CHAPTER 31. Answer to Checkpoint Questions

Transcription

1 CHAPTE 31 INDUCTION AND INDUCTANCE 89 CHAPTE 31 Answer to Checkpont Questons 1. b, then d nd e te, nd then nd c te (zero). () nd (b) te, then (c) (zero) 3. c nd d te, then nd b te 4. b, out; c, out; d, nto; e, nto 5. (d) nd (e) 6. (), 3, 1 (zero); (b), 3, 1 7. nd b te, then c Answer to Questons 1. () ll te (zero); (b) ll te (nonzero); (c) 3, then te of 1 nd (zero). () ll te (zero; (b), then te of 1 nd 3 (zero) 3. out 4. (), 6; (b) 4; (c) 1, 3, 5, 7 5. () nto; (b) counterclockwse; (c) lrger 6. () clockwse; (b) ncresng nd then decresng 7. () leftwrd; (b) rghtwrd 8. clockwse 9. c,, b 10. d nd c te, then b, 11. () 1, 3, ; (b) 1 nd 3 te, then 1. c, b, 13. (), then te of (b) nd (c) 14. (c), (), (b) 15. () more; (b) sme; (c) sme; (d) sme (zero)

2 830 CHAPTE 31 INDUCTION AND INDUCTANCE 16. () ll te (zero); (b) 1 nd te, then 3; (c) ll te (zero) 17., b 4, c 1, d () ll te (ndependent of the current); (b) 1, then nd 3 te Solutons to Exercses & Problems 1E B Z BdA BA cos 57 (4: 10 6 T)(:5 m )(cos 57 ) 5: Wb : E E d B d(ba) A 0 n 0! cos!t : A db A d ( 0n) A 0 n d ( 0 sn!t) 3E The mgnetc eld s norml to the plne of the loop nd s unform over the loop. Thus t ny nstnt the mgnetc ux through the loop s gven by B AB r B, where A ( r ) s the re of the loop. Accordng to Frdy's lw the mgntude of the emf n the loop s E d B r db (0:055 m) (0:16 T/s) 1: V : 4E E d B A db r d (B 0e t ) r B 0 e t : 5E () jej d B d (6:0t + 7:0t) 1t + 7:0 1(:0) + 7:0 31 mv :

3 CHAPTE 31 INDUCTION AND INDUCTANCE 831 (b) Use Lenz's lw to determne the drecton of the current ow. It s from rght to left. 6E Use E d B r db. () For 0 < t < :0 s: E r db (b) :0 s < t < 4:0 s : E / db 0: (c) 4:0 s < t < 6:0 s : E r db 0:5 T (0:1 m) 1:1 10 V : :0 s (0:1 0:5 T m) 6:0 s 4:0 s 1:1 10 V : 7E () The mgntude of the verge nduced emf s (b) E v d B B t BA t (:0 T)(0:0 m) 0:0 s v E v 0:40 V A : 0:40 V : 8E () L A (1: m) (0:10 m) (: ) 4 (b) Use jej jd B j (r )jdbj: Thus db r (10 A)(1: ) 1:4 T/s : (0:10 m) 4 1: : 9P () The emf s functon of tme s gven by E N d B N d(ba) NA d d ( 0n) N 4 d N 4 The plot s shown n the next pge. 0 n d 0 n d (3:0t + 1:0t ) 1 4 d N 0 n(3:0 + :0t) :

4 83 CHAPTE 31 INDUCTION AND INDUCTANCE ε (mv) t (s) (b) j t:0 s Ej t:0 s d N 0 n(3:0 + :0t) 4 (:1 10 m) (130)(1: Tm/A)(: turns/m)[3:0 + (:0)(:0)]( A/s) 4(0:15) 5:8 10 A : 10P The mgntude of the mgnetc eld nsde the solenod s B 0 n s, where n s the number of turns per unt length nd s s the current. The eld s prllel to the solenod xs, so the ux through cross secton of the solenod s B A s B 0 r s n s, where A s ( r s) s the cross-sectonl re of the solenod. Snce the mgnetc eld s zero outsde the solenod ths s lso the ux through the col. The emf n the col hs mgntude E Nd 0 r s Nn d s nd the current n the col s c E 0r s Nn d s ; where N s the number of turns n the col nd s the resstnce of the col. Accordng to Smple Problem 31{1 the current chnges lnerly by 3:0 A n 50 ms, so d s (3:0 A)( s) 60 A/s. Thus c ( Tm/A)(0:016 m) (10)(0 10 m 1 ) 5:3 3:0 10 A :

5 CHAPTE 31 INDUCTION AND INDUCTANCE P E d B d(ba) A d ( 0n) 0 nr d (1: T m/a)(1:00 turns/m)()( m) 0:50 A 1:0 A s 1: 10 3 V : Note tht snce B only ppers nsde the solenod the re A should be the cross-sectonl re of the solenod, not the (lrger) loops. 1P Consder the cross-secton of the torod. The mgnetc ux through the shded re s shown s d B B(r)dA B(r)hdr. Thus from Eq B Z b 0Nh B(r)hdr ln b : Z b 0 0 Nh dr r O b r r+dr cross-sectonl vew h r 13P 0 0 N B BA r A ( Tm/A)(0:800 A)(500)(5:00 10 m) (0:150 m + 0:0500 m) 1: Wb : 14P E d B d(ba) B da B d(r ) rb dr (0:1 m)(0:800 T)( 0:750 m/s) 0:45 V : 15P The mgnetc ux B through the loop s gven by B B(r )(cos 45 ) r B p. Thus E d B d r B p r B p t (3:7 10 m) T p 4: s 5:1 10 V :

