Inductance. * An asterisk indicates a question or problem new to this edition.

Transcription

1 3 Inductance CHAPTER OUTLINE 3.1 Self-Induction and Inductance 3. RL Circuits 3.3 Energy in a Magnetic Field 3.4 Mutual Inductance 3.5 Oscillations in an LC Circuit 3.6 The RLC Circuit * An asterisk indicates a question or problem new to this edition. ANSWERS TO OBJECTIVE QUESTIONS OQ3.1 (i) Answer (a). 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. OQ3. OQ3.3 (ii) Answer (c). 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. Answer (c). The fine wire has considerable resistance, so a few seconds is many time constants. The final current depends on the resistance of the wire, which has not changed; the current is not affected by the inductance of the coil because the current is not changing. Answer. The inductance of a solenoid is proportional to the number of turns squared, so cutting the number of turns in half makes the inductance four times smaller. Doubling the current would by itself make the stored energy ( 1 Li )four times larger, to just compensate. 468

2 Chapter OQ3.4 OQ3.5 OQ3.6 The ranking is ΔV L > ΔV 1 00 Ω > 1.0 V > ΔV 1 Ω. Just before the switch is thrown, the voltage across the 1-Ω resistor is very nearly 1 V (we assume the resistance of the inductor is small). Just after the switch is thrown, the current is nearly the same, maintained by the inductor, but this current is diverted through the 1 00-Ω resistor; thus, the voltage across the 1 00-Ω resistor is much more than 1 V, about 1 00 V, because the same current in the 1-Ω resistor now passes through a resistor 100 times as large. By Kirchhoff s loop rule, the voltage across the coil is larger still. Answer (d). The inductance of a solenoid is proportional to the number of turns squared (N ), to the cross-sectional area (A), and to the reciprocal of the length of its axis (L). Coil A has twice as many turns with the same length of wire, so its circumference must be half as large as that of coil B: therefore, its radius is half as large and its area one quarter as large. For coil A the inductance will be different by the factor N A/L ~ [ (1/4)]/ = 1/. Answer (a). The energy stored in the magnetic field of an inductor is proportional to the square of the current. Doubling I makes U B = 1 Li get four times larger. OQ3.7 Answer (d). The emf across an inductor is zero whenever the current is constant (unchanging), large or small. ANSWERS TO CONCEPTUAL QUESTIONS CQ3.1 (a) 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. The idea of a field was 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.

3 470 Inductance Henry precipitated a basic change if he did not cause it. A list today of what makes up the Universe might include quarks, electrons, muons, tauons, and neutrinos of matter; photons of electric and magnetic fields; W and Z particles; gluons; energy; charge; baryon number; three different lepton numbers; upness; downness; strangeness; charm; topness; and bottomness. Alternatively, the relativistic interconvertibility 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, charge,... bottomness as properties of matter, rather than as things with their own existence. CQ3. (a) The inductance of a coil is determined by (a) the geometry of the coil and 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. CQ3.3 No. The inductance of a coil is proportional to the flux through the coil per unit current, Φ/I, and the flux is proportional to the current I, so the inductance is independent of the current. 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. CQ3.4 (i) (a) The bulb glows brightly right away, and then more and more faintly as the capacitor charges up. The bulb gradually gets brighter and brighter, changing rapidly at first and then more and more slowly. (c) The bulb immediately becomes bright. (d) The bulb glows brightly right away, and then more and more faintly as the inductor starts carrying more and more current (the inductor eventually acts as a short). (ii) (a) The bulb goes out immediately because current stops immediately (charge ceases to flow). The bulb glows for a moment as a spark jumps across the switch. (c) The bulb stays

4 Chapter lit for a while, gradually getting fainter and fainter as the capacitor discharges through the bulb. (d) The bulb suddenly glows brightly. Then its brightness decreases to zero, changing rapidly at first and then more and more slowly. CQ3.5 (a) 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.) CQ3.6 (a) 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. CQ3.7 (a) The instant after the switch is closed, the capacitor acts as a closed switch, and the inductor acts to maintain zero current in itself. The situation is as shown in the circuit diagram of ANS. FIG. CQ3.7(a). The requested quantities are: I L = 0, I C = ε 0 R, I = ε 0 R R ΔV L = ε 0, ΔV C = 0, ΔV R = ε 0 ANS. FIG. CQ3.7(a) After the switch has been closed a long time, the capacitor acts as an open switch. The steady-state conditions shown in ANS. FIG. CQ3.7 will exist. The currents and voltages are: I L = 0, I C = 0, I R = 0 ΔV L = 0, ΔV C = ε 0, ΔV R = 0 ANS. FIG. CQ3.7 CQ3.8 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

5 47 Inductance CQ3.9 CQ3.10 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. According to Equations 3.31 and 3.3, the oscillator is overdamped if R > R C = 4L C : it will not oscillate. If R < R C, then the oscillator is underdamped and can go through several cycles of oscillation before the current falls below background noise. 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. The capacitor s energy is proportional to the charge it stores, the coil s energy is proportional to the current it holds. SOLUTIONS TO END-OF-CHAPTER PROBLEMS Section 3.1 *P3.1 Self-Induction and Induction The magnitude of the average induced emf for this coil is ε = L Δi Δt = H = 19.5 mv = V 1.50 A 0.00 A 0.00 s *P3. Treating the telephone cord as a solenoid, we have: L = µ 0 N A = 1.36 µh 70.0 = 4π 10 7 T m A π m m P3.3 The self-induced emf at any instant is ε L = L di Its average value is ε L,ave = L I f I i t = +100 V A = (.00 H) s V s/a 1 H

6 Chapter P3.4 (a) The inductance of the solenoid is L = µ 0 N A ( 4π 10 7 T m A ) 400 = 0.00 m = H = 1.97 mh π m From ε = L( Δi Δt), P3.5 From ε = L Δi Δt, we have ΔI Δt = ε L = V H = A/s = 38.0 ma s L = From L = NΦ B i ε ( Δi Δt) = V = H 10.0 A s, we have 4.00 A Φ B = Li N = H 500 = 19. µt m P3.6 (a) B = µ 0 ni = 4π 10 7 T m A Φ B = BA = Bπ m m A = T m = 188 µt (c) L = NΦ B i = 450Φ B = mh A (d) B and Φ B are proportional to current; L is independent of current. P3.7 From ε = L Δi Δt, we have L = ε Δi Δt = ε Δt Δi = H = 4.00 mh s = V.00 A 3.50 A

