 2 and a f (a) 1 2 + M2 j dIdt 2 b FIG. P32.45 g [1] g [2] = ∆V L1 − M . dI 2 ∆V M dI 1 = + dt L 2 L 2 dt By substitution, − leads to e− L L + M2 j dIdt = ∆V bL 2 Adding [1] to [2], e− L L + M2 j dIdt = ∆V bL + L2 − 2 M . So, Leq = − L1 L 2 − M 2 ∆V = . dI dt L1 + L 2 − 2 M Section 32.5 1 2 1 2 1 1 −M . g Oscillations in an LC Circuit b g At different times, U C I max = a f C ∆V L max (b) 2 dI 2 M ∆V M dI 2 + + = ∆V dt L1 L1 dt e− L L P32.46 j = max b g = UL max so LM 1 Ca∆V f OP N2 Q 2 = max 1.00 × 10 −6 F 40.0 V = 0.400 A . 10.0 × 10 −3 H a f FG 1 LI IJ H2 K 2 max Chapter 32 P32.47 P32.48 LM 1 Ca∆V f OP N2 Q 2 = max FG 1 LI IJ H2 K 2 max b g so ∆VC max = 251 L 20.0 × 10 −3 H I max = 0.100 A = 20.0 V C 0.500 × 10 −6 F a f When the switch has been closed for a long time, battery, resistor, and coil carry constant current I max = ε . When the switch is opened, R current in battery and resistor drops to zero, but the coil carries this same current for a moment as oscillations begin in the LC loop. We interpret the problem to mean that the voltage amplitude of these 1 1 2 2 . oscillations is ∆V , in C ∆V = LI max 2 2 a f e0.500 × 10 Fja150 Vf a250 Ωf C a ∆V f C a ∆V f R = = Then, L = ε I a50.0 Vf 2 2 2 2 max P32.49 2 P32.51 P32.52 2 = 0.281 H . This radio is a radiotelephone on a ship, according to frequency assignments made by international treaties, laws, and decisions of the National Telecommunications and Information Administration. 1 The resonance frequency is f0 = . 2π LC 1 1 Thus, C= = = 608 pF . 2 2 −6 6 2π f 0 L 2π 6.30 × 10 Hz 1.05 × 10 H b P32.50 2 −6 2 FIG. P32.50 f= 1 2π LC L= : 1 (a) f= (b) Q = Qmax (c) I= (a) f= (b) (c) 1 b 2π f g C 2 = e j e 1 a f e8.00 × 10 j 2π 120 2 1 −6 −6 2π j = 0.220 H = 135 Hz b0.082 0 Hge17.0 × 10 Fj cos ω t = b180 µCg cosb847 × 0.001 00g = 2π LC a fa f a 119 µC f dQ = −ω Qmax sin ω t = − 847 180 sin 0.847 = −114 mA dt 1 = 1 a0.100 Hfe1.00 × 10 Fj Q = Cε = e1.00 × 10 Fja12.0 V f = 12.0 µC 2π LC −6 2π = 503 Hz −6 1 2 1 2 Cε = LI max 2 2 I max = ε (d) = g FIG. P32.52 C 1.00 × 10 −6 F = 12 V = 37.9 mA L 0.100 H At all times U= ja 1 2 1 Cε = 1.00 × 10 −6 F 12.0 V 2 2 e f 2 = 72.0 µJ . 252 P32.53 Inductance ω= 1 LC 1 = a3.30 Hfe840 × 10 Q = Q max cos ω t , I = 2 e 105 × 10 = e −6 je cos 1.899 × 10 4 rad s 2.00 × 10 −3 s UC = Q 2C (b) UL = 2 Qmax sin 2 ω t 1 2 1 2 LI = Lω 2 Qmax sin 2 ω t = 2 2 2C (c) Section 32.6 (a) e105 × 10 Cj = j 2 e 2e840 × 10 jj 2 = 6.03 J b g je sin 2 1.899 × 10 4 rad s 2.00 × 10 −3 s −12 F j j= 0.529 J U total = UC + U L = 6.56 J The RLC Circuit ωd = (b) Rc = (a) ω0 = (b) ωd = (c) e 2 840 × 10 −12 b g FG IJ H K 1 R − 2L LC ∆ω ω0 2 fd = Therefore, P32.56 j (a) UL P32.55 F = 1.899 × 10 4 rad s dQ = −ω Qmax sin ω t dt −6 P32.54 −12 = e 1 je 2.20 × 10 −3 1.80 × 10 −6 F 7.60 −G j GH 2e2.20 × 10 −3 I J j JK 2 = 1.58 × 10 4 rad s ωd = 2.51 kHz . 