v = = ˆ ˆ ˆ B x y x y x y y x = C m s 0.15 T m s T

Transcription

1 Chaptr 8 1. (a) Equation 8-3 lads to F v = = sinφ (b) Th kintic nrgy of th proton is N = ms C T sin 3. 0 c hc h kg m s mv J ( )( ) K = = =, which is quivalnt to K = ( J) / ( J/V) = 835 V.. Th forc associatd with th magntic fild must point in th j dirction in ordr to cancl th forc of gravity in th j dirction. y th right-hand rul, points in th k dirction (sinc i k = j). Not that th charg is positiv; also not that w nd j to assum y = 0. Th magnitud z is givn by Eq. 8-3 (with φ = 90 ). Thrfor, with 4 5 m = kg, v =.0 10 m/s, and q = C, w find ˆ ˆ ˆ zk mg = = k = ( T)k. qv 3. (a) Th forc on th lctron is F = qv = q v ˆi+ v ˆj ˆi+ j = q v v kˆ ( ) ( ) ( ) y y y y 6 6 ( ) ( )( ) ( )( ) 14 ( N) k. ˆ = C.0 10 m s 0.15 T m s T = Thus, th magnitud of F is N, and F points in th positiv z dirction. (b) This amounts to rpating th abov computation with a chang in th sign in th charg. Thus, F has th sam magnitud but points in th ngativ z dirction, namly, 14 = N k. ˆ F ( ) 4. (a) W us Eq. 8-3: 1097

2 1098 CHAPTER 8 F = q v sin φ = ( C) (550 m/s) (0.045 T) (sin 5 ) = N. (b) Th acclration is a = F /m = ( N) / ( kg) = m/s. (c) Sinc it is prpndicular to v, F dos not do any work on th particl. Thus from th work-nrgy thorm both th kintic nrgy and th spd of th particl rmain unchangd. 5. Using Eq. 8- and Eq. 3-30, w obtain d i d b g F = q v v k = q v 3 v y y y whr w us th fact that y = 3. Sinc th forc (at th instant considrd) is F z k whr F z = N, thn w ar ld to th condition Fz q( 3 v vy) = Fz =. q v v i k ( 3 y) Substituting v =.0 m/s, v y = 4.0 m/s, and q = C, w obtain Fz N (3 y) ( C)[3(.0 m/s) 4.0 m] = = =.0 T. q v v 6. Th magntic forc on th proton is F = qv whr q = +. Using Eq this bcoms ( )i^ + ( )j^ = [(0.03v y + 40)i^ + (0 0.03v )j^ (0.0v v y )k^] with SI units undrstood. Equating corrsponding componnts, w find (a) v = m/s, and (b) v y = m/s. d i to solv for E : 7. W apply F = q E + v = ma

3 1099 E ma = + v q = c hd i kg m s i b μ Tg i b. km sg j b. km sgk C = 114. i j k V m. 8. Ltting F = qde + v i= 0, w gt j vsinφ = E. W not that (for givn valus of th filds) this givs a minimum valu for spd whnvr th sin φ factor is at its maimum valu (which is 1, corrsponding to φ = 90 ). So 3 E V/m 3 vmin = = = m/s T 9. Straight-lin motion will rsult from zro nt forc acting on th systm; w ignor gravity. Thus, F = qde + v i= 0. Not that v so v = v. Thus, obtaining th spd from th formula for kintic nrgy, w obtain 3 E E 100 V /(0 10 m) = = = = v K / m V C / kg 3 31 ( ) ( ) ( ) T. In unit-vctor notation, 4 = ( T)kˆ. 10. (a) Th nt forc on th proton is givn by 3 ( C) ( 4.00 V m) k+ ˆ ( 000 m s) ˆ F = F j ( T) ˆ E + F = qe+ qv = i = N k. ˆ 18 ( ) (b) In this cas, w hav F = F + F = qe+ qv E ( ) ( ) ( ) ( ) ( N) k. ˆ = C 4.00 V m kˆ m s ˆj.50 mt ˆi = (c) In th final cas, w hav

