Physics II Chapter H Solutions

Chapter H - Problems Problem H.1 Refer to Problem G.9 where a b µ b square flat coil with N turns sits outside of a long solenoid with n turns per length and a circular cross-section radius a. What is the mutual inductance between the solenoid and the flat coil? b Solenoid a # of turns n= length b Flat Coil N = # of turns Solution to H.1 In Problem G.9 we had the induced EMF in the flat coil as „I 2 = -N m0 n p a2 1 „t where I am now referring to the EMF in the coil as 2 . Using the definition of mutual inductance 2 = -M „I1 „t ï M = N m0 n p a2 . 2 Chapter H - Problems Problem H.2 Refer to Problem G.7 where we discussed a long wire next to a rectangular conduction loop. What is the mutual inductance between the long wire and the loop? b a  I Solution to H.2 We saw in Problem G.7 that =- m0  „F b „I =ln µ . a „t „t 2p Take the current of the long wire to be I1 and the induced EMF in the loop to be 2 then use the definition of mutual inductance: 2 = -M „I1 „t and 2 = - m0  2p ln m0  b „I1 b µ ï M= ln . a a „t 2p Problem H.3 What is the average induced EMF when the current through a coil with a 50 mH inductance varies from 5 A to 3 A in 0.04 s? Solution to H.3 †§ = L £ DI 2 ß = 0.050 µ = 2.5 V Dt 0.04 Problem H.4 The current through a 15 mH inductor varies as IHtL = Imax sin w t where Imax = 6 A and w = 200 s-1 . What is the EMF across the inductor as a function of time? Chapter H - Problems 3 Solution to H.4 „I „ = -L Imax sin w t= -L w Imax cos w t „t „t = -0.015 µ 200 µ 6 cos w t= -H18 VL cosAI200 s-1 M tE  = -L Problem H.5 A solenoid has 250 turns, a length of 40 cm and a diameter of 3 cm. The current through the solenoid varies as IHtL = 15 t3 - 40 t + 30 (in SI units). (a) What is the inductance of the solenoid? (b) What is the voltage drop across the solenoid at t = 2 s? (c) What is the energy stored in the solenoid at t = 2 s? Solution to H.5 (a) This is a long solenoid. The inductance of a long solenoid is: L = m0 N2 2502 A = 4 p µ 10-7 Ip 0.0152 M = 1.3879 µ 10-4 = 0.139 mH  0.40 (b) The voltage drop is V = L „ I ê „ t. We need to differentiate the current. „IHtL „I = 45 t2 - 40 ï H2L = 45 µ 22 - 40 = 140 „t „t „I ï V=L = L µ 14 = 0.0194 V „t (c) The energy stored in an inductor is U = 1 L I2. 2 IH2L = 15 µ 23 - 40 µ 2 + 30 = 70 ï U = 1 1 L I 2 = L µ 702 = 0.340 J 2 2 Problem H.6 A typical electric field at the earth's surface is 100 V ê m and a typical magnetic field value is 5 µ 10-5 T. What are the electric and magnetic energy density, ue and um . Solution to H.6 ue = 1 1 J e0 E 2 = 8.85 µ 10-12 µ 1002 = 4.43 µ 10-8 2 2 m3 2 um = I0.5µ10-4 M B2 J = = 9.95 µ 10-4 2 m0 m3 2µ4 pµ10-7 4 Chapter H - Problems Problem H.7 A 2 mF capacitor is given an initial charge of 6 mC and is then connected across a 15 kW resistor to discharge it. (a) What is the charge on the capacitor 40 ms after the resistor is connected? (b) What is the current through the resistor 40 ms after it is connected? (c) What is the maximum magnitude current through the resistor? Solution to H.7 The time constant is t = R C = 15 kW µ 2 mF = 30 ms (a) For a discharging capacitor: Q HtL = Q0 e-têt = H6 mCL e-40ê30 = 1.58 mC (b) To get the current differentiate the charge. IHtL = „Q „t =- Q0 t e-têt = - 6 mC 30 ms e-40ê30 = -5.27 µ 10-5 A (c) The maximum current is at t = 0. Imax = Q0 t = 6 mC 30 ms = 2 µ 10-4 A Problem H.8 A 2 MW resistor and a 3 mC capacitor are connected across a 12 V battery. What is the timeconstant of the circuit. Express the charge and currents as functions of time. Solution to H.8 R = 2 µ 106 W , C = 3 µ 10-6 F and  = 12 V t=RC= 6s QHtL =  C I1 - e-têt M = 36 mC I1 - e-têH6 sL M IHtL = „  C -têt  -têt QHtL = e = e = 6 mA e-têH6 sL „t RC R Problem H.9 If a charging capacitor reaches 60% of its maximum charge after 0.90 s then what is the time constant of the circuit? Solution to H.9 Q HtL = Qmax I1 - e-têt M ï 0.6 = Q Qmax = I1 - e-têt M ï 0.4 = e-têt ï -t ln 0.4 = t = 0.9 s ï t = 0.982 s Chapter H - Problems Q HtL = Qmax I1 - e-têt M ï 0.6 = Q Qmax 5 = I1 - e-têt M ï 0.4 = e-têt ï -t ln 0.4 = t = 0.9 s ï t = 0.982 s Problem H.10 A 2.5 H inductor is connected in a loop circuit with a DC voltage source and a resistor. If the current increases to 90% of its final value in 3 s then what is the resistance in the circuit? Solution to H.10 For such a circuit the current is I HtL = Imax I1 - e-têt M 0.90 = I = 1 - ‰-3êt Imax t= ï ‰-3êt = 0.10 ï t = - L 2.5 = R R 3 = 1.3029 lnH0.1L ï R = 1.92 W Problem H.11 An inductance L and a capacitance C both have the same time constant when combined with the same resistance. (a) What is the resistance and what is the value of the time constant? (b) If L = 12 mH and C = 30 mF then what is R and what is the value of the time constant? Solution to H.11 (a) The time constants are t = R C and t = L ê C. L =t=RC ï R= R t=RC=C L = C L C LC (b) Using the values L = 12 mH and C = 30 mF gives R= L = C 12µ10-3 30µ10-6 = 20 W and t = L C = 0.6 ms. Problem H.12 A 1 mF capacitor is fully charged across a 40 V battery. It is then disconnected from the battery and connected actoss a 10 mH inductor inductor. What is the maximum current through the inductor during the LC oscillations? Solution to H.12 The initial charge on the capacitor is the maximum charge of the oscillation. It is 6 Chapter H - Problems The initial charge on the capacitor is the maximum charge of the oscillation. It is Qmax = C V = 10-6 µ 40 = 4 µ 10-5 C. The angular frequency of the oscillation is w= 1 = LC 1 = 104 10-6 µ0.010 rad . s For an LC circuit the charge varies as: QHtL = Qmax cosHw t + fL. The current is the derivative of the charge. IHtL = „ Q ê „ t = -w Qmax sinHw t + fL. Since sine varies between +1 and -1 we get Imax = w Qmax = 0.4 A Problem H.13 A 840 nF capacitor with an initial 105 mC is connected across a 3.30 mH inductor at t = 0. (a) What is the energy in the capacitor at t = 2 ms? (b) What is the energy in the inductor at t = 2 ms? (c) What is the total energy? Solution to H.13 For the charge as a function of time we want our sinusoidal function to be cosine since the charge has its maximum value at t = 0. Q HtL = Qmax cos w t ï IHtL = where w = 1 1 = LC 3.3µ10-3 µ840µ10-9 „ Q HtL = -w Qmax sin w t. „t = 18 993 and Qmax = Q0 = 105 µ 10-6 At t = 2 ms , QH2 msL = Qmax cosIw 2 µ 10-3 M = 1.00683 µ 10-4 IH2 msL = -w Qmax sinIw 2 µ 10-3 M = -0.56597 Q2 HaL UC = = 2C 2 I1.00683µ10-4 M 1 = 6.03 m J 2µ840µ10-9 1 HbL UL = L I 2 = 3.3 µ 10-3 µ 0.565972 = 0.529 m J 2 2 Q2max 2 I105µ10-6 M HcL Utotal = = = 6.56 m J 2C 2µ840µ10-9 Note that Utotal = UC +UL .