7 474 Inductance P3.8 (a) In terms of its cross-sectional area (A), length ( ), and number of turns (N), the self inductance of a solenoid is given as L = µ 0 N A. Thus, for the given solenoid, L = µ 0N ( πd 4) ( 4π 10 7 T m A) 580 = 0.36 m 4 π m = H = 5.90 mh ( A s) ε = L Δi Δt = H = V = 3.6 mv = ( ) d ( 1.00t 6.00t) = ( 90.0) (.00t 6.00), where ε P3.9 ε = L di is in millivolts (mv) and t in seconds. (a) At t = 1.00 s, ε = 360 mv At t = 4.00 s, ε = 180 mv (c) ε = ( 90.0) ( t 6) = 0 when t = 3.00 s P3.10 The inductance is L = µ 0 N A with A = πr. The induced emf as a function of time is ε L = L di. By substitution we have ε L = L di = µ N 0 πr di ε L and r = µ 0 N π di/ 1/ ( V)(0.160 m) Then r = (4π 10 7 N/A )(40) π( 0.41 A/s) P3.11 The emf is given by ε = ε 0 e kt = L di from which we obtain di = ε 0 L e kt 1/ = 9.77 mm

8 Chapter If we require i 0 as t, the solution is i = ε 0 Lk e kt = dq, so Q = i = ε 0 Lk e kt = ε 0 P3.1 The inductance of a solenoid is L = µ N 0 A. 0 Lk Q = ε 0 Lk The long solenoid is bent into a circle of radius R, so its length π R; therefore, the inductance of the toroid is L = µ 0 N A µ 0N ( πr ) = 1 π R µ r 0N R ANS. FIG. P3.1 P3.13 Using the definition of self-inductance, ε = L di, we obtain ε = L d ( I i sinω t) = Lω ( I i cosω t) [ π ( 60.0) ] 5.00 = cosω t ε = 18.8cos10πt, where ε is in volts and t is in seconds. P3.14 The current change is linear, so ε = L di Δi = L Δt. t = 0 to 4 ms: t = 4 to 8 ms: t = 8 to 10 ms:.00 ma ε = ( 4.00 mh) = +.00 mv 4.00 ms ma ε = ( 4.00 mh) = 5.00 mv 4.00 ms ε = 4.00 mh 0 = 0.00 mv.00 ms

9 476 Inductance t = 10 to 1 ms: 3.00 ma ε = ( 4.00 mh) = mv.00 ms ANS. FIG. P3.14 Section 3. RL Circuits P3.15 (a) The inductance of a solenoid is L = µ 0 N A = µ 0 N πr = µ = H = mh m π m The time constant of the circuit is P3.16 (a) At time t, where τ = L R = H.50 Ω = ε R i t τ ( 1 e t ) = s = ms τ = L R =.00 H 10.0 Ω = 0.00 s After a long time, I i = ε R ( 1 e ) = ε R ANS. FIG. P3.16

10 Chapter At i(t) = 0.500I i so ( 0.500) ε R = ε τ 1 e t R = 1 e t τ Isolating the constants on the right, e t τ = = ln ( 0.500) [ ] = ( 0.00 s) [ ln ( 0.500) ] = s ln e t τ t = τ ln Similarly, to reach 90% of I i, = 1 e t τ e t τ =0.100 and t = τ ln ( 0.100) Thus, t = ( 0.00 s)ln ( 0.100) = s P3.17 (a) Using τ = RC = L, we get R R = L C = 3.00 H F = Ω = 1.00 kω. The time constant is ( F) τ = RC = Ω = s = 3.00 ms P3.18 The current builds exponentially according to: = ε R i t τ ( 1 e t ) = 1.0 V 4.0 Ω 1 e t τ = e t τ where current I is in amperes (A) and time t is in seconds (s). The current increases from 0 to asymptotically approach A. In case (a) the current jumps up essentially instantaneously. In case it increases with a longer time constant, and in case (c) the increase is still slower.

11 478 Inductance (a) With essentially zero inductance, we take τ = L = ANS. R FIG. P3.18(a) graphs I(t) for this case. ANS. FIG. P3.18(a) We take τ = L = 1. ANS. FIG. P3.18 graphs I(t) for this case. R ANS. FIG. P3.18 (c) We take τ = L = 10. ANS. FIG. P3.18(c) graphs I(t) for this case. R ANS. FIG. P3.18(c)

12 Chapter P3.19 (a) The two resistors are in parallel. Their resistance values 450 Ω and R are related to their equivalent resistance R eq by 1 = 1 R eq R Ω and the equivalent resistance is related to the time constant of the circuit by Thus, τ = L R eq 1 R eq = τ L 1 = 1 R eq R Ω = τ L Solving for R, 1 R = τ L Ω = s H Ω which gives R = 1 90 Ω = 1.9 kω The current will immediately begin to die from the value it had just before the switch was thrown to position b. Before the switch position was changed, the current was constant in time, so there was no emf induced in the inductor. The current was just i = ΔV = ΔV τ R eq L = ( 4.0 V) s H = A = 7.0 ma *P3.0 The current increases as a function of time as i = I i ( 1 e t τ ) Substituting, = 1 e τ = e τ ANS. FIG. P3.0 Solving for the time constant gives τ = ln( ) = s

13 480 Inductance and since τ = L R, L = τ R = ( s) ( 10.0 Ω) = 7.67 mh *P3.1 For the increasing current i = ε R condition on Δt is R ( 1 e Lt ). The final value is ε R, so the ε R = ε ( R 1 e LΔt R ) e LΔt R = 0.00 e +LΔt R = 5.00 LΔt R Δt = = ln 5.00 Rln 5.00 L At the moment when the battery is removed, the current in the coil is quite precisely ε R. During the decrease, i = I i e Lt R = ε R e Lt R. (a) at t = Δt = at t = Δt, Rln 5.00, L i I i = e LΔt R = 0.00 = 0.0% P3. Taking τ = L R, and i = I i e t τ : i = e LΔt R = e LΔt R = ( 0.00) = = 4.00% I i di = I i e t τ 1 τ 1 τ ir + L di = 0 will be true if I ir e t τ + L I i e t τ We have agreement because τ = L R. P3.3 The current at this time is given by = 0 i = ε τ ( 1 e t ) = 10 V R 9.00 Ω 1 ( e ) = 3.0 A Then, ΔV R = ir = ( 3.0) ( 9.00) = 7. V and ΔV L = ε ΔV R = = 9.8 V