6 834 CHAPTE 31 INDUCTION AND INDUCTANCE The drecton of the nduced current s clockwse when vewed long the drecton of B. 16P Snce B does not chnge, E d B 0. 17P () Use B ' 0, where s the rdus of the lrge loop. Here (t) 0 + kt, where 0 00 A nd k ( 00 A 00 A)1:00 s 400 A/s. Thus Bj t0 0 0 ( Tm/A)(00 A) (1:00 m) 1: T ; nd Bj t0:500 s 0( 0 + kt) ( Tm/A)[00 A (400 A/s)(0:500 s)] (1:00 m) 0 ; Bj t1:00 s 0( 0 + kt) 1: T : ( Tm/A)[00 A (400 A/s)(1:00 s)] (1:00 m) (b) Let the re of the smll loop be. Then B B, nd db B t 1: (: m T ) E(t) d B d(b) 1: T 1:00 s 5: V : 18P () In the regon of the smller loop the mgnetc eld produced by the lrger loop my be tken to be unform nd equl to ts vlue t the center of the smller loop, on the xs. Eq. 30{9, wth z x nd much greter thn, gves B 0 x 3 for the mgntude. The eld s upwrd n the dgrm. The mgnetc ux through the smller loop s the product of ths eld nd the re (r ) of the smller loop: B 0r x 3 :

7 CHAPTE 31 INDUCTION AND INDUCTANCE 835 (b) The emf s gven by Frdy's lw: E d B 0 r d 1 x 3 0 r 3 dx 3 0r v : x 4 x 4 (c) The eld of the lrger loop s upwrd nd decreses wth dstnce wy from the loop. As the smller loop moves wy the ux through t decreses. The nduced current wll be drected so s to produce mgnetc eld tht s upwrd through the smller loop, n the sme drecton s the eld of the lrger loop. It wll be counterclockwse s vewed from bove, n the sme drecton s the current n the lrger loop. 19P () The emf nduced round the loop s gven by Frdy's lw: E d B nd the current n the loop s gven by E (1)(d B ). The chrge tht psses through the resstor from tme zero to tme t s gven by the ntegrl q Z t 0 1 Z t 0 d B 1 Z B (t) B (0) d B 1 [ B(0) B (t)] : All tht mtters s the chnge n the ux, not how t ws chnged. (b) If B (t) B (0) then q 0. Ths does not men tht the current ws zero for ny extended tme durng the ntervl. If B ncreses nd then decreses bck to ts orgnl vlue there s current n the resstor whle B s chngng. It s n one drecton t rst, then n the opposte drecton. When equl chrge hs pssed through the resstor n opposte drectons the net chrge s zero. 0P From the result of the lst problem q 1 [ B(0) B (t)] A [B(0) B(t)] 1: m [1:60 T ( 1:60 T)] :95 10 C : 13:0 1P q(t) N B N [BA cos 70 ( BA cos 70 )] (1000)(0: T)(0:100 m) (cos 30 ) 85: W BA cos 30 1: C : Note tht the xs of the col s t 30, not 70, from the mgnetc eld of the Erth.

8 836 CHAPTE 31 INDUCTION AND INDUCTANCE P () Let L be the length of sde of the squre crcut. Then the mgnetc ux through the crcut s B L B nd the nduced emf hs mgntude E d B L db : Now B 0:04 0:870t nd db 0:870 T/s. Thus E (:00 m) (0:870 T/s) 1:74 V : The mgnetc eld s out of the pge nd decresng so the nduced emf s counterclockwse round the crcut, n the sme drecton s the emf of the bttery. The totl emf s E + E 0:0 V + 1:74 V 1:7 V. (b) The current s n the sense of the totl emf, counterclockwse. 3P () The mgnetc ux B through the loop s Z BdA 1 4 r B : Thus jej d B d 1 4 r B r 4 db 1 4 (0:10 m) (3: T/s) : V : (b) From c to b (followng Lenz's lw). 4P () nd (b) Let A r. The emf s functon of tme s gven by E d B d (BA) BA d cos(b; A) BA d cos(ft + 0) 9f)BA sn(ft + 0 ) : Thus the frequency of E s f nd the mpltude s E m fba (f)b(r ) r Bf :

9 CHAPTE 31 INDUCTION AND INDUCTANCE 837 5P () The re of the col s A b. Suppose tht t some nstnt of tme the norml to the loop mkes the ngle wth the mgnetc eld. The mgnetc ux through the loop s then B NbB cos nd the emf nduced round the col s E d B d d (NbB cos ) (NbB sn ) : In terms of the frequency of rotton f nd the tme t, s gven by ft nd d f. The emf s therefore E fnbb sn(ft) : Ths cn be wrtten E E 0 sn(ft), where E 0 fnbb. (b) You wnt fnbb 150 V. Ths mens Nb E 0 fb 150 V (60:0 rev/s)(0:500 T) 0:796 m : Any loop for whch Nb 0:796 m wll do the job. An exmple s N 100 turns, b 8:9 cm. 6P Use the result obtned n 4P, prt(b): E m (fba)n ( s)(3:50 T)(0:500 m)(0:300 m)(100) 5: V : 7P () It s cler tht the mgnetc ux through res 1 nd s shown cncel out. Thus B B3 b Z 0 r b Z B(r)(bdr) bdr 0b ln b ; O 3 b- b- -b nd r b

10 838 CHAPTE 31 INDUCTION AND INDUCTANCE E d B 0b ln b d 0 b ln b 0b ln b d (4:50t 10:0t) 4:50 0bt (4:50)(1: T m/a)(0:160 m)(3:00 s) 5: V : ln d ln b 1:0 cm 16:0 cm 1:0 cm (b) From Lenz's lw, the nduced current n the loop s counterclockwse. 8P Use Frdy's lw to nd n expresson for the emf nduced by the chngng mgnetc eld. Frst nd n expresson for the mgnetc ux through the loop. Snce the eld depends on y but not on x, dvde the re nto strps of length L nd wh dy, prllel to the x xs. Here L s the length of one sde of the squre. At tme t the ux through strp wth coordnte y s d B BL dy 4:0Lt y dy nd the totl ux through the squre s B Z L 0 4:0Lt y dy :0L 3 t : Accordng to Frdy's lw the mgntude of the emf round the squre s E d B d :0L 3 t 4:0L 3 t : At t :5 s ths s 4:0(0:00 m) 3 (:5 s) 8: V. The externlly-produced mgnetc eld s out of the pge nd s ncresng wth tme. The nduced current produces eld tht s nto the pge, so t must be clockwse. The nduced emf s lso clockwse. 9P () Anlogous to 7P, prt (), we hve B Z loop Z r+ b r b B(r)dA Z r+ b r b 0 r dr 0 B(r)dr r + b ln r b :