2π 4L = 69.9 Ω C 1 LC = 1 a0.500fe0.100 × 10 j −6 FG IJ H K 1 R − 2L LC 2 = 4.47 krad s = 4.36 krad s = 2.53% lower Choose to call positive current clockwise in Figure 32.21. It drains charge from the capacitor dQ . A clockwise trip around the circuit then gives according to I = − dt dI Q + − IR − L = 0 C dt Q dQ d dQ R+L + + = 0 , identical with Equation 32.28. C dt dt dt Chapter 32 *P32.57 The period of damped oscillation is T = capacitor is Q = Qmax e − RT 2L = Qmax e −2π R after one oscillation it is U = U 0 e e 2π R Lω d = 1 0.99 b Lω d = 2 Lω d . The energy is proportional to the charge squared, so = 0.99U 0 . Then g F GH R2 2π 2 Ω 1 = 1 250 Ω = L − 2 LC 4L 0.001 005 1.563 × 10 6 Ω 2 = 2π . After one oscillation the charge returning to the ωd − 2π R Lω d 2π 2 Ω = ln 1.010 1 = 0.001 005 Lω d a f 2Ω L − C 4 I JK 12 2 L = 1.563 × 10 6 Ω 2 C We are also given 1 ω = 2π × 10 3 s = LC = LC 1 e2π × 10 sj 3 2 = 2.533 × 10 −8 s 2 Solving simultaneously, C = 2.533 × 10 −8 s 2 L L2 = 1.563 × 10 6 Ω 2 L = 0.199 H 2.533 × 10 −8 s 2 2.533 × 10 −8 s 2 C= = 127 nF = C 0.199 H P32.58 (a) Q = Qmax e − Rt 2 L cos ω d t so I max ∝ e − Rt 2 L 0.500 = e − Rt 2 L and Rt = − ln 0.500 2L so Q = 0.500Qmax = 0.707Qmax t=− (b) a f t=− a f FG IJ H K 2L 2L ln 0.500 = 0.693 R R 2 U 0 ∝ Qmax and U = 0.500U 0 a f FG IJ H K 2L 2L ln 0.707 = 0.347 R R 253 (half as long) 254 Inductance Additional Problems *P32.59 (a) Let Q represent the magnitude of the opposite charges on the plates of a parallel plate capacitor, the two plates having area A and separation d. The negative plate creates electric Q Q2 toward itself. It exerts on the positive plate force F = field E = toward the 2 ∈0 A 2 ∈0 A Q . The energy density is negative plate. The total field between the plates is ∈0 A uE = Q2 Q2 1 1 ∈0 E 2 = ∈0 2 2 = . Modeling this as a negative or inward pressure, we 2 2 ∈0 A 2 ∈0 A 2 have for the force on one plate F = PA = (b) Q2 , in agreement with our first analysis. 