14 Chapter P3.4 (a) τ = L R = H 4.00 Ω (c) i = ε τ ( 1 e t ) = 6.00 V R 4.00 Ω 1 e I i = ε R = 6.00 V 4.00 Ω = 1.50 A = s =.00 ms = A (d) = 1 e t.00 ms t = (.00 ms)ln ( 0.00) = 3. ms ANS. FIG. P3.4 P3.5 Name the currents as shown in ANS. FIG. P3.5. By Kirchhoff s laws: i 1 = i + i 3 [1] V 4.00i i = 0 [] V 4.00i i 3 ( 1.00) di 3 = 0 [3] From [1] and [], ANS. FIG. P i i i 3 = 0 i 1 = 0.500i A Then [3] becomes 10.0 V 4.00( 0.500i A) 8.00i 3 ( 1.00) di 3 = 0 ( 1.00 H) di 3 + ( 10.0 Ω)i 3 = 5.00 V or 5.00 V ( 10.0 Ω)i H di 3 = 0

15 48 Inductance which can be compared to the general form (Equation 3.6) ε ir L di = 0 which has the solution (from Equation 3.7) i = ε L ( 1 e Rt ). R Thus, we have: (a) i 3 = 5.00 V 10.0 Ω 1 e 10.0 Ω t 1.00 H = ( A s ) 1 e 10t i 1 = I 3 = 1.50 A ( 0.50 A)e 10t s P3.6 Refer to ANS. FIG. P3.5 above. Name the currents as shown. By Kirchhoff s laws: From [1] and [], i 1 = i + i 3 [1] ε Ri 1 Ri = 0 [] ε Ri 1 Ri 3 L di 3 = 0 [3] ε Ri 1 R( i 1 i 3 ) = 0 ε Ri 1 Ri 1 + Ri 3 = 0 i 1 = 1 i 3 + ε R Then [3] becomes ε R 1 i + ε 3 R L di 3 +.5Ri = ε 3 ε.5ri L di 3 3 = 0 Ri 3 L di 3 = 0 which can be compared to the general form (Equation 3.6) ε ir L di = 0 which has the solution (from Equation 3.7) i = ε R L ( 1 e Rt )

16 Chapter Thus, we have: (a) i 3 = ε.5r 1 e.5rt L = ε 5R ε L ( 1 e 5Rt ) i 1 = 1 i + ε 3 R = 1 5R 1 e 5Rt L + ε R i 1 = ε 10R 1 e 5Rt L + 5ε 10R = ε 10R 6 e 5Rt L P3.7 (a) When i =.00 A, the voltage across the resistor is ΔV R = ir = (.00 A) ( 8.00 Ω) = 16.0 V Kirchhoff s loop rule tells us that the sum of the changes in potential around the loop must be zero: so and ε ΔV R ε L = 36.0 V 16.0 V ε L = 0 ε L = 0.0 V ΔV R ε L = 16.0 V 0.0 V = Similarly, for i = 4.50 A, ΔV R = ir = ( 4.50 A) ( 8.00 Ω) = 36.0 V and ε ΔV R ε L = 36.0 V 36.0 V ε L = 0 so ε L = 0 P3.8 For t 0, the current in the inductor is zero. For 0 t 00 µs, there will be current i R in the resistor and I L in the inductor so that i = i R + i L = I i = 10.0 A. Assuming both currents are downward in ANS. FIG. P3.8, we apply Kirchhoff s loop rule going counterclockwise around the loop, and we find that ANS. FIG. P3.8 i R R + L di L = 0 Using I i = i R + i L i R = I i i L, we have ( I i i L )R + L di L = 0

17 484 Inductance Then, L di L = ( I I max L )R I t di L = R 0 ( I i I L ) L 0 ln I i i L I i = R L t which gives i L = I i ( 1 e Rt L ) We see that t = 0, i L = 0 as we expect because of the back emf induced in the inductor. With the time constant we have τ = L 10.0 mh = R 100 Ω i L = I i ( 1 e t τ ) = 10.0 A At t = 00 µs, i = A = s 1 e 1 e.00 decays. The loop rule gives the same result, i R R + L di L = 0 but now i R + i L = 0 i R = i L, so we have i L R + L di L = 0 L di L = i LR t s 0 t 00 µs = 8.65 A; thereafter, the current I Ii di L i L = t 00 µs ln i L = R I i L R L For t = 00 µs, i i = 8.65 A, and for t 00 µs, ( t 00 µs ) i = I L i t 00 µs i = ( 8.65 A)e s t s+.00 = ( 8.65 A)e ( t 00µs ) e R L = ( 8.65e.00 A)e t s = ( 63.9 A)e t s t 00 µs

18 Chapter P3.9 From Equation 3.7, i = I i ( 1 e t τ ). Therefore, where Then, (a) At t = 0, di = I i e t τ 1 τ τ = L R = 15.0 H 30.0 Ω = s di = R L I i e t τ At t = 1.50 s, with I i = ε R di = R L I i e0 = ε L = 100 V 15.0 H = 6.67 A s di = ε L e t τ = ( 6.67 A s)e = ( 6.67 A s)e 3.00 = 0.33 A s P3.30 (a) For a series connection, both inductors carry equal currents at every instant, so di is the same for both. The voltage across the pair is di L eq = L di 1 + L di L eq = L 1 + L For a parallel connection, the voltage across each inductor is the same for both. di L eq = L di 1 1 = L di = ΔV L where the currents are related by i = i 1 + i. Therefore, di = di 1 + di ΔV L = ΔV L + ΔV L L eq L 1 L 1 L eq = 1 L L

19 486 Inductance (c) di L eq + R i = L di eq 1 + ir + L di 1 + ir Now i and di are independent quantities under our control, so functional equality requires both L eq = L 1 + L and R eq = R 1 + R. di (d) Yes. The relations ΔV = L eq + R i = L di 1 eq 1 + R i = L di R i, where i = i 1 + i and di = di 1 + di, must always be true. We may choose to keep the currents constant in time. Then, from i = i 1 + i, we have ΔV L R eq = ΔV L R 1 + ΔV L R 1 R eq = 1 R R We may choose to make the current oscillate so that at a given moment it is zero. Then, from di = di 1 + di, as in part, we 1 have = L eq L 1 L P3.31 (a) The equation for current buildup is obtained by combining Equations 3.7 and 3.8: ε i = 1 e Rt/L R We proceed step-by-step to solve for t in terms of the other quantities, all of which are given: so ir/ε = 1 e Rt/L e Rt/L = 1 ir/ε and Rt/L = ln(1 ir/ε ) then, t = (L/R) ln(1 ir/ε ) t = (0.140 H/4.90 Ω ) ln[1 (0.0 A)(4.90 Ω )/6.00 V] = ( s) ln(0.80) t = ( s) ( 0.198) = 5.66 ms