11 CHAPTE 31 INDUCTION AND INDUCTANCE 839 (b) Now Thus E d 1 r + b dr r E bv (4r b ) : 0bv (4r b ) : r + b ln r b 30P () Suppose ech wre hs rdus nd the dstnce between ther xes s. Consder sngle wre nd clculte the ux through rectngulr re wth the xs of the wre long one sde. Tke ths sde to hve length L nd the other dmenson of the rectngle to be. The mgnetc eld s everywhere perpendculr to the rectngle. Frst consder the prt of the rectngle tht s nsde the wre. The eld dstnce r from the xs s gven by B 0 r nd the ux through the strp of length L nd wh dr t tht dstnce s ( 0 r )L dr. Thus the ux through the re nsde the wre s dr L n Z 0 0 L r dr 0L 4 : Now consder the regon outsde the wre. There the eld s gven by B 0 r nd the ux through n nntesmlly thn strp s ( 0 r)l dr. The ux through the whole regon s Z 0 L dr out r 0L ln : The totl ux through the re bounded by the dotted lnes s the sum of the two contrbutons: h 0L 1 + ln : 4 Now nclude the contrbuton of the other wre. Snce the currents re n the sme drecton the two contrbutons hve the sme sgn. They lso hve the sme mgntude, so totl 0L h 1 + ln :

12 840 CHAPTE 31 INDUCTION AND INDUCTANCE The totl ux per unt length s h 1 + ln totl L 0 1: Wb/m : ( Tm/A)(10 A) 0 mm 1 + ln 1:5 mm (b) Agn consder the ux of sngle wre. The ux nsde the wre tself s gn n 0 L4. The ux nsde the regon of the other wre s out Z 0 L dr r 0L ln Double ths to nclude the ux of the other wre (nsde the rst wre) nd dvde by L to obtn the ux per unt length. The totl ux per unt length tht s nsde the wres s wres L ln ( Tm/A)(10 A) : Wb/m : 1 + ln The frcton of the totl ux tht s nsde the wres s : Wb/m 1: Wb/m 0:17 : : 0 mm 0 mm 1:5 mm (c) The contrbutons of the two wres to the totl ux hve the sme mgntudes but now the currents re n opposte drectons, so the contrbutons hve opposte sgns. Ths mens totl 0. 31E U therml P therml t E t db 1 t 1 A B t A B t t : 3E Therml energy s generted t the rte E, where E s the emf n the wre nd s the resstnce of the wre. The resstnce s gven by LA, where s the resstvty of copper, L s the length of the wre, nd A s the cross-sectonl re of the wre. The resstvty cn be found n Tble 7{1. Thus (1: m)(0:500 m) (0: m) 1: :

13 CHAPTE 31 INDUCTION AND INDUCTANCE 841 Frdy's lw s used to nd the emf. If B s the mgntude of the mgnetc eld through the loop, then E A db, where A s the re of the loop. The rdus r of the loop s r L nd ts re s r L 4 L 4. Thus E L 4 db The rte of therml energy generton s (0:500 m) (10: T/s) 1: V : 4 P E (1: V) 1: : W : 33E () The ux chnges becuse the re bounded by the rod nd rls ncreses s the rod moves. Suppose tht t some nstnt the rod s dstnce x from the rght-hnd end of the rls nd hs speed v. Then the ux through the re s B BA BLx, where L s the dstnce between the rls. Accordng to Frdy's lw the mgntude of the emf nduced s E d B BL (dx) BLv (0:350 T)(0:50 m)(0:550 m/s) 4:81 10 V. (b) Use Ohm's lw. If s the resstnce of the rod then the current n the rod s E (4:81 10 V)(18:0 ) : A. (c) The rte t whch therml energy s generted s P (: A) (18:0 ) 1: W : 34E () Let x be the dstnce from the rght end of the rls to the rod. The re enclosed by the rod nd rls s Lx nd the mgnetc ux through the re s B BLx. The emf nduced s E d B BL dx BLv, where v s the speed of the rod. Thus E (1: T)(0:10 m)(5:0 m/s) 0:60 V. (b) If s the resstnce of the rod, the current n the loop s E (0:60 V)(0:40 ) 1:5 A. Snce the rod moves to the left n the dgrm, the ux ncreses. The nduced current must produce mgnetc eld tht s nto the pge n the regon bounded by the rod nd rls. To do ths the current must be clockwse. (c) The rte of generton of therml energy by the resstnce of the rod s P E (0:60 V) (0:40 ) 0:90 W. (d) Snce the rod moves wth constnt velocty the net force on t must be zero. Ths mens the force of the externl gent hs the sme mgntude s the mgnetc force but s n the opposte drecton. The mgntude of the mgnetc force s F B LB (1:5 A)(0:10 m)(1: T) 0:18 N. Snce the eld s out of the pge nd the current s upwrd through the rod the mgnetc force s to the rght. The force of the externl gent must be 0:18 N, to the left.