4 1100 CHAPTER 8 F = F + F = qe+ qv E ( ) ( ) ( ) ( ) 19 ( N) ˆ i+ ( N) k. ˆ = C 4.00 V m ˆi+ 000 m s ˆj.50 mt ˆi = 11. Sinc th total forc givn by F = de + v i vanishs, th lctric fild E must b prpndicular to both th particl vlocity v and th magntic fild. Th magntic fild is prpndicular to th vlocity, so v has magnitud v and th magnitud of th lctric fild is givn by E = v. Sinc th particl has charg and is acclratd through a potntial diffrnc V, mv /= V and v= V m. Thus, 3 ( )( ) 7 ( kg) V C V E = = ( ) = m 5 1. T V m. 1. (a) Th forc du to th lctric fild ( F = qe ) is distinguishd from that associatd with th magntic fild ( F = qv ) in that th lattr vanishs whn th spd is zro and th formr is indpndnt of spd. Th graph shows that th forc (y-componnt) is ngativ at v = 0 (spcifically, its valu is N thr), which (bcaus q = ) implis that th lctric fild points in th +y dirction. Its magnitud is E F q C nt, y.0 10 N = = = 1.5 N/C = 1.5 V/m. (b) W ar told that th and z componnts of th forc rmain zro throughout th motion, implying that th lctron continus to mov along th ais, vn though magntic forcs gnrally caus th paths of chargd particls to curv (Fig. 8-11). Th cption to this is discussd in Sction 8-3, whr th forcs du to th lctric and magntic filds cancl. This implis (Eq. 8-7) = E/v = T. For F = qv to b in th opposit dirction of F = qe w must hav v in th opposit dirction from E, which points in th +y dirction, as discussd in part (a). Sinc th vlocity is in th + dirction, thn (using th right-hand rul) w conclud that th magntic fild must point in th +z dirction ( i^ k^ = j^ ). In unit-vctor notation, w hav = ( T)kˆ. 13. W us Eq. 8-1 to solv for V: V ( 3A)( 0.65 T) i = = = nl 8 3 ( m )( 150μm)( C) V.

5 For a fr charg q insid th mtal strip with vlocity v w hav F = qde + v i. W st this forc qual to zro and us th rlation btwn (uniform) lctric fild and potntial diffrnc. Thus, E V V d v = = y y c V = = ms. 3 c Thc mh 15. (a) W sk th lctrostatic fild stablishd by th sparation of chargs (brought on by th magntic forc). With Eq. 8-10, w dfin th magnitud of th lctric fild as E = v = 0.0 m/s T = V/m. ( )( ) Its dirction may b infrrd from Figur 8-8; its dirction is opposit to that dfind by v. In summary, E =(0.600V m)kˆ which insurs that F = qde + v i vanishs. (b) Equation 8-9 yilds V = Ed = (0.600 V/m)(.00 m) = 1.0 V. 16. W not that must b along th ais bcaus whn th vlocity is along that ais thr is no inducd voltag. Combining Eq. 8-7 and Eq. 8-9 lads to V V d = = E v whr on must intrprt th symbols carfully to nsur that d, v, and ar mutually prpndicular. Thus, whn th vlocity if paralll to th y ais th absolut valu of th voltag (which is considrd in th sam dirction as d ) is 0.01 V, and 0.01 V d = d z = = 0.0 m. (3.0 m/s)(0.00 T) On th othr hand, whn th vlocity is paralll to th z ais th absolut valu of th appropriat voltag is V, and h Thus, our answrs ar V d = d y = = 0.30 m. (3.0 m/s)(0.00 T)