20 Chapter (c) We now make the general equation refer to a different instant. The current after ten seconds is i = = 1. A 6.00 V 4.90 Ω 1 e ( 35.0 s 1 )(10.0 s) (1 e 350 ) = 1. A The equation for current decrease after the battery is removed is i = ε R e Rt/L. We solve for t: Then, ir ε = e Rt/L or ε ir = e+rt/l ln( ε ir) = Rt L and t = ( L R) ln( ε ir) Substituting, t = (0.140 H/4.90 Ω) ln[6.00 V/(0.160 A 4.90 Ω)] = ( s) ( ln 7.65) = 58.1 ms Section 3.3 Energy in a Magnetic Field P3.3 The inductance of the solenoid is L = N Φ B i The energy stored is = Wb 1.75 A = H U B = 1 Li = 1 ( H) ( 1.75 A) = J = 64.8 mj P3.33 For a solenoid of length, the inductance is L = µ N 0 A. Thus, since U B = 1 Li = µ N 0 Ai, the stored energy is U B = (4π 10 7 N/A )(68) π( m) (0.770 A) ( ) m = J

21 488 Inductance P3.34 From Equation 3.7, i = ε L ( 1 e Rt ): R (a) The maximum current, after a long time t, is i = I i = ε =.00 A. R At that time, the inductor is fully energized and P = i( ΔV) = (.00 A) ( 10.0 V) = 0.0 W (c) (d) P lost = i R = (.00 A) ( 5.00 Ω) = 0.0 W = 0 The inductor has no resistance: P inductor = i ΔV drop U B = 1 Li = 1 ( 10.0 H) (.00 A) = 0.0 J P3.35 (a) The energy density stored by the electric field is E u E = 0 = 8.85 C 10 1 N m = J m 3 = 44.3 nj m3 ( 100 V m) The energy density stored by the magnetic field is P3.36 We compute the integral: ( u B = B T) = µ 0 4π 10 7 T m/a 0 N = m m m = e Rt L = L R e Rt L R L 0 N/A m T J C V N m m 3 = 995 µj m 3 = L R e Rt L = L ( R e e 0 ) = L R ( 0 1) = L R N m J *P3.37 The current in the circuit at time t is i = I i ( 1 e t τ ), where I i = ε R, and the energy stored in the inductor is U B = 1 Li. 0 (a) As t, I I i = ε R = 4.0 V = 3.00 A, and 8.00 Ω U B = 1 LIi = 1 ( 4.00 H) ( 3.00 A) = 18.0 J

22 Chapter At t = τ, I = I i ( 1 e 1 ) = ( 3.00 A) ( ) = 1.90 A, and U B = 1 Li = 1 ( 4.00 H) ( 1.90 A) = 7.19 J P3.38 (a) P = iδv = ( 3.00 A) (.0 V) = 66.0 W (c) P = iδv R = i R = ( 3.00 A) ( 5.00 Ω) = 45.0 W METHOD 1: We treat the real inductor as an ideal inductor (with no resistance) in series with an ideal resistor (with no inductance). When the current is 3.00 A, Kirchhoff s loop rule reads +.0 V 3.00 A ΔV L = 7.00 V ( 5.00 Ω) ΔV L = 0 The power being stored in the inductor is iδv L = ( 3.00 A) ( 7.00 V) = 1.0 W METHOD : We do not treat the real inductor as an ideal inductor in series with an ideal resistor. We wish to find the rate at which energy is being delivered to the inductor. As discussed in Section 3.3, U B = 1 Li du B = Li di. We know L ( H) and i (3.00 A); we need to evaluate the term di. From Equations 3.7 and 3.8 (or Equation 3.9), i = ε τ ( 1 e t ) di R = ε L e t τ because τ = L R. Also, i = ε τ ( 1 e t ) e t τ = 1 ir R ε Therefore, du B When i = 3.00 A, du B = Li di = Li ε L e t τ = iεe t τ = iε 1 ir ε = i ε ir = i( ε ir) = ( 3.00 A).0 V ( 3.00 A) ( 5.00 Ω) = 1.0 W [ ]

23 490 Inductance (d) (e) (f) The power supplied by the battery is equal to the sum of the power delivered to the internal resistance of the coil and the power stored in the magnetic field. Yes. Just after t = 0, the current is very small, so the power delivered to the internal resistance of the coil (ir ) is nearly zero, but the rate of the change of the current is large, so the power delivered to the magnetic field (Ldi/) is large, and nearly all the battery power is being stored in the magnetic field. Long after the connection is made, the current is not changing, so no power is being stored in the magnetic field, and all the battery power is being delivered to the internal resistance of the coil. P3.39 (a) The magnetic energy density is given by u B = B µ 0 = ( 4.50 T) 4π 10 7 T m A = J m 3 = 8.06 MJ The magnetic energy stored in the field equals u times the volume of the solenoid (the volume in which B is non-zero) m U B = u B V = J m 3 = 6.3 kj π m Section 3.4 Mutual Inductance P3.40 We use Equation 3.17, ε = M di 1 mutual inductance:, from which we obtain the M = ε di 1 = V = H = 80.0 mh 1.0 A s

24 Chapter P3.41 Let the changing current in coil 1 induce an emf in coil. Then, ε = M di 1 d 10.0sin ( t) = = ( )( 10.0) ( )cos( t) = ( 1.00)cos t Therefore, the peak emf is ( ε ) max = 1.00 V. P3.4 The current is given by i = I i e αt sinω t, with I i = 5.00, α = , and ω = 10π. Then, di = d I i e α t sinω t = I i ( α e α t )sinω t + I i e α t ω cosω t = I i e α t α sinω t + ω cosω t where di is in amperes per second, I i is in amperes, and t in seconds. At t = s, di = ( 5.00)e { ( )sin π = A s Thus, from ε = M di 1, [ ] M = ε di 1 = +3.0 V = 1.73 mh A s P3.43 (a) The mutual inductance of the coils is M = N BΦ BA i A = Wb 3.50 A + 10π cos[ 0.800( 10π )]} = 18.0 mh (c) The inductance of coil A is L A = Φ A i A = Wb 3.50 A The emf induced in the other coil is = 34.3 mh ε B = M di A = ( 18.0 mh) ( A s) = 9.00 mv