14 84 CHAPTE 31 INDUCTION AND INDUCTANCE (e) As the rod moves n nntesml dstnce dx the externl gent does work dw F dx, where F s the force of the gent. The force s n the drecton of moton, so the work done by the gent s postve. The rte t whch the gent does work s dw F dx F v (0:18 N)(5:0 m/s) 0:90 W, the sme s the rte t whch therml energy s generted. The energy suppled by the externl gent s converted completely to therml energy. 35P Let F net BL mg 0 nd solve for : so mg BL jej 1 d B B da B(v tl) ; v t mg B L : 36P () At tme t the re of the closed trngulr loop s A(t) 1 (vt)(vt) v t. Thus (b) B j t3:00 s BA(t) (0:350 T)[(5:0 m/s)(3:00 s)] 85: Tm : E d B d(ba) B da B d(v t ) (0:350 T)(5:0 m/s) (3:00 s) 56:8 V : Bv t (c) From prt (b) bove we see tht E Bv t / t 1. Thus n 1. 37P From we obtn P v E v [sn (ft)] v 1 T E 0 [sn (ft)] v Z T 0 sn (ft) 1 E 0 (150 V) (4:0 ) 68 W : 38P () Apply Newton's second lw to the rod: m dv BL :

15 CHAPTE 31 INDUCTION AND INDUCTANCE 843 Integrte to obtn v BLt m : v ponts wy from the genertor G. (b) When the current n the rod becomes zero, the rod wll no longer be ccelerted by force F BL nd wll therefore rech constnt termnl velocty. Ths hppens when je nduced j E,.e., je nduced j d B d(ba) B da BvL E ; or v EBL, to the left. (c) In cse () bove electrc energy s suppled by the genertor nd s trnsferred nto the knetc energy of the rod. In the current cse the bttery ntlly supples electrc energy to the rod, cusng ts knetc energy to ncrese to mxmum vlue of 1 mv 1 (EBL). Afterwrds, there s no further energy trnsfer from the bttery to the rod, nd the knetc energy of the rod remns constnt. 39P () Let x be the dstnce from the rght end of the rls to the rod nd nd n expresson for the mgnetc ux through the re enclosed by the rod nd rls. The mgnetc eld s not unform but vres wth dstnce from the long strght wre. The eld s norml to the re nd hs mgntude B 0 r, where r s the dstnce from the wre nd s the current n the wre. Consder n nntesml strp of length x nd wh dr, prllel to the wre nd dstnce r from t. The re of ths strp s A x dr nd the ux through t s d B ( 0 xr)dr. The totl ux through the re enclosed by the rod nd rls s B 0x Z +L dr r 0x ln + L Accordng to Frdy's lw the emf nduced n the loop s E d 0 dx + L ln 0v ( Tm/A)(100 A)(5:00 m/s) : V : + L ln 1:00 cm + 10:0 cm ln 1:00 cm (b) If s the resstnce of the rod then the current n the conductng loop s ` E (: V)(0:400 ) 6: A. Snce the ux s ncresng the mgnetc eld produced by the nduced current must be nto the pge n the regon enclosed by the rod nd rls. Ths mens the current s clockwse. (c) Therml energy s beng generted t the rte P ` (6: A) (0:400 ) 1: W. :

16 844 CHAPTE 31 INDUCTION AND INDUCTANCE (d) Snce the rod moves wth constnt velocty the net force on t s zero. The force of the externl gent must hve the sme mgntude s the mgnetc force nd must be n the opposte drecton. The mgntude of the mgnetc force on n nntesml segment of the rod, wth length dr nd dstnce r from the long strght wre, s df B `B dr ( 0 `r) dr. The totl mgnetc force on the rod hs mgntude F B 0` Z +L dr r 0` ln + L ( Tm/A)(6: A)(100 A) : N : 1:00 cm + 10:0 cm ln 1:00 cm Snce the eld s out of the pge nd the currentn the rod s upwrd n the dgrm, ths force s towrd the rght. The externl gent must pply force of : N, to the left. (e) The externl gent does work t the rte P F v (: N)(5:00 m/s) 1: W. Ths s the sme s the rte t whch therml energy s generted n the rod. All the energy suppled by the gent s converted to therml energy. 40E () The eld pont s nsde the solenod, so Eq pples. nduced electrc eld s The mgntude of the E 1 db r 1 (6: T/s)(0:00 m) 7: V/m : (b) Now the eld pont s outsde the solenod nd Eq. 31{9 pples. The mgntude ofthe nduced eld s E 1 db r 1 (6: T/s) (0:0600 m) (0:080 m) 1: V/m : 41E I 1 Eds d B1 d (B 1A 1 ) A 1 db 1 r db 1 1 (0:00 m) ( 8: T/s) 1: V ; I Eds d B r db (0:300 m) ( 8: T/s) : V ;

17 CHAPTE 31 INDUCTION AND INDUCTANCE 845 nd I Eds I 3 1 Eds I Eds 1: V ( : V) 1: V : 4P The mgnetc eld B cn be expressed s B(t) B 0 + B 1 sn(!t + 0 ) ; where B 0 (30:0 T + 9:6 T) 9:8 T nd B 1 (30:0 T 9:6 T) 0:00 T. Then from Eq E 1 db r r d [B 0 + B 1 sn(!t + 0 )] 1 B 1!r cos(!t + 0 ) : Thus E mx 1 B 1(f)r 1 (0:00 T)()(15 Hz)(1:6 10 m 0:15 V/m : 43P The nduced electrc eld E s functon of r s gven by E(r) (r)(db). So c ee m er db m (1: C)(5:0 10 m)( T/s) (9: kg) 4: m/s : ponts to the rght nd c ponts to the left. At pont b we hve b / r b 0: 44P H Use Frdy's lw n the form Eds (d B ). Integrte round the dotted pth shown n Fg. 31{61. At ll ponts on the upper nd lower sdes the electrc eld s ether perpendculr to the sde or else t vnshes. Assume t vnshes t ll ponts on the rght sde (outsde the cpctor). On the left sde H t s prllel to the sde nd hs constnt mgntude. Thus drect ntegrton yelds E ds EL, where L s the length of the left sde of the rectngle. The mgnetc eld s zero nd remns zero, so d B 0. Frdy's lw leds to contrdcton: EL 0, but nether E nor L s zero. There must be n electrc eld long the rght sde of the rectngle.

