(b) This is about one-sixth the magnitude of the Earth s field. It will affect the compass reading.

Transcription

1 Chapter 9 (a) The magntude of the magnetc feld due to the current n the wre, at a pont a dstance r from the wre, s gven by r Wth r = ft = 6 m, we have c4 T m AhbAg 6 33 T 33 T m b g (b) Ths s about one-sxth the magntude of the Earth s feld It wll affect the compass readng Equaton 9- s maxmzed (wth respect to angle) by settng = 9º ( = / rad) Its value n ths case s ds dmax 4 R From Fg 9-35(b), we have max 6 T We can relate ths max to our d max by settng ds equal to 6 m and R = 5 m Ths allows us to solve for the current: = 375 A Pluggng ths nto Eq 9-4 (for the nfnte wre) gves = 3 T 3 THINK The magnetc feld produced by a current-carryng wre can be calculated usng the ot-savart law EXPRESS The magntude of the magnetc feld at a dstance r from a long straght wre carryng current s, usng the ot-savart law, r ANALYZE (a) The feld due to the wre, at a pont 8 cm from the wre, must be 39 T and must be drected due south Therefore, 6 b gc Th r m 39 4 T m A 6A (b) The current must be from west to east to produce a feld that s drected southward at ponts below t LEARN The drecton of the current s gven by the rght-hand rule: grasp the element n your rght hand wth your thumb pontng n the drecton of the current The drecton of 53

2 54 CHAPTER 9 the feld due to the current-carryng element corresponds to the drecton your fngers naturally curl 4 The straght segment of the wre produces no magnetc feld at C (see the straght sectons dscusson n Sample Problem 9 Magnetc feld at the center of a crcular arc of current ) Also, the felds from the two semcrcular loops cancel at C (by symmetry) Therefore, C = 5 (a) We fnd the feld by superposng the results of two sem-nfnte wres (Eq 9-7) and a semcrcular arc (Eq 9-9 wth = rad) The drecton of s out of the page, as can be checked by referrng to Fg 9-7(c) The magntude of at pont a s therefore a 4 R R R (5 m) 7 (4 Tm/A)( A) 3 T upon substtutng = A and R = 5 m (b) The drecton of ths feld s out of the page, as Fg 9-7(c) makes clear (c) The last remark n the problem statement mples that treatng b as a pont mdway between two nfnte wres s a good approxmaton Thus, usng Eq 9-4, b R R (5 m) 7 (4 Tm/A)( A) 4 8 T (d) Ths feld, too, ponts out of the page 6 Wth the usual x and y coordnates used n Fg 9-38, then the vector r pontng from a current element to P s r s ˆ R ˆj Snce ds ds î, then ds r Rds Therefore, wth r s R, Eq 9-3 gves R ds d 4 ( s R ) 3/ (a) Clearly, consdered as a functon of s (but thnkng of ds as some fnte-szed constant value), the above expresson s maxmum for s = Its value n ths case s d ds R max /4 (b) We want to fnd the s value such that d d / max Ths s a nontrval algebra exercse, but s nonetheless straghtforward The result s s = /3 R If we set R cm, then we obtan s = 38 cm