6 110 CHAPTER 8 (a) d = 5 cm (which w arriv at by limination, sinc w alrady hav figurd out d y and d z ), (b) d y = 30 cm, and (c) d z = 0 cm. 17. (a) Using Eq. 8-16, w obtain rq m C 10. T v = = = 7 m 400. u 4. 00u kg u α c hc hb g b gc h (b) T = πr/v = π( m)/( m/s) = s. 6 = ms. (c) Th kintic nrgy of th alpha particl is 1 K = mα v = 7 6 b4. 00 ugc kg uhc m sh JV c h 5 = V. (d) ΔV = K/q = V/ = V. 18. With th pointing out of th pag, w valuat th forc (using th right-hand rul) at, say, th dot shown on th lft dg of th particl s path, whr its vlocity is down. If th particl wr positivly chargd, thn th forc at th dot would b toward th lft, which is at odds with th figur (showing it bing bnt toward th right). Thrfor, th particl is ngativly chargd; it is an lctron. (a) Using Eq. 8-3 (with angl φ qual to 90 ), w obtain F v = = m s. (b) Using ithr Eq or Eq. 8-16, w find r = m. (c) Using Eq (in ithr its first or last form) radily yilds T = s. 19. Lt ξ stand for th ratio ( m/ q ) w wish to solv for. Thn Eq can b writtn as T = πξ/. Noting that th horizontal ais of th graph (Fig. 8-36) is invrs-fild (1/) thn w conclud (from our prvious prssion) that th slop of th lin in th graph must b qual to πξ. W stimat that slop is T. s, which implis ξ 9 = m/ q = kg/c.

7 Combining Eq with nrgy consrvation (V = 1 m v in this particular application) lads to th prssion r = m V m which suggsts that th slop of th r vrsus V graph should b m / 8-37, w stimat th slop to b in SI units. Stting this qual to and solving, w find = T.. From Fig. m / 1. (a) From K = 1 mv w gt K V v = = kg m 3 c hc V Jh 7 = ms. (b) From r = m v/ q w gt mv = = qr (c) Th orbital frquncy is 31 7 c kghc m sh = C m c hc h 4 T. f 7 v m s = = = π r π m ( ) Hz. (d) T = 1/f = ( Hz) 1 = s.. Using Eq. 8-16, th radius of th circular path is mv r = = q mk q whr K = mv / is th kintic nrgy of th particl. Thus, w s that K = (rq) /m q m 1. (a) ( ) ( ) ( ) ( ) K = q q m m K = 1 4 K = K = 1.0MV; α α p p α p p p

8 1104 CHAPTER 8 (b) ( ) ( ) ()( ) K = q q m m K = 1 1 K = 1.0 MV = 0.50MV. d d p p d p p 3. From Eq. 8-16, w find mv r 31 6 ( kg)( m s) 5 = = = T. ( C)( m) 4. (a) Th acclrating procss may b sn as a convrsion of potntial nrgy V into kintic nrgy. Sinc it starts from rst, 1 mv = V and (b) Equation 8-16 givs ( )( ) V C 350 V v = = = 31 m kg r mv 31 7 ( kg)( m s) 3 ( C)( T) m s. 4 = = = m. 5. (a) Th frquncy of rvolution is 6 q c f = = Thc Ch 5 = Hz. 31 p p kg m c h (b) Using Eq. 8-16, w obtain c hb gc h c hc h 31 mv mk kg 100V J V r = = = 6 q q C T = m. 6. W considr th point at which it ntrs th fild-filld rgion, vlocity vctor pointing downward. Th fild points out of th pag so that v points lftward, which indd sms to b th dirction it is pushd ; thrfor, q > 0 (it is a proton). (a) Equation 8-17 bcoms T = π m /, or p 9 ( ) = 7 π ( ) ( )

9 1105 which yilds = 0. 5T. (b) Doubling th kintic nrgy implis multiplying th spd by. Sinc th priod T dos not dpnd on spd, thn it rmains th sam (vn though th radius incrass by a factor of ). Thus, t = T/ = 130 ns. 7. (a) W solv for from m = q /8V (s Sampl Problm Uniform circular motion of a chargd particl in a magntic fild ): Vm = 8. q W valuat this prssion using =.00 m: = c hc h V kg c Chb00. mg = T. (b) Lt N b th numbr of ions that ar sparatd by th machin pr unit tim. Th currnt is i = qn and th mass that is sparatd pr unit tim is M = mn, whr m is th mass of a singl ion. M has th valu Sinc N = M/m w hav i qm = = m kg 8 M = = kg s. 3600s 19 8 c hc kg sh C kg = A. (c) Each ion dposits nrgy qv in th cup, so th nrgy dpositd in tim Δt is givn by For Δt = 1.0 h, iqv E = NqV Δt = Δt = iv Δ t. q c hc hb g 3 6 E = A V 3600s = J. To obtain th scond prssion, i/q is substitutd for N. 8. Using F = mv / r (for th cntriptal forc) and th rlation K = mv /, w can asily driv