18 846 CHAPTE 31 INDUCTION AND INDUCTANCE 45E Snce N B L, where N s the number of turns, L s the nductnce, nd s the current, B L N (8: H)(5: A) 400 1: Wb : 46E () B NBA NB(r ) (30:0)(: T)()(0:100 m) : Wb : (b) L B : Wb 3:80 A 6: H/m : 47E () N :0 m:5 mm 800. (b) Ll 0 n A ( Tm/A)(800:0 m) ()(0:040 m) 4 : H: 48P If the solenod s long nd thn, then when t s bent nto torod (b L torod 0N h ln b 0N h ln 1 + b b 0N h(b ) b ) 1. Thus Snce A h(b ) s the cross-sectonl re nd l b s the length of the torod, we my rewrte the expresson for L torod bove s : L torod l 0N A l 0 n A ; whch ndeed reduces to tht of long solenod. Note tht the pproxmton ln(1+x) x ws used for jxj 1. 49P () Suppose we dvde the one-turn solenod nto N smll crculr loops plced long the wh W of the copper strp. Ech loop crres current N. Then the mgnetc eld nsde the solenod s B 0 n 0 (NW )(N) 0 W: (b) L B B ( 0 W ) 0 W :

19 CHAPTE 31 INDUCTION AND INDUCTANCE P The re of ntegrton for the clculton of the mgnetc ux s bounded by the two dotted lnes nd the boundres of the wres. If the orgn s tken to be on the xs of the rght-hnd wre nd r mesures dstnce from tht xs, t extends from r to r d. Consder the rght-hnd wre rst. In the regon of ntegton the eld t produces s nto the pge nd hs mgntude B 0 r. Dvde the regon nto strps of length l nd wh dr, s shown. The ux through the strps dstnce r from the xs of the wre s d Bl dr dn the ux through the entre regon s 0l Z d dr r 0l d ln The other wre produces the sme result, so the totl ux through the dotted rectngle s totl 0l d ln : : dr d r l The nductnce s totl dveded by : L totl 0l d ln : 51E () Snce E enhnces, must be decresng. (b) From E L d we get L E d 17 V :5 ka/s 6: H : 5E Snce E L(d), we my obtn the desred nduced emf by settng d E L 60 V 1 H 5:0 A/s : You mght, for exmple, unformly reduce the current from :0 A to zero n 40 ms.

20 848 CHAPTE 31 INDUCTION AND INDUCTANCE 53E () L l 0 n A ( Tm/A)(100 turnscm) ()(1:6 cm) 0:10 H/m : (b) The nduced emf per meter s E L d (0:10 H/m)(13 As) 1:3 V/m : 54E () L E d 3: V 5:0 A/s 6: H : (b) From L N B we get N L B (8:0 A)(6: H) Wb 10 : 55P Use E L d. () For 0 < t < ms E L t (4:6 H)(7:0 A 0) : s 1: V : (b) For ms < t < 5 ms E L t (4:6 H)(5:0 A 7:0 A) (5:0 :0)10 3 s 3: V : (c) For 5 ms < t < 6 ms E L t (4:6 H)(0 5:0 A) (6:0 5:0)10 3 s :3 104 V : 56P () L eq totl L 1 + L :

21 CHAPTE 31 INDUCTION AND INDUCTANCE 849 (b) If the seprton s not lrge enough then the mgnetc eld n one nductor my produce ux n the other, whch would render the expresson L nvld. (c) For N nductors n seres L eq totl 1 NX n1 n NX n1 n NX n1 L n : 57P () In ths cse totl 1 +. But V totl L eq nd 1; L 1; so L eq L 1 + L, or : L eq L 1 L (b) The reson s the sme s n prt (b) of the prevous problem. (c) For N nductors n prllel you cn esly generze the expresson bove to 1 L eq NX n1 1 L n : 58E Suppose tht (t 0 ) 0 3 t t t 0. Wrte (t 0 ) 0 (1 e t 0 L ), whch gves L t 0 ln(1 0 ) 5:00 s ln(1 13) 1:3 s : 59E Strtng wth zero current t tme t 0, when the swtch s closed, the current n n L seres crcut t lter tme t s gven by E 1 e t L ; where L s the nductve tme constnt, E s the emf, nd s the resstnce. You wnt to clculte the tme t for whch 0:9990E. Ths mens 0:9990 E E 1 e t L ; so 0: e t L

22 850 CHAPTE 31 INDUCTION AND INDUCTANCE or e t L 0:0010 : Tke the nturl logrthm of both sdes to obtn (t) ln(0:0010) 6:91. Thus t 6:91 L. Tht s, 6:91 nductve tme constnts must elpse. 60E The current n the crcut s gven by 0 e t L ; where 0 s the ntl current (t tme t 0) nd L ( L) s the nductve tme constnt. Solve for L. Dvde by 0 nd tke the nturl logrthm of both sdes of the resultng equton to obtn Ths yelds L t ln [ 0 ] Thus L L (10 H)(0:17 s) 46. ln 0 t L : 1:0 s ln [( A)(1:0 A)] 0:17 s : 61E Wrte 0 e t L nd note tht 10% 0. Solve for t: t L ln 0 L ln 0 :00 H 3:00 ln 0 10:0% 0 1:54 s : 6E () Immedtely fter the swtch s closed E E L. But 0 t ths nstnt so E L E. (b) E L (t) Ee t L Ee :0 L L Ee :0 0:135E: (c) Solve E L (t) Ee t L for t L : t L ln(ee L ) ln : So t L ln 0:693 L. 63E () If the bttery s swtched nto the crcut t tme t 0, then the current t lter tme t s gven by E 1 e t L ; where L L. You wnt to nd the tme for whch 0:800E. Ths mens 0:800 1 e t L