2 ∈0 A 2 The lower of the two current sheets shown creates µ J above it magnetic field B = 0 s − k . Let A and w 2 represent the length and width of each sheet. The upper sheet carries current J s w and feels force e j Js Js y x µ J µ wAJ s2 F = IA × B = J s wA 0 s i × − k = 0 j. 2 2 e j The force per area is P = (c) z µ 0 J s2 F = . Aw 2 FIG. P32.59(b) µ 0 Js µ 0 Js −k + − k = µ 0 J s k , with 2 2 . Outside the space they enclose, the fields of the separate sheets are Between the two sheets the total magnetic field is magnitude B = µ 0 J s e j e j in opposite directions and add to zero . P32.60 1 B2 = µ 02 J s2 µ 0 J s2 = 2µ 0 2 (d) uB = (e) This energy density agrees with the magnetic pressure found in part (b). 2µ 0 With Q = Qmax at t = 0 , the charge on the capacitor at any time is Q = Q max cos ω t where ω = The energy stored in the capacitor at time t is then 2 Q 2 Qmax cos 2 ω t = U 0 cos 2 ω t . = 2C 2C 1 1 cos ω t = and When U = U 0 , 4 2 U= Therefore, The inductance is then: t LC = π 3 or 1 ω t = π rad . 3 t2 π 2 = . LC 9 L= 9t 2 . π 2C 1 LC . Chapter 32 P32.61 a ε L = −L (b) Q = Idt = (c) z za t t 0 0 f 20.0t dt = 10.0t 2 −Q −10.0t 2 = = − 10.0 MV s 2 t 2 C 1.00 × 10 −6 F e e e jt j j 2 −10.0t 2 Q2 1 2 1 2 ≥ LI , or When ≥ 1.00 × 10 −3 20.0t , −6 2C 2 2 2 1.00 × 10 e then 100 t 4 ≥ 400 × 10 −9 P32.62 f d 20.0t dI = − 1.00 mH = −20.0 mV dt dt (a) ∆VC = fa (a) ε L = −L (b) I= 2 a f dI d = −L Kt = − LK dt dt dQ , dt z z t t 0 0 Q = Idt = Ktdt = so ∆VC = When f . The earliest time this is true is at t = 4.00 × 10 −9 s = 63.2 µs . b g 1 C ∆VC 2 2 = 1 2 Kt 2 −Q Kt 2 = − C 2C F GH I JK 1 K 2t 4 1 = L K 2t2 C 2 2 2 4C 1 2 LI , 2 e j t = 2 LC Thus P32.63 ja j e and (c) FG IJ H K 1 Q2 1 Q = 2 C 2C 2 2 + 1 2 LI 2 so The flux through each turn of the coil is I= 3Q 2 . 4CL ΦB = LI Q = N 2N 3L C where N is the number of turns. P32.64 B= (a) 255 µ 0 NI 2π r z Φ B = BdA = L= z b a FG IJ HK z µ 0 NI µ NIh b dr µ 0 NIh b = hdr = 0 ln a 2π r 2π a r 2π FG IJ HK NΦ B µ0N2h b ln = I a 2π a f b0.010 0g lnFG 12.0 IJ = 91.2 µH H 10.0 K 2π µ N F A I µ a500 f F 2.00 × 10 = G J = 2π GH 0.110 2π H R K µ 0 500 (b) L= (c) Lappx FIG. P32.64 0 2 2 0 2 −4 m2 I= JK 90.9 µH , only 0.3% different. 256 P32.65 Inductance (a) B= So the coil creates flux through itself Φ B = BA cos θ = When the current it carries changes, ε L = −N a f 2π r = 3 0.3 m e 2 2 R +0 L≈ so (b) Nµ 0 IR 2 At the center, j 2 3 2 = Nµ 0 I . 2R Nµ 0 I π π R 2 cos 0° = Nµ 0 IR . 2R 2 FG IJ H K dΦ B π dI dI ≈ −N Nµ 0 R = − L 2 dt dt dt π 2 N µ0R . 