3 55 7 (a) Recallng the straght sectons dscusson n Sample Problem 9 Magnetc feld at the center of a crcular arc of current, we see that the current n the straght segments collnear wth P do not contrbute to the feld at that pont Usng Eq 9-9 (wth = ) and the rght-hand rule, we fnd that the current n the semcrcular arc of radus b contrbutes 4 b (out of the page) to the feld at P Also, the current n the large radus arc contrbutes 4 a (nto the page) to the feld there Thus, the net feld at P s -7 (4p Tm A)(4A)(74 /8 ) 4 b a 4 7m 35m -7 T (b) The drecton s out of the page 8 (a) Recallng the straght sectons dscusson n Sample Problem 9 Magnetc feld at the center of a crcular arc of current, we see that the current n segments AH and JD do not contrbute to the feld at pont C Usng Eq 9-9 (wth = ) and the rghthand rule, we fnd that the current n the semcrcular arc H J contrbutes 4R (nto the page) to the feld at C Also, arc D A contrbutes 4R (out of the page) to the feld there Thus, the net feld at C s (4p Tm A)(8A) 4 R R 4 35m 78m -7 (b) The drecton of the feld s nto the page T 9 THINK The net magnetc feld at a pont half way between the two long straght wres s the vector sum of the magnetc felds due to the currents n the two wres EXPRESS Snce the magntude of the magnetc feld at a dstance r from a long straght wre carryng current s gven by r, at a pont half way between the two sres, the magnetc feld s, where r (assumng the two wres to be r apart) The drectons of and are determned by the rght-hand rule ANALYZE (a) The currents must be opposte or ant-parallel, so that the resultng felds are n the same drecton n the regon between the wres If the currents are parallel, then the two felds are n opposte drectons n the regon between the wres Snce the currents are the same, the total feld s zero along the lne that runs halfway between the wres (b) The total feld at the mdpont has magntude = /r and

4 56 CHAPTER 9 m3 6 T pr p 4 3A -7 4p Tm A LEARN For two parallel wres carryng currents n the opposte drectons, a pont that s a dstance d from one wre and r d from the other, the magntude of the magnetc feld s d ( r d) d r d (a) Recallng the straght sectons dscusson n Sample Problem 9 Magnetc feld at the center of a crcular arc of current, we see that the current n the straght segments collnear wth C do not contrbute to the feld at that pont Equaton 9-9 (wth = ) ndcates that the current n the semcrcular arc contrbutes 4R to the feld at C Thus, the magntude of the magnetc feld s -7 (4p Tm A)(348A) -7 8 T 4R 4(96m) (b) The rght-hand rule shows that ths feld s nto the page (a) and /r where = 65 A and r = d + d = 75 cm + 5 cm = 5 cm, P /r where r = d = 5 cm From P = P we get P r 5 cm 65A 43A r 5 cm (b) Usng the rght-hand rule, we see that the current carred by wre must be out of the page (a) Snce they carry current n the same drecton, then (by the rght-hand rule) the only regon n whch ther felds mght cancel s between them Thus, f the pont at whch we are evaluatng ther feld s r away from the wre carryng current and s d r away from the wre carryng current 3, then the cancelng of ther felds leads to (3 ) 6 cm d r 4 cm r ( d r) 4 4 (b) Doublng the currents does not change the locaton where the magnetc feld s zero 3 Our x axs s along the wre wth the orgn at the mdpont The current flows n the postve x drecton All segments of the wre produce magnetc felds at P that are out of

5 57 the page Accordng to the ot-savart law, the magntude of the feld any (nfntesmal) segment produces at P s gven by d sn dx 4 r where (the angle between the segment and a lne drawn from the segment to P ) and r (the length of that lne) are functons of x Replacng r wth x R and sn wth R r R x R, we ntegrate from x = L/ to x = L/ The total feld s R dx R x L 4p 4p R pr L 4R L L L 3 L x R x R -7 p p3 4 T m A 58 A 8m m (8m) 4(3m) 8 53 T 4 We consder Eq 9-6 but wth a fnte upper lmt (L/ nstead of ) Ths leads to L / R ( L/ ) R In terms of ths expresson, the problem asks us to see how large L must be (compared wth R) such that the nfnte wre expresson (Eq 9-4) can be used wth no more than a % error Thus we must solve = Ths s a nontrval algebra exercse, but s nonetheless straghtforward The result s R L L 4 R 4 R 5 (a) As dscussed n Sample Problem 9 Magnetc feld at the center of a crcular arc of current, the radal segments do not contrbute to P and the arc segments contrbute accordng to Eq 9-9 (wth angle n radans) If k desgnates the drecton out of the page then 4 A rad ˆ 8 A / 3rad ˆ 6 k k (7 T)kˆ 4 5 m 4 4 m or 6 7 T (b) The drecton s ˆk, or nto the page