10 (a) If v is th spd of th positron thn v sin φ is th componnt of its vlocity in th plan that is prpndicular to th magntic fild. Hr φ is th angl btwn th vlocity and th fild (89 ). Nwton s scond law yilds v sin φ = m (v sin φ) /r, whr r is th radius of th orbit. Thus r = (m v/) sin φ. Th priod is givn by 31 ( ) πr πm π kg T = = = = v sin φ C 0.100T ( )( ) s. Th quation for r is substitutd to obtain th scond prssion for T. (b) Th pitch is th distanc travld along th lin of th magntic fild in a tim intrval of on priod. Thus p = vt cos φ. W us th kintic nrgy to find th spd: K = 1 mv mans 3 ( )( ) K V J V v = = = 31 m kg m s. Thus, ( )( ) p = = m s s cos m. (c) Th orbit radius is 31 7 ( )( ) mvsin kg m s sin89 φ R = = = ( C)( T) 34. (a) Equation 3-0 givs φ = cos 1 (/19) = m. (b) No, th magntic fild can only chang th dirction of motion of a fr (unconstraind) particl, not its spd or its kintic nrgy. (c) No, as rfrnc to Fig should mak clar. (d) W find v = v sin φ = 61.3 m/s, so r = mv / = 5.7 nm. 35. (a) y consrvation of nrgy (using qv for th potntial nrgy, which is convrtd into kintic form) th kintic nrgy gaind in ach pass is 00 V. (b) Multiplying th part (a) rsult by n = 100 givs ΔK = n(00 V) = 0.0 kv. (c) Combining Eq with th kintic nrgy rlation (n(00 V) = m p v / in this particular application) lads to th prssion

11 1109 m K 1 r = = q m q Km. For th avrag nrgy r = c hc hc h V J V kg c Chb157. Tg = m. Th total distanc travld is about nπr = (104)(π)(0.375) =.4 10 m. 38. (a) Using Eq. 8-3 and Eq. 8-18, w find f ( C)( 1.0T) q = = = kg ( ) osc 7 π mp π (b) From r = m v q = m k q w hav p P ( rq) ( 0.500m)( C)( 1.0T) 7 ( )( ) Hz. K = = = m kg J V p V. 39. (a) Th magnitud of th magntic forc on th wir is givn by F = il sin φ, whr i is th currnt in th wir, L is th lngth of th wir, is th magnitud of th magntic fild, and φ is th angl btwn th currnt and th fild. In this cas φ = 70. Thus, 6 F = 5000A 100m T sin 70 = 8. N. b gb gc h (b) W apply th right-hand rul to th vctor product is to th wst. 40. Th magntic forc on th (straight) wir is ( ) ( ) ( ) ( ) F il = to show that th forc F = ilsin θ = 13.0A 1.50T 1.80m sin 35.0 = 0.1N. 41. (a) Th magntic forc on th wir must b upward and hav a magnitud qual to th gravitational forc mg on th wir. Sinc th fild and th currnt ar prpndicular to ach othr th magnitud of th magntic forc is givn by F = il, whr L is th lngth of th wir. Thus,