2 r ≈ 0.14 m π L ≈ 1 2 4π × 10 −7 T ⋅ m A 0.14 m = 2.8 × 10 −7 H 2 L ~ 100 nH so ja e je (c) P32.66 (a) −7 L 2.8 × 10 V ⋅ s A = = 1.0 × 10 −9 s R 270 V A L ~ 1 ns R If unrolled, the wire forms the diagonal of a 0.100 m (10.0 cm) rectangle as shown. The length of this rectangle is L′ = a9.80 mf − a0.100 mf 2 2 f 9.80 m 0.100 m L′ . FIG. P32.66(a) The mean circumference of each turn is C = 2π r ′ , where r ′ = radius of each turn. The number of turns is then: a a f a f f 2 2 N= 9.80 m − 0.100 m L′ = = 127 . C 2π 24.0 + 0.644 2 × 10 −3 m (b) R= −8 ρA 1.70 × 10 Ω ⋅ m 10.0 m = = 0.522 Ω 2 A π 0.322 × 10 −3 m (c) L= µN 2 A 800 µ 0 L ′ = A′ A′ C L= ja e e e 800 4π × 10 −7 0.100 m 24.0 + 0.644 mm is the mean 2 f j FG IJ π ar ′f H K j FG a9.80 mf − a0.100 mf IJ GH π a24.0 + 0.644f × 10 m JK 2 L = 7.68 × 10 −2 H = 76.8 mH 2 2 2 −3 2 π LMFG 24.0 + 0.644 IJ × 10 NH 2 K −3 m OP Q 2 Chapter 32 P32.67 257 From Ampere’s law, the magnetic field at distance r ≤ R is found as: b g e j B 2π r = µ 0 J π r 2 = µ 0 F I GH π R 2 I eπ r j , or B = µ Ir JK 2π R 2 0 2 . The magnetic energy per unit length within the wire is then z F I GH JK z R µ I2 R µ I 2 R4 µ0I2 U B2 = = . 2π rdr = 0 4 r 3 dr = 0 4 A 0 2µ 0 4 16π 4π R 0 4π R b g This is independent of the radius of the wire. P32.68 The primary circuit (containing the battery and solenoid) is an RL circuit with R = 14.0 Ω , and jb e ge j 2 −7 12 500 1.00 × 10 −4 µ N 2 A 4π × 10 L= 0 = = 0.280 H . A 0.070 0 (a) The time for the current to reach 63.2% of the maximum value is the time constant of the circuit: τ= (b) L 0.280 H = = 0.020 0 s = 20.0 ms . R 14.0 Ω The solenoid’s average back emf is where f Thus, (c) FIG. P32.68 FG ∆I IJ = LFG I − 0 IJ H ∆t K H ∆t K F ∆V IJ = 0.632FG 60.0 V IJ = 2.71 A . = 0.632G I = 0.632 I HRK H 14.0 Ω K F 2.71 A I = 37.9 V . ε = a0.280 H fG H 0.020 0 s JK f εL = L max L The average rate of change of flux through each turn of the overwrapped concentric coil is the same as that through a turn on the solenoid: jb a f e ga fe 4π × 10 −7 T ⋅ m A 12 500 0.070 0 m 2.71 A 1.00 × 10 −4 m 2 ∆Φ B µ 0 n ∆I A = = 0.020 0 s ∆t ∆t j = 3.04 mV (d) The magnitude of the average induced emf in the coil is ε L = N FG ∆Φ IJ and magnitude of H ∆t K the average induced current is I= FG H IJ K ε L N ∆Φ B 820 = = 3.04 × 10 −3 V = 0.104 A = 104 mA . R R ∆t 24.0 Ω e j B 258 P32.69 Inductance b g ε − b I + I gR ε − I + I 2 R1 − I 2 R2 = 0 . Left-hand loop: Outside loop: 2 1 −L dI =0. dt dI =0. dt This is of the same form as Equation 32.6, so its solution is of the same form as Equation 32.7: ε′ It = 1 − e − R ′t L . R′ Eliminating I 2 gives ε ′ − IR ′ − L af But R ′ = b b af It = g g j. ε 1 − e − R ′t L R1 e When switch is closed, steady current