6 58 CHAPTER 9 (c) If the drecton of s reversed, we then have 4 A rad ˆ 8 A / 3rad ˆ 6 k k (67 T)kˆ 4 5 m 4 4 m or 6 67 T (d) The drecton s ˆk, or nto the page 6 Usng the law of cosnes and the requrement that = nt, we have cos 44, where Eq 9- has been used to determne (68 nt) and (5 nt) 7 THINK We apply the ot-savart law to calculate the magnetc feld at pont P An ntegral s requred snce the length of the wre s fnte EXPRESS We take the x axs to be along the wre wth the orgn at the rght endpont The current s n the +x drecton All segments of the wre produce magnetc felds at P that are out of the page Accordng to the ot-savart law, the magntude of the feld any (nfntesmal) segment produces at P s gven by d sn dx 4 r where (the angle between the segment and a lne drawn from the segment to P ) and r (the length of that lne) are functons of x Replacng r wth x R and sn wth R r R x R, we ntegrate from x = L to x = ANALYZE The total feld s R dx R x L 4 4 R 4R L R L 3 L x R x R 4 T m A 693 A 36 m m (36 m) (5 m) 7 3 T LEARN In calculatng at P, we could have chosen the orgn to be at the left endpont Ths only changes the ntegraton lmt, but the result remans the same:

7 59 R dx R x L 4 4 R 4R L R L L 3 x R x R 8 In the one case we have small + bg = 475 T, and the other case gves small bg = 575 T (cautonary note about our notaton: small refers to the feld at the center of the small-radus arc, whch s actually a bgger feld than bg!) Dvdng one of these equatons by the other and cancelng out common factors (see Eq 9-9) we obtan (/ rsmall ) (/ rbg ) ( rsmall / rbg ) 3 (/ r ) (/ r ) ( r / r ) small bg small bg The soluton of ths s straghtforward: r small = r bg / Usng the gven fact that the r 4 cm, then we conclude that the small radus s r small cm bg 9 The contrbuton to net from the frst wre s (usng Eq 9-4) 7 ˆ (4 Tm/A)(3 A) ˆ 6 ˆ k k (3 T)k r ( m) The dstance from the second wre to the pont where we are evaluatng net s r = 4 m m = m Thus, 7 ˆ (4 Tm/A)(4 A) ˆ 6 ˆ (4 T) r ( m) and consequently s perpendcular to The magntude of net s therefore net (3 T) (4 T) 5 T (a) The contrbuton to C from the (nfnte) straght segment of the wre s The contrbuton from the crcular loop s C R C Thus, R 3 4 Tm A578 A R m 7 C C C 53 T

8 6 CHAPTER 9 C ponts out of the page, or n the +z drecton In unt-vector notaton, C (53 T)k (b) Now, 7 ˆ so C C p Tm A578 A 7 C C C T R 89 m and C ponts at an angle (relatve to the plane of the paper) equal to In unt-vector notaton, C C tan tan 766 ˆ ˆ ˆ ˆ C T(cos766 sn766k) (9 T) (6 T)k Usng the rght-hand rule (and symmetry), we see that net ponts along what we wll refer to as the y axs (passng through P), consstng of two equal magnetc feld y- components Usng Eq 9-7, net sn r where = 4 A, r = r d d / 4 5 m, and Therefore, d 4 m 4 d / 6 m / 3 tan tan tan 53 r (4 Tm A)(4 A) ( m) net sn sn T The fact that y = at x = cm mples the currents are n opposte drectons Thus, y 4 ( L x) x L x x usng Eq 9-4 and the fact that 4 To get the maxmum, we take the dervatve wth respect to x and set equal to zero Ths leads to 3x Lx L =, whch factors and becomes (3x + L)(x L) =, whch has the physcally acceptable soluton: x = L Ths produces the maxmum y : o /L To proceed further, we must determne L