23 CHAPTE 31 INDUCTION AND INDUCTANCE 851 or e t L 0:00 : Tke the nturl logrthm of both sdes to obtn (t L ) ln(0:00) 1:609. Thus t 1:609 L 1:609L 1:609(6: H) 1: : s : (b) At t 1:0 L the current n the crcut s E 1 1:0 e 14:0 V 1: e 1:0 7: A : 64E () L Wb5:5 A 4: H: (b) Use Eq to solve for t: t L ln 1 4: H 0:75 L E ln 1 ln 1 E (:5 A)(0:75 ) 6:0 V : s : 65P Apply the loop theorem E(t) L(d) nd solve for E(t): E(t) L d + L d (3:0 + 5:0t) + (3:0 + 5:0t) (6:0 H)(5:0 A/s) + (3:0 A + 5:0t)(4:0 ) (4 + 0t) V ; where t s n seconds. 66P () nd (b) Wrte V L (t) Ee t l. Consder the rst two dt ponts, (V L1 ; t 1 ) nd (V L ; t ), stsfyng V L Ee t L ( 1; ). We hve V L1 V L Ee (t 1 t ) L ; whch gves L t 1 t 1:0 ms :0 ms 3:6 ms : ln(v V 1 ) ln(13:818:) So E V L1 e t 1 L (18: V)e 1:0 ms3:6 ms 4 V. You cn esly check tht the vlues of L nd E re consstent wth the rest of the dte ponts.

24 85 CHAPTE 31 INDUCTION AND INDUCTANCE 67P Tke the tme dervtve of both sdes of Eq : d d E 1 e t E T L e L 45:0 V 50: H e (180 )(1:010 3 s)(50:010 3 H) 1:0 A/s : 68P () Snce the nner crcumference of the torod s l (10 cm) 6:8 cm, the number fo turns of the torod s roughly N 6:8 cm1:0 mm 68. Thus L 0N H : H : ln b ( Tm/A)(68) (0:1 m 0:10 m) 1 cm ln 10 cm (b) Snce the totl length l of the wre s l (68)4(:0 cm) 50 m, the resstnce of the wre s (50 m)(0:0 m) 1:0 : Thus L L : H 1:0 : s : 69P () The nductor prevents fst buld-up of the current through t, so mmedtely fter the swtch s closed the current n the nductor s zero. Ths mens 1 E V 10:0 + 0:0 3:33 A : (b) A long tme lter the current reches stedy stte nd no longer chnges. The emf cross the nductor s zero nd the crcut behves s f t were replced by wre. The current n 3 s 1. Krchho's loop rule gves nd E E 1 1 ( 1 ) 3 0 : Solve these smultneously for 1 nd. The results re E( + 3 ) (100 V)(0:0 + 30:0 ) (10:0 )(0:0 ) + (10:0 )(30:0 ) + (0:0 )(30:0 ) 4:55 A

25 CHAPTE 31 INDUCTION AND INDUCTANCE 853 nd E (100 V)(30:0 ) (10:0 )(0:0 ) + (10:0 )(30:0 ) + (0:0 )(30:0 ) :73 A : (c) The left-hnd brnch s now broken. If ts nductnce s zero the current mmedtely drops to zero when the swtch s opened. Tht s, 1 0. The current n 3 chnges only slowly becuse there s n nductor n ts brnch. Immedtely fter the swtch s opened t hs the sme vlue s t hd just before the swtch ws opened. Tht vlue s 4:55 A :73 A 1:8 A. The current n s the sme s tht n 3, 1:8 A. (d) There re no longer ny sources of emf n the crcut, so ll currents eventully drop to zero. 70P () When swth S s just closed (cse I), V 1 E nd 1 E 1 10 V5:0 :0 A. After long tme (cse II) we stll hve V 1 E, so 1 :0 A: (b) Cse I: snce now E L E, 0; cse II: snce now E L 0, E 10 V10 1:0 A. (c) Cse I: 1 + :0 A + 0 :0 A; cse II: 1 + :0 A + 1:0 A 3:0 A. (d) Cse I: snce E L E, V E E L 0; cse II: snce E L 0, V E E L E 10 V: (e) Cse I: E L E 10 V; cse II: E L 0: (f) Cse I: d E L L EL 10 V5:0 H :0 A/s; cse II: d E L L 0. 71P For t < 0, no current goes through L, so 0 nd 1 E. As the swtch s opened there wll be very bref sprkng cross the gp. 1 drops whle ncreses, both very quckly. The loop rule cn be wrtten s E 1 L 1 d 1 L L d 0 ; where the ntl vlue of 1 t t 0 s gven by E nd tht of t t 0 s 0. Now consder the stuton shortly fter t 0. Snce the sprkng s very bref, we cn resonbly ssume tht both 1 nd get equlzed quckly, before they cn chnge pprecbly from ther respectve ntl vlues. Thus the loop rule requres tht L 1 (d 1 ), whch s lrge nd negtve, must roughly cncel L (d ), whch s lrge nd postve: L 1 d 1 L d :

26 854 CHAPTE 31 INDUCTION AND INDUCTANCE Let the common vlue reched by 1 nd be, then d 1 1 t E t nd The equtons bove yeld or L 1 d t 0 t : E EL 1 L 1 + L 1 L ( 0) ; L 1 E L 1 + L : 7P () Before the fuse blows, the current through the resstor remns zero. Apply the loop theorem to the bttery-fuse-nductor loop: E L d 0. So EtL. As the fuse blows t t t 0, 0 3:0 A. Thus t 0 0L E (3:0 A)(5:0 H) 10 V 1:5 s : (b) 3.0 A 0.67 A 1.5 s (fuse blows) t 73P () Assume s from left to rght through the closed swtch. Let 1 be the current n the resstor nd tke t to be downwrd. Let be the current n the nductor nd lso tke t to be downwrd. The juncton rule gves 1 + nd the loop rule gves 1 L(d ) 0.