I 0 = 1.20 A . When the switch is opened after being closed a long time, the current in the right loop is I = I 0 e − R2 t so e Rt L = Therefore, P32.71 j ε ′ ε R 2 R1 + R 2 ε = . = R ′ R1 R 2 R1 + R 2 R1 R1 R 2 R2ε and ε ′ = , so R1 + R 2 R1 + R 2 Thus P32.70 e FIG. P32.69 (a) I0 I L a FG IJ H K I Rt = ln 0 . L I and fa f 1.00 Ω 0.150 s R2 t = = 0.095 6 H = 95.6 mH . L= ln I 0 I ln 1.20 A 0.250 A b g b FIG. P32.70 g While steady-state conditions exist, a 9.00 mA flows clockwise around the right loop of the circuit. Immediately after the switch is opened, a 9.00 mA current will flow around the outer loop of the circuit. Applying Kirchhoff’s loop rule to this loop gives: a f e j + ε 0 − 2.00 + 6.00 × 10 3 Ω 9.00 × 10 −3 A = 0 + ε 0 = 72.0 V with end b at the higher potential (b) FIG. P32.71(b) (c) After the switch is opened, the current around the outer loop decays as I = I max e − Rt L with I max = 9.00 mA , R = 8.00 kΩ , and L = 0.400 H . Thus, when the current has reached a value I = 2.00 mA , the elapsed time is: t= FG L IJ lnFG I IJ = FG 0.400 H IJ lnFG 9.00 IJ = 7.52 × 10 H R K H I K H 8.00 × 10 Ω K H 2.00 K max 3 −5 s = 75.2 µs . Chapter 32 P32.72 (a) IL = 0 ∆V = ε 0 + – L The instant after the switch is closed, the situation is as shown in the circuit diagram of Figure (a). The requested quantities are: I L = 0 , IC = ε0 ε , IR = 0 R R ∆VR = ε 0 ∆VC= 0 ∆VL = ε 0 , ∆VC = 0 , ∆VR = ε 0 (b) I R = ε 0 /R Q=0 +ε – 0 I C = ε 0 /R After the switch has been closed a long time, the steady-state conditions shown in Figure (b) will exist. The currents and voltages are: Figure (a) IL = 0 ∆V = 0 + – L I L = 0 , IC = 0 , I R = 0 IR = 0 Q = Cε 0 ∆VL = 0 , ∆VC = ε 0 , ∆VR = 0 ∆VR = 0 ∆VC= ε 0 +ε – 0 Figure (b) FIG. P32.72 P32.73 When the switch is closed, as shown in figure (a), the current in the inductor is I: 12.0 − 7.50 I − 10.0 = 0 → I = 0.267 A . When the switch is opened, the initial current in the inductor remains at 0.267 A. a0.267 AfR ≤ 80.0 V IR = ∆V : R ≤ 300 Ω (a) (b) FIG. P32.73 P32.74 jb e ge j 2 (a) −7 −4 2 µ N 2 A 4π × 10 T ⋅ m A 1 000 1.00 × 10 m L1 = 0 1 = = 2.51 × 10 −4 H = 251 µH A1 0.500 m (b) M= M= (c) b e4π × 10 −7 jb ga fe T ⋅ m A 1 000 100 1.00 × 10 −4 m 2 0.500 m ε 1 = −M j = 2.51 × 10 dI 2 dI dQ1 M dI 2 , or I 1 R1 = − M 2 and I 1 = =− dt dt dt R1 dt M Q1 = − R1 Q1 = g N 2 Φ 2 N 2 Φ 1 N 2 BA N 2 µ 0 N 1 A 1 I 1 A µ 0 N 1 N 2 A = = = = I1 I1 I1 I1 A1 z tf dI 2 = − 0 e2.51 × 10 −5 MI 2 i M M I 2 f − I 2i = − 0 − I 2i = R1 R1 R1 d ja H 1.00 A 1 000Ω i