12 1110 CHAPTER 8 ( kg)( 9.8m s ) mg il = mg i = = = A. L ( 0.60 m)( T) (b) Applying th right-hand rul rvals that th currnt must b from lft to right. 4. (a) From symmtry, w conclud that any -componnt of forc will vanish (valuatd ovr th ntirty of th bnt wir as shown). y th right-hand rul, a fild in th k dirction producs on ach part of th bnt wir a y-componnt of forc pointing in th j dirction; ach of ths componnts has magnitud F = i sin 30 = (.0 A)(.0 m)(4.0 T)sin 30 = 8 N. y Thrfor, th forc on th wir shown in th figur is ( 16j) ˆ N. (b) Th forc rtd on th lft half of th bnt wir points in th k dirction, by th right-hand rul, and th forc rtd on th right half of th wir points in th + k dirction. It is clar that th magnitud of ach forc is qual, so that th forc (valuatd ovr th ntirty of th bnt wir as shown) must ncssarily vanish. 43. W stablish coordinats such that th two sids of th right triangl mt at th origin, and th y = 50 cm sid runs along th +y ais, whil th = 10 cm sid runs along th + ais. Th angl mad by th hypotnus (of lngth 130 cm) is θ = tan 1 (50/10) =.6, rlativ to th 10 cm sid. If on masurs th angl countrclockwis from th + dirction, thn th angl for th hypotnus is = Sinc w ar only askd to find th magnituds of th forcs, w hav th frdom to assum th currnt is flowing, say, countrclockwis in th triangular loop (as viwd by an obsrvr on th +z ais. W tak to b in th sam dirction as that of th currnt flow in th hypotnus. Thn, with = = T, = cosθ = T, = sinθ = 0.088T. (a) Equation 8-6 producs zro forc whn L so thr is no forc rtd on th hypotnus of lngth 130 cm. (b) On th 50 cm sid, th componnt producs a forc i y k, and thr is no contribution from th y componnt. Using SI units, th magnitud of th forc on th y sid is thrfor 4. 00A m T N. b gb gb g= y

13 1111 (c) On th 10 cm sid, th y componnt producs a forc i k, and thr is no contribution from th componnt. Th magnitud of th forc on th sid is also b gb gb g= 4. 00A 10. m T N. y (d) Th nt forc is i k + i k =0, y y kping in mind that < 0 du to our initial assumptions. If w had instad assumd wnt th opposit dirction of th currnt flow in th hypotnus, thn > 0, but y < 0 and a zro nt forc would still b th rsult. 44. Considr an infinitsimal sgmnt of th loop, of lngth ds. Th magntic fild is prpndicular to th sgmnt, so th magntic forc on it has magnitud df = i ds. Th horizontal componnt of th forc has magnitud df = ( i cos θ ) ds h and points inward toward th cntr of th loop. Th vrtical componnt has magnitud df = ( isin θ ) ds y and points upward. Now, w sum th forcs on all th sgmnts of th loop. Th horizontal componnt of th total forc vanishs, sinc ach sgmnt of wir can b paird with anothr, diamtrically opposit, sgmnt. Th horizontal componnts of ths forcs ar both toward th cntr of th loop and thus in opposit dirctions. Th vrtical componnt of th total forc is F i ds ai 3 3 v = sinθ = p sinθ= π(0.018 m)( A)( T)sin 0 = N. W not that i,, and θ hav th sam valu for vry sgmnt and so can b factord from th intgral. 45. Th magntic forc on th wir is F = il = ilˆi ˆj+ kˆ = il ˆj+ kˆ ( ) ( ) y z z y ( ) ( ) ( ) ˆ ( ) = 0.500A 0.500m T j T kˆ ( ) 3 ˆ 3 ˆ = j k N.