27 CHAPTE 31 INDUCTION AND INDUCTANCE 855 Accordng to the juncton rule, (d 1 ) obtn (d ). Substtute nto the loop equton to L d : Ths equton s smlr to Eq , nd ts soluton s the functon gven s Eq : 1 0 e tl ; where 0 s the current through the resstor t t 0, just fter the swtch s closed. Now just fter the swtch s closed the nductor prevents the rpd buld-up of current n ts brnch, so t tht tme 0 nd 1. Thus 0, so 1 e tl nd 1 1 e tl : (b) When 1, so e tl 1 e tl ; e tl 1 : Tke the nturl logrthm of both sdes nd use ln(1) ln to obtn (tl) ln or t L ln : 74P () Use U B 1 L : L U B (5: J) 13:9 H: (60: A) (b) Snce U B /, for U B to ncrese by fctor of 4 must ncrese by fctor of,.e., should be ncresed to (60:0 ma) 10 ma. 75E Let U B (t) 1 L (t) 1 U B(t! 1) 1 4 L 0. Ths gves (t) 0 p. But (t) 0 (1 e t L ), so 1 e t L 1 p. Solve for t: t L ln 1 1 p 1:3 L :

28 856 CHAPTE 31 INDUCTION AND INDUCTANCE 76E () 0 E 100 V10 10 A. (b) U B 1 L 0 1 (:0 H)(10 A) 100 J: 77E () (b) (c) du B d 1 L L d L 0 L e t L L 0 1 e t L 1 e t L d h 0 1 e t L LE (L) e tl 1 e t L (100 h V) 10 e (10 )(0:10 s):0 H 1 e (10 )(0:10 s):0 H :4 10 W : 1 e t E L P therml 0 (100 V) h1 e (10 )(0:10 s):0 H 10 1:5 10 W : 1 e tl P bttery E E 0 1 e t L E 1 e tl (100 V) h1 e (10 )(0:10 s):0 H 10 3:9 10 W : Note tht the lw of conservton of energy requrement P bttery du B + P therml s stsed. 78P Use the results of the lst problem. Let P therml du B, or nd solve for t: E 1 e tl E e tl 1 e tl t L ln L ln (37:0 ms) ln 5:6 ms :

29 CHAPTE 31 INDUCTION AND INDUCTANCE P () If the bttery s ppled t tme t 0 the current s gven by E 1 e t L ; where E s the emf of the bttery, s the resstnce, nd L s the nductve tme constnt. In terms of nd the nductnce L, L L. Solve the current equton for the tme constnt. Frst obtn e t L 1 E ; then tke the nturl logrthm of both sdes to obtn Snce ln 1 ln 1 E t ln 1 L E : (: A)(10: ) :5108 ; 50:0 V the nductve tme constnt s L t0:5108 (5: s)(0:5108) 9: s nd the nductnce s (b) The energy stored n the col s L L (9: s)(10: ) 97:9 H : U B 1 L 1 (97:9 H)(: A) 1: J : 80P () (b) E bttery Z t 0 Z t E P bttery 1 e tl E 0 :00 s + (5:50 H) e (6:70 )(:00 s)5:50 H 1 (10:0 V) 6:70 18:7 J : 6:70 t + L e tl 1 E nductor U B (t) 1 L (t) 1 E L (1 e tl ) 1 e (6:70 )(:00 s)5:50 H 1 10:0 V (5:50 H) 6:70 5:10 J :

30 858 CHAPTE 31 INDUCTION AND INDUCTANCE (c) E dsspted E bttery E nductor 18:7 J 5:10 J 13:6 J: 81P () The nductnce of the solenod s Thus L 0 n Al ( Tm/A)(3000 turns0:800 m) ()(5:00 10 m) (0:800 m) 0:111 H : U B (t) 1 L (t) 1 E L (0:111 H) 1:0 V 10:0 1:05 10 J : 1 e tl h 1 e (10:0 )(5: s)0:111 H (b) E bttery E t + L e tl 1 (1:0 V) 10:0 5: s + 0:111 H h e 10:0 (10:0 )(5: s)0:111 H 1 1:41 10 J : 8P Suppose tht the swtch hs been n poston for long tme, so the current hs reched the stedy-stte vlue 0. The energy stored n the nductor s U B 1 L 0. Now the swtch s thrown to poston b t tme t 0. Therefter the current s gven by 0 e t L ; where L s the nductve tme constnt, gven by L. The rte t whch therml energy s generted n the resstor s gven by P 0 e t L : Over long tme perod the energy dsspted s Z 1 Z 1 E P 0 e t L Le t L L :

31 CHAPTE 31 INDUCTION AND INDUCTANCE 859 Substtute L L to obtn E 1 L 0 ; the sme s the totl energy orgnlly stored n the nductor. 83E () At ny pont the mgnetc energy densty s gven by u B B 0, where B s the mgntude of the mgnetc eld t tht pont. Insde solenod B 0 n, where n s the number of turns per unt length nd s the current. For the solenod of ths problem n (950)(0:850 m) 1: m 1. The mgnetc energy densty s u B 1 0n 1 ( Tm/A)(1: m 1 ) (6:60 A) 34: J/m 3 : (b) Snce the mgnetc eld s unform nsde n del solenod, the totl energy stored n the eld s U B u B V, where V s the volume of the solenod. V s clculted s the product of the cross-sectonl re nd the length. Thus U B (34: J/m 3 )(17: m )(0:850 m) 4:94 10 J : 84E The mgnetc energy stored n the torod s gven by U B 1 L, where L s ts nductnce nd s the current. The energy s lso gven by U B u B V, where u B s the verge energy densty nd V s the volume. Thus r s ub V L (70:0 J/m 3 )(0:000 m 3 ) 5:58 A : 90: H 85E Let u E 1 0E u B 1 B 0 nd solve for E: E B p 0 0 0:50 T p (8: F/m)( Tm/A) 1:5 108 V/m: 86E Use 1 ly 9: m. U B V u B V B 0 (9: ) 3 (10 10 T) ( Tm/A) J :