f = 2.51 × 10 b −8 g C = 25.1 nC −5 H = 25.1 µH 259 260 P32.75 Inductance It has a magnetic field, and it stores energy, so L = (b) Every field line goes through the rectangle between the conductors. (c) Φ = LI so L= Φ 1 = I I z I2 is non-zero. w− a BdA y=a F µ I + µ I I = 2 µ Ix dy = 2µ x ln y GH 2π y 2π bw − yg JK I z 2π y 2π µ x F w − aI L= lnG H a JK . π L= 1 I z w−a xdy 0 0 0 0 a w−a . a 0 Thus P32.76 2U (a) For an RL circuit, af I t = I max e b g − RL t af It R − RL t = 1 − 10 −9 = e b g ≅ 1 − t I max L : e3.14 × 10 je10 j = = b2.50 yrge3.16 × 10 s yrj −8 R t = 10 −9 L so Rmax −9 7 3.97 × 10 −25 Ω . (If the ring were of purest copper, of diameter 1 cm, and cross-sectional area 1 mm 2 , its resistance would be at least 10 −6 Ω ). P32.77 a fe 1 2 1 LI = 50.0 H 50.0 × 10 3 A 2 2 j 2 = 6.25 × 10 10 J (a) UB = (b) Two adjacent turns are parallel wires carrying current in the same direction. Since the loops have such large radius, a one-meter section can be regarded as straight. B= Then one wire creates a field of µ0I . 2π r This causes a force on the next wire of F = IAB sin θ F = IA giving Evaluating the force, e F = 4π × 10 µ0I µ AI 2 sin 90° = 0 . 2π r 2π r a1.00 mf 50.0 × 10 A j 2π ea0.250 mf j 3 −7 N A 2 2 = 2 000 N . Chapter 32 P32.78 P 1.00 × 10 9 W = = 5.00 × 10 3 A ∆V 200 × 10 3 V P = I∆V I= From Ampere’s law, B 2π r = µ 0 I enclosed or B = (a) 261 b g µ 0 I enclosed . 2π r I enclosed = 5.00 × 10 3 A At r = a = 0.020 0 m , FIG. P32.73 (b) (c) −7 je T ⋅ m A 5.00 × 10 3 A B= At r = b = 0.050 0 m , I enclosed = I = 5.00 × 10 3 A and e4π × 10 B= z U = udV = z b a e4π × 10 U= (d) e4π × 10 and −7 b −7 2 0 2µ 0 b 2π 0.050 0 m 2 4π je T ⋅ m A 5.00 × 10 3 A 4π je T ⋅ m A 5.00 × 10 3 A a f b2π rAdr g = µ I A Br g 2π 0.020 0 m g j = 0.050 0 T = 50.0 mT . j = 0.020 0 T = 20.0 mT . FG IJ HK z b b dr µ 0 I 2 A = ln a r 4π a j e1 000 × 10 mj lnFG 5.00 cmIJ = 2.29 × 10 H 2.00 cm K 2 3 6 J = 2.29 MJ The magnetic field created by the inner conductor exerts a force of repulsion on the current in the outer sheath. The strength of this field, from part (b), is 20.0 mT. Consider a small rectangular section of the outer cylinder of length A and width w. F w I e5.00 × 10 AjGH 2π b0.050 0 mg JK 3 It carries a current of and experiences an outward force e5.00 × 10 Ajw Ae20.0 × 10 Tj sin 90.0° . 2π b0.050 0 mg F F e5.00 × 10 A je 20.0 × 10 Tj = = The pressure on it is P = = A wA 2π b0.050 0 mg 3 F = IAB sin θ = −3 3 −3 318 Pa . 