14 1113 which w diffrntiat (with rspct to θ) and st th rsult qual to zro. This provids a dtrmination of th angl: Consquntly, min b sg b g 1 θ = tan μ = tan = 31. ( )( ) kg 9.8m s = = 0.10T. ( 50A)( 1.0m)( cos sin 31 ) 1 1 (b) As shown abov, th angl is ( ) ( ) θ = tan μ s = tan 0.60 = W us df = idl, whr dl = d i and = i + j. Thus, F idl f f = = id ˆi ˆi+ ˆj = i d kˆ ( ) y y i i ( ) 3.0 ( ) ( d) ( ) 1.0 = 5.0A 8.0 m mt k ˆ = ( 0.35N)k. ˆ 49. Th applid fild has two componnts: > 0 and z > 0. Considring ach straight sgmnt of th rctangular coil, w not that Eq. 8-6 producs a nonzro forc only for th componnt of that is prpndicular to that sgmnt; w also not that th quation is ffctivly multiplid by N = 0 du to th fact that this is a 0-turn coil. Sinc w wish to comput th torqu about th hing lin, w can ignor th forc acting on th straight sgmnt of th coil that lis along th y ais (forcs acting at th ais of rotation produc no torqu about that ais). Th top and bottom straight sgmnts princ forcs du to Eq. 8-6 (causd by th z componnt), but ths forcs ar (by th right-hand rul) in th ±y dirctions and ar thus unabl to produc a torqu about th y ais. Consquntly, th torqu drivs compltly from th forc rtd on th straight sgmnt locatd at = m, which has lngth L = 0.10 m and is shown in Figur 8-44 carrying currnt in th y dirction. Now, th z componnt will produc a forc on this straight sgmnt which points in th dirction (back towards th hing) and thus will rt no torqu about th hing. Howvr, th componnt (which is qual to cosθ whr = 0.50 T and θ = 30 ) producs a forc qual to NiL that points (by th right-hand rul) in th +z dirction. Sinc th action of this forc is prpndicular to th plan of th coil, and is locatd a distanc away from th hing, thn th torqu has magnitud ( NiL )( ) NiL θ ( )( )( )( )( ) τ = = cos = A 0.10 m m 0.50 T cos 30 = N m. Sinc τ = r F, th dirction of th torqu is y. In unit-vctor notation, th torqu is 3 τ = ( N m)j ˆ. y

15 1114 CHAPTER 8 An altrnativ way to do this problm is through th us of Eq W do not show thos dtails hr, but not that th magntic momnt vctor (a ncssary part of Eq. 8-37) has magnitud μ = NiA = A m bgb gc h and points in th z dirction. At this point, Eq may b usd to obtain th rsult for th torqu vctor. 50. W us τ = μ = μ= iπ r, and not that i = qf = qv/πr. So ma ma τ ma qv 1 1 ( C)( m/s)( m)( = pr= qvr= T) p r = N m. 51. W us Eq whr μ is th magntic dipol momnt of th wir loop and is th magntic fild, as wll as Nwton s scond law. Sinc th plan of th loop is paralll to th inclin th dipol momnt is normal to th inclin. Th forcs acting on th cylindr ar th forc of gravity mg, acting downward from th cntr of mass, th normal forc of th inclin F N, acting prpndicularly to th inclin through th cntr of mass, and th forc of friction f, acting up th inclin at th point of contact. W tak th ais to b positiv down th inclin. Thn th componnt of Nwton s scond law for th cntr of mass yilds mg sin θ f = ma. For purposs of calculating th torqu, w tak th ais of th cylindr to b th ais of rotation. Th magntic fild producs a torqu with magnitud μ sinθ, and th forc of friction producs a torqu with magnitud fr, whr r is th radius of th cylindr. Th first tnds to produc an angular acclration in th countrclockwis dirction, and th scond tnds to produc an angular acclration in th clockwis dirction. Nwton s scond law for rotation about th cntr of th cylindr, τ = Iα, givs fr μ sin θ = Iα. Sinc w want th currnt that holds th cylindr in plac, w st a = 0 and α = 0, and us on quation to liminat f from th othr. Th rsult is mgr = μ. Th loop is rctangular with two sids of lngth L and two of lngth r, so its ara is A = rl and th dipol momnt is μ = NiA = Ni( rl). Thus, mgr = NirL and b gc h b gb gb g mg 0. 50kg 9. 8m s i = = = 45. A. NL m T 5. Th insight cntral to this problm is that for a givn lngth of wir (formd into a rctangl of various possibl aspct ratios), th maimum possibl ara is nclosd whn