25 49 Inductance P3.44 (a) Solenoid S 1 creates a nearly uniform field everywhere inside it, given by B 1 = µ 0 N 1 i/. The flux through one turn of solenoid S is µ 0 πr N 1 i/ The emf induced in solenoid S is (µ 0 πr N 1 N / )(di/) The mutual inductance is M 1 = µ 0 πr N 1 N / Solenoid S creates a nearly uniform field everywhere inside it, given by B = µ 0 N i / and nearly zero field outside. The flux through one turn of solenoid 1 is µ 0 πr N i / The emf induced in solenoid 1 is (µ 0 πr N 1 N / )(di /) The mutual inductance is M 1 = µ 0 πr N 1 N /. (c) They are the same. P3.45 Assume the long wire carries current I. Then the magnitude of the magnetic field it generates at distance x from the wire is B = µ 0 I π x, and this field passes perpendicularly through the plane of the loop. The flux through the loop is Φ B = B d A = BdA = B dx = µ 0 I π 1.70 ln = µ 0 I π 1.70 mm mm The mutual inductance between the wire and the loop is then M = N Φ 1 I 1 = N µ 0 I π I = 1 4π 10 7 T m A π ln ( m) M = H = 781 ph dx x ln

26 Chapter P3.46 (a) A current i in the large loop of radius R produces a magnetic field of magnitude B = µ 0i at its center. Because the radius of the small R loop r<< R, we may treat the flux through the small loop as being approximately Φ B = BAcos0.00 = µ i 0 R A = µ 0πr i. The mutual R inductance of the loops is then M = Φ B i M = µ 0πr R = 3.95 nh = µ 0πr R π m = 4π 10 7 T m/a 0.00 m = H Section 3.5 Oscillations in an LC Circuit P3.47 When the switch has been closed for a long time, battery, resistor, and coil carry constant current I i = 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 oscillations is ΔV, in 1 C( ΔV) = 1 LIi. Then, L = C( ΔV) I i = C( ΔV) R ε = 0.81 H = 81 mh 150 V = F 50.0 V ( 50 Ω) P3.48 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 f 0 = π LC. 1 Thus, C = ( π f 0 ) L = 1 π ( Hz) H ANS. FIG. P3.47 = 608 pf

27 494 Inductance P3.49 At different times, ( U C ) max = ( U L ) max, so Then, 1 C ΔV = 1 max LI i I i = C L ( ΔV) = F max H ( 40.0 V) = A = 400 ma P3.50 From the angular frequency of oscillation of the circuit, we have ω = 1 LC = π f Solving for the inductance gives 1 L = C π f = ( F) [ π ( 10 Hz) ] = 0.0 H P3.51 At different times, the maximum energy stored in the capacitor is equal to the maximum energy stored in the inductor. so 1 C ΔV = 1 max LI i ( ΔV C ) max = L C I = H i F ( A) = 0.0 V P3.5 Find the energy stored in the circuit from Equation 3.7: U = Q max C = C F 1 = J = 400 µj If the energy is split equally between the capacitor and inductor at some instant, the energy would be half this value, or 00 µj. Therefore, there would be no time when each component stores 50 µj. P3.53 (a) The frequency of oscillation of the circuit is 1 f = π LC = 1 π H F = 503 Hz

28 Chapter The maximum charge on the capacitor is Q = Cε = ( F) ( 1.0 V) = 1.0 µc (c) To find the maximum current I i, we equate 1 Cε = 1 LI i Then solve for I i to obtain I i = ε C L = 1.0 V F H = 37.9 ma (d) The total energy the circuit possesses at t = 3.00 s and at all times is U = 1 Cε = 1 ( F) ( 1.0 V) = 7.0 µj P3.54 At t = 0 the capacitor charge is at its maximum value, so φ = 0 in Q = Q max cos(ωt + φ) = Q max cos t LC Substituting the given information, the charge at ms is Q = ( C)cos = ( C)cos( 38.0 rad) = C s (3.30 H)( F) (a) (c) Then the energy in the capacitor is U C = Q C = ( C) F = 6.03 J The constant total energy is that originally of the capacitor: = 6.56 J U = Q max C = C F So the inductor s energy is the remaining U L = 6.56 J 6.03 J = 0.59 J

29 496 Inductance P3.55 (a) The angular frequency of oscillations is ω = 1 LC = H ( F) = 847 rad/s = π f so f = 135 Hz The charge on the capacitor is Q = Q max cosω t = ( 180 µc)cos 847 rad/s = 119 µc s (c) The current in the circuit is given by Equation 3.3: i = dq = ω Q max sinω t ( 180 µc) sin ( 847 Hz) s = 847 rad/s = 114 ma Section 3.6 The RLC Circuit P3.56 We choose to call positive current clockwise in Figure It drains charge from the capacitor according to i = dq. A clockwise trip around the circuit then gives or + q C ir L di = 0 + q C + dq R + L d dq = 0, identical to Equation 3.8. P3.57 (a) The frequency of damped oscillations is given by Equation 3.3: ω d = = 1 LC R L 1 ( H) ( F) 7.60 ( H) = rad s Therefore, f d = ω d π = rad π s =.51 khz.

30 Chapter Critical damping occurs when ω d = 0, or when R c = 4L C = 4 ( H) = 69.9 Ω F P3.58 (a) The angular frequency of undamped oscillations is ω 0 = 1 LC = H = 4.47 krad s = rad s F The frequency of the damped oscillations is ω d = 1 LC R L (c) = H Ω ( H) F = 4.36 krad s Δω 100% = ω ω d % = 100% =.53% ω 0 ω P3.59 (a) The charge on the capacitor is given by Equation 3.31: q = Q max e Rt L Rt L cosω d t so I i e When the amplitude of the oscillation falls to 50.0% of its initial value, we have Then, = e Rt L and Rt = ln L t = L R = L R ln The initial energy of the circuit is U 0 Q max. When U = 0.500U 0, Then, q = 0.500Q max = 0.707Q max t = L R = L R ln (half as long as part (a))

31 498 Inductance Additional Problems P3.60 (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 an electric field E = F = Q 0 A Q 0 A plates is toward itself. It exerts on the positive plate force toward the negative plate. The total field between the Q. The energy density is 0 A Q u E = 1 0 E = A = Q. Modeling this as a negative or 0 A inward pressure, we have for the force on one plate Q F = PA =, in agreement with our first analysis. 0 A The lower of the two current sheets shown creates above it. Let and w represent the length magnetic field B = µ 0J s ˆk and wih of each sheet. The upper sheet carries current J s w and feels force F = I B = J s w î µ J 0 s ˆk = µ w J 0 s ĵ The force per area is P = F w = µ 0J s. (c) ANS. FIG. P3.60 Between the two sheets, each sheet contributes the same field, so + µ 0J s = µ 0 J s ˆk, with the total magnetic field is µ 0J s ˆk ˆk magnitude B = µ 0 J s. Outside the space they enclose, the fields of the separate sheets are in opposite directions and add to zero.