262 P32.79 Inductance jb e f ga 4π × 10 −7 T ⋅ m A 1 400 2.00 A µ 0 NI = = 2.93 × 10 −3 T upward 1.20 m A (a) B= (b) 2.93 × 10 −3 T B2 u= = = 3.42 J m3 2 µ 0 2 4π × 10 −7 T ⋅ m A e e j 2 j e b jFGH 1 N1 J⋅ m IJK = 3.42 N m 2 g = 3.42 Pa (c) To produce a downward magnetic field, the surface of the superconductor must carry a clockwise current. (d) The vertical component of the field of the solenoid exerts an inward force on the superconductor. The total horizontal force is zero. Over the top end of the solenoid, its field diverges and has a radially outward horizontal component. This component exerts upward force on the clockwise superconductor current. The total force on the core is upward . You can think of it as a force of repulsion between the solenoid with its north end pointing up, and the core, with its north end pointing down. (e) a fLMN e j OPQ = F = PA = 3.42 Pa π 1.10 × 10 −2 m 2 1.30 × 10 −3 N Note that we have not proven that energy density is pressure. In fact, it is not in some cases; Equation 21.2 shows that the pressure is two-thirds of the translational energy density in an ideal gas. ANSWERS TO EVEN PROBLEMS P32.2 1.36 µH P32.4 240 nWb P32.6 19.2 µWb P32.8 (a) 360 mV ; (b) 180 mV ; (c) t = 3.00 s P32.10 (a) 15.8 µH ; (b) 12.6 mH P32.12 P32.26 (a) 1.00 A ; (b) ∆V12 = 12.0 V , ∆V1 200 = 1.20 kV , ∆VL = 1.21 kV ; (c) 7.62 ms P32.28 (a), (b), and (c) see the solution; (d) yes; see the solution P32.30 (a) 8.06 MJ m3 ; (b) 6.32 kJ P32.32 (a) 27.8 J ; (b) 18.5 ms see the solution P32.34 see the solution P32.14 1.92 Ω P32.36 (a) 20.0 W ; (b) 20.0 W ; (c) 0;(d) 20.0 J P32.16 see the solution P32.18 92.8 V P32.20 30.0 mH P32.22 7.67 mH P32.24 (a) 1.00 kΩ; (b) 3.00 ms P32.38 2π B02 R 3 µ0 P32.40 1.00 V P32.42 781 pH P32.44 M= = 2.70 × 10 18 J N1 N 2 πµ 0 R12 R 22 e 2 x 2 + R12 j 32 Chapter 32 P32.46 400 mA P32.48 281 mH P32.50 220 mH P32.52 (a) 503 Hz ; (b) 12.0 µC ; (c) 37.9 mA ; (d) 72.0 µJ P32.54 (a) 2.51 kHz; (b) 69.9 Ω P32.56 see the solution P32.58 (a) 0.693 P32.60 P32.62 P32.64 (a) see the solution; (b) 91.2 µH ; (c) 90.9 µH , 0.3% smaller P32.66 (a) 127; (b) 0.522 Ω; (c) 76.8 mH P32.68 (a) 20.0 ms; (b) 37.9 V; (c) 3.04 mV; (d) 104 mA P32.70 95.6 mH P32.72 (a) I L = 0 , I C = P32.74 (a) 251 µH ; (b) 25.1 µH ; (c) 25.1 nC P32.76 3.97 × 10 −25 Ω P32.78 (a) 50.0 mT; (b) 20.0 mT; (c) 2.29 MJ; (d) 318 Pa FG 2L IJ ; (b) 0.347FG 2L IJ H RK H RK 9t 2 ε0 ε , IR = 0 , R R ∆VL = ε 0 , ∆VC = 0 , ∆VR = ε 0 ; (b) I L = 0 , I C = 0 , I R = 0 , ∆VL = 0 , ∆VC = ε 0 , ∆VR = 0 2 π C (a) ε L = − LK ; (b) ∆Vc = (c) t = 2 LC − Kt 2 ; 2C 263