16 1115 th ratio of hight to width is 1 (that is, whn it is a squar). Th maimum possibl valu for th width, th problm says, is = 4 cm (this is whn th hight is vry clos to zro, so th total lngth of wir is ffctivly 8 cm). Thus, whn it taks th shap of a squar th valu of must b ¼ of 8 cm; that is, = cm whn it ncloss maimum ara (which lads to a maimum torqu by Eq and Eq. 8-37) of A = (0.00 m) = m. Sinc N = 1 and th torqu in this cas is givn as Nm, thn th aformntiond quations lad immdiatly to i = A. 53. W rplac th currnt loop of arbitrary shap with an assmbly of small adjacnt rctangular loops filling th sam ara that was nclosd by th original loop (as narly as possibl). Each rctangular loop carris a currnt i flowing in th sam sns as th original loop. As th sizs of ths rctangls shrink to infinitsimally small valus, th assmbly givs a currnt distribution quivalnt to that of th original loop. Th magnitud of th torqu Δ τ rtd by on th nth rctangular loop of ara ΔA n is givn by Δτ = Ni sin θδa. Thus, for th whol assmbly n n n n n τ = Δ τ = Ni Δ A = NiAsin θ. 54. (a) Th kintic nrgy gaind is du to th potntial nrgy dcras as th dipol swings from a position spcifid by angl θ to that of bing alignd (zro angl) with th fild. Thus, K = U U = μcosθ μcos 0. Thrfor, using SI units, th angl is F HG i 1 K 1 θ = cos 1 cos 1 μ f I F I = KJ HG b0. 00gb0. 05g n b g KJ = 77. (b) Sinc w ar making th assumption that no nrgy is dissipatd in this procss, thn th dipol will continu its rotation (similar to a pndulum) until it rachs an angl θ = 77 on th othr sid of th alignmnt ais. 55. (a) Th magnitud of th magntic momnt vctor is (b) Now, ( ) ( ) ( ) μ= ia n n = πri 1 1+ πri = π 7.00A 0.00m m =.86A m. n ( ) ( ) ( ) μ= πri πri 1 1= π 7.00A 0.300m 0.00m = 1.10A m. b g b g 56. (a) μ = NAi = pr i = p m. 60A = A m.

17 1117 (a) y using th right-hand rul, w s that μ is in th y dirction. Thus, w hav ˆ 3 ˆ μ = ( NiA)( j) = (3)(.00 A)( m )j =(0.040 A m )j ˆ. Th corrsponding orintation nrgy is U =μ = μ = = y y 3 5 ( A m )( T) J (b) Using th fact that ˆˆ j i = 0, ˆ j ˆ j = 0, and ˆ j k ˆ = ˆ i, th torqu on th coil is τ = μ = μ ˆiμ kˆ y z y = ˆ 5 ˆ 5 = ( N m)i + ( N m)k. ˆ 3 3 ( A m )( T)i ( A m )( T)k. ˆ Not: Th orintation nrgy is highst whn μ is in th opposit dirction of, and lowst whn μ lins up with. 6. Looking at th point in th graph (Fig. 8-50(b)) corrsponding to i = 0 (which mans that coil has no magntic momnt) w ar ld to conclud that th magntic 5 momnt of coil 1 must b μ 1 =.0 10 A m. Looking at th point whr th lin crosss th ais (at i = 5.0 ma) w conclud (sinc th magntic momnts cancl thr) 5 that th magnitud of coil s momnt must also b μ =.0 10 A m whn i = A, which mans (Eq. 8-35) A m = = = m. i A NA μ Now th problm has us considr th dirction of coil s currnt changd so that th nt momnt is th sum of two (positiv) contributions, from coil 1 and coil, spcifically for th cas that i = A. W find that total momnt is μ = ( A m ) + (N A i ) = A m. 63. Th magntic dipol momnt is μ = μ 060. i080. jj, whr μ = NiA = Niπr = 1(0.0 A)π(0.080 m) = A m. Hr i is th currnt in th loop, N is th numbr of turns, A is th ara of th loop, and r is its radius.