32 Chapter (d) u B = 1 B = µ 0 J s = µ 0J s µ 0 µ 0 (e) The energy density found in part (d) agrees with the magnetic pressure found in part. P3.61 (a) The voltage across the inductor is given by d( 0.0t) ε L = L di = 1.00 mh The charge that flows into the capacitor is t q = I = ( 0.0t) = 10.0t 0 t 0 = 0.0 mv Going across the capacitor in the directon of the current, the potential drops from the positive to the negative side, so ΔV C = q C = 10.0t F = 10.0t where ΔV C is in megavolts and t is in seconds. (c) When q C 1 Li, or ( 10.0t ) ( ) 0.0t, then 100t 4 ( )t The earliest time this is true is at t = s = 63. µs P3.6 (a) The voltage across the inductor is given by ε L = L di = L d ( Kt) = LK The current into the capacitor is i = dq, so the charge that flows into the capacitor is t q = i = Kt = 1 Kt 0 t 0

33 500 Inductance Going across the capacitor in the direction of the current, the potential drops from the positive to the negative side, so ΔV C = q C = Kt C (c) When 1 C ( ΔV C ) 1 LI, 1 C K t 4 4C 1 L K t Thus, the energy in the capacitor begins to exceed the energy in the inductor after t = LC. P3.63 The total energy equals the energy in the capacitor and inductor: 1 Q C = 1 C so i = 3Q 4CL Q + 1 Li The flux through each turn of the coil is Φ B = Li N = q N 3L C where N is the number of turns. P3.64 (a) The inductor has no resistance, therefore it has voltage across it. It behaves as a short circuit. The battery sees an equivalent resistance Ω Ω Ω The battery current is 10.0 V 6.67 Ω = 1.50 A 1 = 6.67 Ω The voltage across the parallel combination of resistors is 10.0 V ( 1.50 A) ( 4.00 Ω) = 4.00 V

34 Chapter The current in the 8-Ω resistor and the inductor is (c) 4.00 V = 500 ma 8.00 Ω The energy stored in the inductor for t < 0 is U B = 1 Li = 1 ( 1.00 H) ( A) = 15 mj (d) (e) The energy becomes 15 mj of additional internal energy in the 8-Ω resistor and the 4-Ω resistor in the middle branch. See ANS. FIG. P3.64(e). The current decreases from 500 ma toward zero, showing exponential decay with a time constant τ = L R = 1.00 H = s = 8.33 ms 3( 4.00 Ω) P3.65 The voltages are related as ε ir L di = 0 ANS. FIG. P3.64(e) When the current is increasing: di ε = ir + L = V = (.00 A) R + L (0.500 A/s) [1] When the current is decreasing: (a) 5.00 V = (.00 A) R + L ( A/s) [] Subtracting [] from [1] gives 9.00 V 5.00 V = L (1.00 A/s) L = 4.00 H Substituting the value for L in either equation gives 7.00 V = (.00 A) R R = 3.50 Ω P3.66 (a) Initially, the current is zero because of the emf induced in the coil resists an increase in the current. Just after the circuit is connected, the potential difference across the resistor is 0 and the emf across the coil is 4.0 V.

35 50 Inductance (c) After several seconds, the current has reached a steady value and does not change. After several seconds, the potential difference across the resistor is 4.0 V and that across the coil is 0. The resistor voltage and inductor voltage always add to 4 V. The resistor voltage increases monotonically, so the two voltages are equal to each other, both being 1.0 V, just once. The time is given by V = ir = Rε/R(1 e Rt/L ) Substituting, 1 V = 4 V(1 e 6Ωt/0.005 H ) 0.5 = e 1 00 t 1 00 t = ln t = ms The two voltages are equal to each other, both being 1.0 V, just once, at ms after the circuit is connected. (d) There is now no battery in the circuit, so the current decays to zero. The resistor and inductor are in parallel because they have common connections on each side. As the current decays the potential difference across the resistor is always equal to the emf across the coil. P3.67 (a) At the center, B Nµ 0i R. So the coil creates flux through itself Φ B = BAcosθ = Nµ 0i R π R cos0 = 1 Nµ 0πiR The inductance is L = N Φ B i N Nµ 0π IR i 1 µ 0π N R To find the inductance of the circuit, we compute its radius from π R = 3( m) R = m Then, from the expression found in part(a), the inductance is L 1 µ 0π N R = 1 ( 4π 10 7 T m A)π 1 = H L ~ 10 7 H m

36 Chapter (c) The time constant is τ = L R H 70 Ω = s 10 9 s L R ~ 1 ns P3.68 We calculate the angular frequency of the circuit from Equation 3.3: ω d = = = 0 1 LC R L 1/ 1 ( H) ( F) 16.0 Ω ( H) The fact that the angular frequency at which the circuit oscillates is zero tells you that the circuit is critically damped. There will be no decaying oscillations. The critical resistance is given by R c = 4L C = H F = 16.0 Ω which is just the resistance that you are using for your experiment. P3.69 The emf across the inductor is given by ε = L di Δi Δi = L = 50 Δt Δt where the rate of change of current Δi is in amperes per second (A/s), Δt and the induced emf ε is in millivolts (mv). Between t = 0 and t = 1 ms: Between t = 1 ms and t = ms: Between t = ms and t = 3 ms: Between t = 3 ms and t = 5 ms: Between t = 5 ms and t = 6 ms: Δi Δt = Δi Δt = 0 ε = 0 Δi Δt = 1 Δi Δt = 3 ε = 100 mv ε = 50 mv ε = +75 mv Δi Δt = 0 ε = 0 1/

37 504 Inductance The graph of ε is shown in ANS. FIG. P3.69. P3.70 (a) i 1 = i + i ANS. FIG. P3.69 For the left-hand loop, (c) ε i 1 R 1 i R = 0 For the outside loop, ε i 1 R 1 L di = 0 ANS. FIG. P3.70 (d) Substitute the equation for i 1 from part (a) into the equation in part : ε ( i + i)r 1 i R = 0 i = ε ir 1 R 1 + R Substitute the equation for i 1 from part (a) into the equation in part (c): ε ( i + i)r 1 L di Equate the two expressions for i : ε ir 1 R 1 + R = ε L di i R 1 di ε L = 0 i = i R 1

