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Chapter32solns

Chapter32solns - Chapter 32 Solutions 1 Two coils have a...

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Unformatted text preview: Chapter 32 Solutions 1. Two coils have a mutual inductance of 2.0 H. If current in the first coil is changing at the rate of 60 A/s, what is the emf in the second coil? Solution Iii-om Equation 32-2, 82 = —M(dI;/dt) = —(2 H)x (60 A/s) = -—120 V. (The minus sign, Lenz‘s law. signifies that an induced emf opposes the process which creates it.) 3. The current in one coil is given by I = I? sin 21rft, where I], = 75 mA. f = 60 Hz, and t = time. Find the peak emf in a second coil if the mutual inductance between the coils is 440 mil. Solution Suppose 11 = 1? sin 27rft in Equation 32-2. Then 82 = —M dill/d: = —2'irfll{il'p cos 2rrft. and the peak emf (when cos2nft = :l:1) is 21rfMIp = (2nx 60 Hz)(440 mH)(75 mA) = 12.4 V. 4. Two coils have a mutual inductance of 580 mH. One coil is supplied with a current given by I = 3t2 — 2t + 4, where I is in amperes and t in seconds. What is the induced emf in the other coil at time t. = 2.5 5? Solution Since dfl/dt = 6t — 2 (in A/s). Equation 32-2 gives, for t : 2.5 s. 82 = —(580 mH}(6x2.5 - 2)(A/s) = —7.54 V (see comment in solution to Problem 1). 9. A rectangular loop of length f and width at is located a distance a from a long, straight, wire, as shown in Fig. 32-20. What is the mutual inductance of this arrangement? t 0 —I—_ FIGURE 32-20 Problem 9. Solution When current I1 flows to the left in the wire, the flux through the loop is (932 = (null/2n) f2+w€ drfr : {pullf/21r)ln(1 + w/o) (see Example 31-2). Then Equation 32-1 gives M = (bag/fl = (nag/2n) x 111(1 + w/a). (In calculating the flux‘ the normal to the loop area was taken into the page. so the positive sense of circulation around the loop is CW. This determines the direction of the induced emf 52 in Equation 32-2.) 11. What is the self-inductance of a solenoid 50 cm long and 4.0 cm in diameter that contains 1,000 turns of wire? Solution Equation 324 gives L = ,ugNgA/f : (4nx10'7 H/m)x (103)2rr(2 cm)2/(50 cm) 2 3.16 mil. (The long thin solenoid approximation is valid here.) 13. A 2.0—A current is flowing in a 20-H inductor. A switch is opened. interrupting the current in 1.0 ms. What emf is induced in the inductor? Solution Assume that the current changes uniformly from 2 A to zero in 1 [11.5 (or consider average values). Then dI/dt = —2 Aims. and Equation 32-5 gives 8 = —(20 H)(—20 A/ms) = 40 kV. (The emf opposes the decreasing current.) 22. A coaxial cable consists of an inner conductor of radius a and outer conductor of radius b, as shown in Fig. 32-21. Current flows along one conductor and back along the other. Show that the inductance per unit length of the cable is $3 ln(b/u). Solution 'The magnetic field strength between the conductors (a 5 r 5 b) is like that from a long straight wire. _B = pal/2n.“ {see Section 30-4). If We divide a rectangular area of length E betWeen the conductors into strips of width air. then the flux linking this length of cable is $5 = f: Bf dr : (puff/2n) f: dr/r = (puff/2r) ln(b‘i’u). Therefore, Equation 32-3 gives U5 = (#of'31rllnlb/al- FIGURE 3221 Problem 22 Solution. Chapter 32 Solutions continued 25. The current in a series RL circuit rises to 20% of its final value in 3.] us. If L = 1.8 mH. what is the resistance R? Solution The buildup of current in an RL circuit with a battery is given by Equation 32-8. In) = 100(1 — arm/L), where 1m = fig/R is the final current. Solving for R, one finds R = ~(L/t)ln(1~ I/Im) = —(1.8 mH-:— 3.1 ps)ln(1— 20%) = 130 Q. 27. A 10-H inductor is wound of wire with resistance 2.0 Q. If the inductor is connected across an ideal 12-V battery, how long will it take the current to reach 95% of its final value? Solution Reference to the solution to Problem 25 shonrs that t: —(L/R) ln{1 — 1/130) = —(10 H/2 Q)1n(1 — 0.95} = 15.0 s. (The percentage I/Ix is independent of £0.) 39. What is the current in a 10-an inductor when the stored energy is 50 ,uJ? Solution From Equation 32-10. I = ‘/2L'/ = «2(50 inn/10 mH = 0.1 A. 49. The Alcator fusion experiment at MIT has a 50-T magnetic field. What is the magnetic energy density in Alcator? Solution From Equation 32-11. us = (50 T)2/(8nx 1n-7 N/Az) = 995 MJ/ms. (This is about 2.8% of the energy density content of gasoline; see Appendix C.) ...
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