18 1118 CHAPTER 8 (a) Th torqu is j j b gb g j b gb g j b gb g j τ = μ = μ 060. i 080. j 05. i k = μ i k j i j k = μ 018. j+ 00. k 04. i. Hr i k = j,j i = k, and j k = i ar usd. W also us substitut th valu for μ to obtain ( ˆ ˆ ˆ ) τ = i j k N m. (b) Th orintation nrgy of th dipol is givn by U =μ =μ 060. i 080. j 05. i k Hr i i = 1, i k = 0, j i =0, and j j b gb g = μ = 015. μ = J. jk =0 ar usd. i i =0. Now, w 64. Eq givs U = μ = μ cosφ, so at φ = 0 (corrsponding to th lowst point on th graph in Fig. 8-51) th mchanical nrgy is K + U = K o + (μ) = J + ( J) = J. Th turning point occurs whr K = 0, which implis U turn = J. So th angl whr this taks plac is givn by J φ = cos = 110 μ whr w hav usd th fact (s abov) that μ = J. 65. If N closd loops ar formd from th wir of lngth L, th circumfrnc of ach loop is L/N, th radius of ach loop is R = L/πN, and th ara of ach loop is A= πr = π L πn = L 4 πn. ( ) (a) For maimum torqu, w orint th plan of th loops paralll to th magntic fild, so th dipol momnt is prpndicular (i.., at a 90 angl) to th fild. (b) Th magnitud of th torqu is thn

19 114 CHAPTER 8 E E c = 0.65 T nρ = = ( m )( C)( Ω m) (a) Sinc K = qv w hav Kp Kα ( qα Kp) (b) Similarly, q = K, K / K = α (c) Sinc r = mk q mk q, w hav d d α = as =, or Kp / K α = (.00u) ( 1.00u) mk qr p p K d d p rd = = rp = 10 cm = 14 cm. mk q K p p d p (d) Similarly, for th alpha particl, w hav r α ( 4.00u) Kα ( ) ( ) mk qr p p r α α p = = = 10 cm = 14 cm. mk q 1.00u K p p α 80. (a) Th largst valu of forc occurs if th vlocity vctor is prpndicular to th fild. Using Eq. 8-3, F,ma = q v sin (90 ) = v = ( C) ( m/s) ( T) = N. (b) Th smallst valu occurs if thy ar paralll: F,min = q v sin (0) = 0. (c) y Nwton s scond law, a = F /m = q v sin θ /m, so th angl θ btwn v and is θ = sin F H G I α L M c hd i O P Nc hc hc hq ma KJ = kg m s sin M P = qv C m s T Th contribution to th forc by th magntic fild( = ˆ ˆ i = ( 0.00 T)i ) is givn by Eq. 8-: (( 17000i ˆ ˆi ) ( 11000ˆj ˆi ) ( 7000kˆ ˆi )) ( 0kˆ 140ˆj ) F = qv = q + + = q

20 116 CHAPTER 8 F = qv = + v v + v v + v v ( ) ( ) ( ) ˆ i ( ) ˆ j ( ) k ˆ y z z y z z y y ( ) ((( 4)( 0.008) ( 6) ( 0.004) ) ˆi+ (( 6)( 0.00) ( ) ( 0.008) ) ˆj + (( )( 0.004) ( 4)( 0.00) ) kˆ) 1 ( ) ˆ i ( ) ˆj = = + with SI units undrstood. (b) y dfinition of th cross product, v F. This is asily vrifid by taking th dot (scalar) product of v with th rsult of part (a), yilding zro, providd car is takn not to introduc any round-off rror. (c) Thr ar svral ways to procd. It may b worthwhil to not, first, that if z wr 6.00 mt instad of 8.00 mt thn th two vctors would b actly antiparalll. Hnc, th angl θ btwn and v is prsumably clos to 180. Hr, w us Eq. 3-0: 1 v 1 68 θ = cos = cos = 173 v (a) W ar givn ˆ = ˆ i = (6 10 T)i, so that v = v y k whr vy = m/s. W not that th magntic forc on th lctron is v k and thrfor points in b g th + k dirction, at th instant th lctron ntrs th fild-filld rgion. In ths trms, Eq bcoms mv y r = = m. (b) On rvolution taks T = πr/v y = 0.60 μs, and during that tim th drift of th lctron in th dirction (which is th pitch of th hli) is Δ = v T = m whr v = m/s. (c) Rturning to our obsrvation of forc dirction mad in part (a), w considr how this is prcivd by an obsrvr at som point on th ais. As th lctron movs away from him, h ss it ntr th rgion with positiv v y (which h might call upward ) but pushd in th +z dirction (to his right). Hnc, h dscribs th lctron s spiral as clockwis. y j