38 Chapter Expanding and solving, ε L di = ε ir 1 + i R 1 + R R 1 ε = ir 1 + i R 1 + R R 1 + R R 1 = ε + ir R 1 + R R 1 L di = ε ε + ir R 1 + R R 1 ε ( = R + R 1 ) ( ε + ir )R ε 1 ( = R ) ( ir )R 1 R 1 + R R 1 + R And finally, ε Calling written R R i 1 R L di R 1 + R R 1 + R = 0 ε = ε R ε i R L di = 0 and R = R R 1, the equation for i can be R 1 + R R 1 + R (e) This is of the same form as the Equation 3.6 in the text for a simple RL circuit, so its solution is of the same form as Equation 3.7: where Thus, i = ε R 1 e ε R t L = ε R 1 ε R = R R 1 + R R 1 R R 1 + R i = ε ( 1 e R t L ) where R = R 1R R 1 R 1 + R P3.71 See ANS. FIG The magnetic field inside the toroid is given by B = µ 0Ni π r ANS. FIG. P3.71

39 506 Inductance The magnetic flux through a cross-sectional area A = h(b a) is given by b Φ B = BdA = µ Ni 0 π r hdr Thus, the inductance is L = NΦ B i a = µ 0 N h π ln b a = µ 0 Nih π Substituting numerical values, we obtain b a dr r = µ 0 Nih π ln b a L = µ ( m) 1.0 cm ln π 10.0 cm = 91. µh P3.7 See ANS. FIG. P3.71. B = µ 0 Ni π r inside the toroid. Calculate the flux through a cross-sectional area A = h(b a): b Φ B = BdA = µ Ni 0 π r hdr Thus, the inductance is a = µ 0 Nih π b a dr r = µ 0 Nih π ln b a L = NΦ B i = µ 0 N h π ln b a P3.73 (a) U B = 1 Li = 1 ( 50.0 H) ( A) = J 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. Then one wire creates a field of B = µ 0i π r. This causes a force on the next wire of F = i Bsinθ, giving a force per unit length Evaluating, F = 4π 10 7 T m/a F = i µ i 0 π r sin 90 = µ 0 i π r. ( A ) π 0.50 m = 000 N/m

40 Chapter P3.74 For an RL circuit, i = I i e ( R L)t and i I i = = e R L t 1 R L t so R L t = 10 9 R max = L 10 9 t = Ω ( 10 9 ) = H.50 yr 1 yr s (If the ring were of purest copper, of diameter 1 cm, and cross-sectional area 1 mm, its resistance would be at least 10 6 Ω.) P3.75 Find the current in the cylinder. From Ampère s law, P = iδv i = P ΔV = W V = A B( π r) = µ 0 i enclosed or B = µ i 0 enclosed π r ANS. FIG. P3.75 (a) At r = a = m, i enclosed = A and ( 4π 10 7 T m A) A B = π m At r = b = m, i enclosed = i = A and = T = 50.0 mt ( 4π 10 7 T m A) A B = = T = 0.0 mt π m

41 508 Inductance (c) The energy density is u B = B µ 0 : U B = udv = b a [ B( r) ] π r dr µ 0 = µ 0i 4π ( 4π 10 7 T m A) A U B = 4π = J =.9 MJ b a dr r = µ 0i 4π ln b a ( m) 5.00 cm ln.00 cm (d) 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, is 0.0 mt. Consider a small rectangular section of the outer cylinder of length and wih w. w It carries a current of ( A) π m and experiences an outward force ( F = i Bsinθ = A)w π m The pressure on it is sin T ( T) w w P = F A = F w = A π m = 318 Pa P3.76 (a) The magnetic field inside the solenoid is given by B = µ 0Ni : ( 4π 10 7 T m A) ( 1 400) (.00 A) B = 1.0 m = T =.93 mt ANS. FIG. P3.76

42 Chapter (c) The energy density of the magnetic field is ( u B = B T) = = 3.4 J µ 0 4π 10 7 m3 T m A = 3.4 N m = 3.4 Pa 1 N m The supercurrents must be clockwise to produce a downward magnetic field to cancel the upward field of the current in the windings. 1 J (d) (e) The field of the windings is upward and radially outward around the top of the solenoid. It exerts a force radially inward and upward on each bit of the clockwise supercurrent. The total force on the supercurrents in the bar 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. F = PA = 3.4 Pa π m = N = 1.30 mn Note that we have not proved that energy density is pressure. In fact, it is not in some cases. Chapter 1 proved that the pressure is two-thirds of the translational energy density in an ideal gas. P3.77 From Ampère s law, the magnetic field at distance r R is found as: i B( π r) = µ 0 j( π r ) = µ 0 π R ( π r ),, or B = µ ir 0 π R The magnetic energy per unit length within the wire is then R = B ( π rdr) = µ 0i R r 3 dr µ 0 4π R = µ 0i R 4 4 4π R 4 4 = µ 0i 16π U B 0 This is independent of the radius of the wire. 0 Challenge Problems P3.78 (a) It has a magnetic field, and it stores energy, so L = U B i is non-zero. Every field line goes through the rectangle between the conductors.

43 510 Inductance (c) When the wires carry current i, magnetic flux passes through the rectangle bordered by the wires (surface to surface of the wires): L = Φ i = 1 i w a y=a BdA where y is measured from the center of the lower wire, da is a rectangular area element of length x and wih dy, and B is the magnitude of the net magnetic field generated by the upper and lower wires that passes through da. The inductance is L = 1 i w a a µ 0 i π y + µ 0 i π w y x dy We can simplify this calculation by noting that by the symmetry of the arrangement, each conductor contributes equally to the field that passes through the area between them. Thus, the total inductance of both wires is twice the inductance of one wire. The inductance due to the lower wire is L lower = 1 i w a a µ 0 i π y xdy = µ 0x π ln w a a = µ w a 0x π ln y a [ ] = µ 0x π ln ( w a ) ln a The inductance due to both wires is twice this: L = µ 0x π ln w a a. P3.79 The total magnetic energy is the volume integral of the energy density, u B = B µ 0 Because B changes with position, u B is not constant. For B = B 0 u B = B 0 µ 0 R r 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 dr, so the energy stored in it is du B = u B ( 4π r dr) = π B 0 µ 0 R 4 dr r R r,