32 860 CHAPTE 31 INDUCTION AND INDUCTANCE 87E Use Eq to solve for L: L U 0 l 4 ln b 0l ln b : 88E () The energy needed s U E u E V 1 0E V 1 (8: F/m)(100 kv/m) (10 cm) 3 4: J : (b) The energy needed s U B u B V 1 0 B V (1:0 T) ( Tm/A) (10 cm)3 4:0 10 J : (b) Obvously, snce U B > U E greter mounts of energy cn be stored n the mgnetcc eld. 89E () (b) B 0 ( Tm/A)(100 A) ( m) u B B 0 1: T : (1: T) ( Tm/A) 0:63 J/m3 : 90P () The energy densty t ny pont s gven by u B B 0, where B s the mgntude of the mgnetc eld. The mgntude of the eld nsde torod, dstnce r from the center, s gven by Eq. 30{6: B 0 Nr, where N s the number of turns nd s the current. Thus 0 N u B 1 0 r 0 N 8 r : (b) Evlute the ntegrl U B u B dv over the volume of the torod. A crculr strp wth rdus r, heght h, nd thckness dr hs volume dv rh dr, so Z U B 0 N b dr h 8 r 0 N h b ln : 4

33 CHAPTE 31 INDUCTION AND INDUCTANCE 861 Numerclly, U B ( Tm/A)(0:500 A) (150) ( m) 4 3: J : ln 95 mm 5 mm (c) The nductnce s gven by Eq. 31{37: L 0N h b ln so the energy s gven by U B 1 L 0N h 4 b ln : Ths the exctly the sme s the expresson found n prt (b) nd yelds the sme numercl result. 91P () (b) u B B ( Tm/A)(10 A) 1:0 J/m 3 : 8(: m) u E 1 0E 0 J 0 l 1 (8: F/m)[(10 A)(3: m)] 4: J/m 3 : Here we used J A nd l A 1A to obtn J l: 9P () u B B e 0 ( T) ( Tm/A) 1: J/m 3 : (b) The volume of the shell of thckness h s V 4 e h, where e s the rdus of the Erth. So U B V u B 4(6: m) ( m)(1: J/m 3 ) 8: J :

34 86 CHAPTE 31 INDUCTION AND INDUCTANCE 93E () (b) M E 1 jd 1 j 5:0 mv 15:0 A/s 1:67 mh : M 1 (1:67 mh)(3:60 A) 6:00 mwb : 94E () 1 L 1 1 N 1 (5 mh)(6:0 ma) 100 1:5 Wb : (b) E 1 L 1 d 1 1 M 1 N (5 mh)(4:0 A/s) 100 mv : (3:0 mh)(6:0 ma) nwb : E 1 M d 1 95E Use E M d 1 to nd M: M E d 1 (3:0 mh)(4:0 A/s) 1 mv : V 6:0 A(: s) 13 H : 96P () Assume the current s chngng t the rte d nd clculte the totl emf cross both cols. Frst consder the left-hnd col. The mgnetc eld due to the current n tht col ponts to the left. So does the mgnetc eld due to the current n col. When the current ncreses both elds ncrese nd both chnges n ux contrbute emf's n the sme drecton. Thus the emf n col 1 s E 1 (L 1 + M) d : The mgnetc eld n col due to the current n tht col ponts to the left, s does the eld n col due to the current n col 1. The two sources of emf re gn n the sme drecton nd the emf n col s E (L + M) d : The totl emf cross both cols s E E 1 + E (L 1 + L + M) d : Ths s exctly the emf tht would be produced f the cols were replced by sngle col wth nductnce L eq L 1 + L + M.

35 CHAPTE 31 INDUCTION AND INDUCTANCE 863 (b) everse the leds of col so the current enters t the bck of col rther thn the front s pctured n the dgrm. Then the eld produced by col t the ste of col 1 s opposte the eld produced by col 1 tself. The uxes hve opposte sgns. An ncresng current n col 1 tends to ncrese the ux n tht col but n ncresng current n col tends to decrese t. The emf cross col 1 s Smlrly the emf cross col s The totl emf cross both cols s E 1 (L 1 M) d : E (L M) d : E (L 1 + L M) d : Ths the sme s the emf tht would be produced by sngle col wth nductnce L eq L 1 + L M. 97P M M cs N cs N( 0 s n ) 0 nn: s s As long s the mgnetc eld of the solenod s entrely contned wthn the cross-secton of the col we lwys hve sc B s A s B s, regrdless of the shpe, sze, or possble lck of close-pckng of the col. 98P Use the expresson for the ux over the torod cross-secton derved n Smple Problem 31-6 to obtn M M ct N c ct t N c t 0 t N t h ln b 0N 1 N h ln b ; where N t N 1 nd N c N. 99P Assume the current n solenod 1 s nd clculte the ux lnkge n solenod. The mutul nductnce s ths ux lnkge dvded by. The mgnetc eld nsde solenod 1 s prllel to the xs nd hs unform mgntude B 0 n 1, where n 1 s the number of turns per unt length of the solenod. The cross-sectonl re of the solenod s 1 nd snce the eld s norml to cross secton the ux through cross secton s AB 1 0n 1 :

36 864 CHAPTE 31 INDUCTION AND INDUCTANCE Snce the mgnetc eld s zero outsde the solenod, ths s lso the ux through cross secton of solenod. The number of turns n length l of solenod s N n l nd the ux lnkge s N n l 1 0n 1 : The mutul nductnce s M N 1 l 0n 1 n : M does not depend on becuse there s no mgnetc eld n the regon between the solenods. Chngng does not chnge the ux through solenod, but chngng 1 does. 100P () The ux over the loop cross secton due to the current n the wre s gven by Thus Z +b B wre (r)l dr M N (b) From the formul for M obtned bove Z +b M (100)( Tm/A)(0:30 m) 0 l r dr 0l 1 ln + b N 0l 1 ln + b : ln 1 + 8:0 1:0 : 1: H :