44 Chapter We integrate this expression for r = R to r = to obtain the total magnetic energy outside the sphere. This gives U B = π B 0R 3 µ 0 Substituting numerical values, U B = π B 0R 3 = π T µ 0 4π 10 7 = J ( m) 3 ( T m A) P3.80 (a) 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 going clockwise around this loop gives: +ε ( ) 10 3 Ω ε = 7.0 V ( A) = 0 (c) Starting at point a, the potential rises across the inductor then falls across resistors R and R 1. The positive answer in part (a) means that point b is the higher potential. The currents in R 1 and R are shown in ANS. FIG. P3.80(c).below. After t = 0, the current in R 1 decreases from an initial value of 9.00 ma according to i = I i e Rt/L. Taking the original current direction as positive in each resistor, the current decreases from ma (to the right) to zero in R 1. In R the current jumps from ma (downward) to 9.00 ma (upward) and then decreases in magnitude to zero. The time constant of each decay is 0.4 H/8 000 Ω = 50 µs. Thus we draw each current dropping to 1/e = 36.8% of its original value = 3.3 µa at the 50 µs instant. ANS. FIG. P3.80(c)

45 51 Inductance (d) After the switch is opened, the current around the outer loop decays as i = I i e Rt L with I i = 9.00 ma, R = 8.00 kω, and L = H. Thus, when the current has reached a value i =.00 ma, the elapsed time is: t = L R ln I i i = = s = 75. µs H 9.00 ma Ω ln.00 ma P3.81 When the switch is closed, as shown in ANS. FIG. P3.81(a), the current in the inductor is i: i 10.0 = 0 i = 0.67 A When the switch is opened, as shown in ANS. FIG. P3.81, the initial current in the inductor remains at 0.67 A. Across resistor R and the armature, ir = ΔV and ( 0.67 A)R 80.0 V R 300 Ω *P3.8 (a) After a long time, ANS. FIG. P3.81 i = ε R = 1.0 V 1.0 Ω = 1.00 A With the switch thrown to position b, the emf is no longer part of the circuit. The initial current is 1.00 A: ΔV 1 = ( 1.00 A) ( 1.00 Ω) = 1.0 V ANS. FIG. P3.8 ΔV 1 00 = ( 1.00 A) ( 1 00 Ω) = 1.0 kv ΔV L = ΔV R = 1 00 V V = 1.1 kv

46 Chapter (c) With the switch thrown to position b, ε = 0, so ΔV L = ΔV R = ir eff = i( 1 00 Ω Ω) = i( 1.1 kω) Then, when the voltage across the inductor has reached 1.0 V, i = ΔV L R eff = 1.0 V 1.1 kω = A The current in the inductor decays as i = I i e Rt L. Solving for the time t, t = L R ln I i i = P3.83 With i = i 1 + i, the voltage across the pair is: and di ΔV = L 1 1 M di = L di eq di ΔV = L M di 1 = L di eq.00 H 1.00 A 1.1 kω ln A = s [1] [] So, from [1], we have di 1 = ΔV + M di L 1 L 1 ANS. FIG. P3.83 which, when substituted into [], gives di L + M ( ΔV) + M di L 1 L 1 = ΔV ( L 1 L + M ) di = ΔV ( L 1 M) [3] From [], di = ΔV L + M L di 1,

47 514 Inductance which, when substituted into [1], gives di L M ΔV + M di 1 L L = ΔV ( L 1 L + M ) di 1 = ΔV ( L M) [4] Adding [3] to [4], we have ( L 1 L + M ) di = ΔV ( L + L M) 1 So, L eq = ΔV di = L 1 L M L 1 + L M

48 Chapter ANSWERS TO EVEN-NUMBERED PROBLEMS P μh P3.4 (a) 1.97 mh; 38.0 ma/s P3.6 (a) 188 µt; T m ; (c) mh; (d) B and Φ B are proportional to current: L is independent of current. P3.8 (a) 5.90 mh; 3.6 mv P mm P3.1 See P3.1 for full explanation. P3.14 See ANS. FIG. P3.14. P3.16 (a) s; s P3.18 See ANS. FIGs. P3.18(a),, and (c). P mh P3. See P3. for full explanation. P3.4 (a).00 ms; A; (c) 1.50 A; (d) 3. ms P3.6 (a) ε L ( 1 e 5Rt ); 5R ε 10R 6 e 5Rt L P3.8 For t 0, the current in the inductor is zero; for 0 t 00 µs, t s t s i L = ( 10.0 A) ( 1 e ); for t 00 µs,, ( 63.9 A)e P3.30 (a) See P3.30(a) for full explanation; See P3.30 for full explanation; (c) See P3.30(c) for full explanation; (d) Yes. See P3.30(d) for full explanation. P mj P3.34 (a) 0.0 W; 0.0 W; (c) 0; (d) 0.0 J P3.36 L R P3.38 (a) 66.0 W; 45.0 W; (c) 1.0 W; (d) The power supplied by the battery is equal to the sum of the power delivered to the internal resistance of the coil and the power stored in the magnetic field; (e) Yes; (f) Just after t = 0, the current is very small, so the power delivered to the internal resistance of the coil (ir ) is nearly zero, but the rate of the change of the current is large, so the power delivered to the magnetic field (Ldi/) is large, and nearly all the battery power is being stored in the magnetic field. Long after the connection is made, the current is not changing, so no power is being stored in the

49 516 Inductance P mh P mh magnetic field, and all the battery power is being delivered to the internal resistance of the coil. P3.44 (a) M 1 = µ 0 πr N 1 N / ; M 1 = µ 0 πr N 1 N / ; (c) They are the same. P3.46 (a) See P3.46(a) for full explanation; 3.95 nh P pf P H P3.5 If the energy is split equally between the capacitor and inductor at some instant, the energy would be half this value, or 00 µj. Therefore, there would be no time when each component stores 50 µj. P3.54 (a) 6.03 J; 0.59 J; (c) 6.56 J P3.56 See P3.56 for full explanation. P3.58 (a) 4.47 krad/s; 4.36 krad/s; (c).53% P3.60 (a) See P3.60(a) for full explanation; µ J 0 s ; (c) B = µ 0J s, zero; (d) µ J 0 s ; (e) The energy density found in part (d) agrees with the magnetic pressure found in part. P3.6 (a) LK; Kt C ; (c) LC P3.64 (a) short circuit; 500 ma; (c) 15 mj; (d) The energy becomes 15 mj of additional internal energy in the 8-Ω resistor and the 4-Ω resistor in the middle branch; (e) See ANS FIG P3.64(e). P3.66 (a) Just after the circuit is connected, the potential difference across the register is 0, and the emf across the coil is 4.0 V; After several seconds, the potential difference across the resistor is 4.0 V and that across the coil is 0; (c) The two voltages are equal to each other, both being 1.0 V, just once, at ms after the circuit is connected; (d) As the current decays, the potential difference across the resistor is always equal to the emf across the coil. P3.68 See P3.68 for full explanation.