9 6 Examnaton of the datum at x = cm n Fg 9-5(b) leads (usng our expresson above for y and settng that to zero) to L = 3 cm (a) The maxmum value of y occurs at x = L = 3 cm (b) Wth = 3 A we fnd o /L = nt (c) and (d) Fgure 9-5(b) shows that as we get very close to wre (where ts feld strongly domnates over that of the more dstant wre ) y ponts along the y drecton The rght-hand rule leads us to conclude that wre s current s consequently s nto the page We prevously observed that the currents were n opposte drectons, so wre s current s out of the page 3 We assume the current flows n the +x drecton and the partcle s at some dstance d n the +y drecton (away from the wre) Then, the magnetc feld at the locaton of a proton wth charge q s ( ˆ / d)k Thus, In ths stuaton, v v j F q qv d v k e e j (where v s the speed and s a postve value), and q > Thus, -7 9 qv ˆ ˆ qv ˆ (4p Tm A)(35A)(6 C)(m/s) F j k ˆ pd pd (89 m) -3 ( 775 N) ˆ 4 Intally, we have net,y = and net,x = + 4 = ( o /d) usng Eq 9-4, where d 5 m To obtan the 3º condton descrbed n the problem, we must have net, y net, xtan(3 ) 3 tan(3 ) j d where 3 = o /d and / d Snce tan(3º) = / 3, ths leads to 3 d d 464d 3 (a) Wth d = 5 cm, ths gves d = 7 cm eng very careful about the geometry of the stuaton, then we conclude that we must move wre to x = 7 cm (b) To restore the ntal symmetry, we would have to move wre 3 to x = +7 cm

10 6 CHAPTER 9 5 THINK The magnetc feld at the center of the crcle s the vector sum of the felds of the two straght wres and the arc EXPRESS Each of the sem-nfnte straght wres contrbutes straght 4R (Eq 9-7) to the feld at the center of the crcle (both contrbutons pontng out of the page ) The current n the arc contrbutes a term gven by Eq 9-9: arc, pontng nto the R page ANALYZE The total magnetc feld s 4 R R 4 R straght arc Therefore, = rad would produce zero total feld at the center of the crcle LEARN The total contrbuton of the two sem-nfnte wres s the same as that of an nfnte wre Note that the angle s n radans rather than degrees 6 Usng the Pythagorean theorem, we have R R whch, when thought of as the equaton for a lne n a versus graph, allows us to dentfy the frst term as the y-ntercept ( ) and the part of the second term that multples as the slope (5 ) The latter observaton leads to or R 4 Tm A 5 T /A R R 4 T m A 8 m 894 m 89 mm 5 T /A 3 R The other observaton about the y-ntercept determnes the angle subtended by the arc: (4 Tm A)(5 A) 3 T (33 ) T 4R 4 (894 m) or T rad 8 rad 33 T

11 63 7 We use Eq 9-4 to relate the magntudes of the magnetc felds and to the currents ( and, respectvely) n the two long wres The angle of ther net feld s = tan ( / ) = tan ( / ) = 533º The accomplsh the net feld rotaton descrbed n the problem, we must acheve a fnal angle = 533º º = 333º Thus, the fnal value for the current must be /tan = 63 ma 8 Lettng out of the page n Fg 9-56(a) be the postve drecton, the net feld s R ( R/ ) from Eqs 9-9 and 9-4 Referrng to Fg 9-56, we see that = when = 5 A, so (solvng the above expresson wth set equal to zero) we must have = 4( / ) = 4(5/) = rad (or 573º) 9 THINK Our system conssts of four long straght wres whose cross secton form a square of length a The magnetc feld at the center of the square s the vector sum of the felds of the four wres EXPRESS Each wre produces a feld wth magntude gven by = /r, where r s the dstance from the corner of the square to the center Accordng to the Pythagorean theorem, the dagonal of the square has length a, so r a and a The felds due to the wres at the upper left (wre ) and lower rght (wre 3) corners both pont toward the upper rght corner of the square The felds due to the wres at the upper rght (wre ) and lower left (wre 4) corners both pont toward the upper left corner ANALYZE The horzontal components of the felds cancel and the vertcal components sum to 4 T m A A net 4 cos 45 a a m 5 8 T In the calculaton, cos 45 was replaced wth The total feld ponts upward, or n the +y drecton Thus, 5 (8 T)j ˆ net LEARN In the fgure to the rght, we show the contrbutons from the ndvdual wres The drectons of the felds are deduced usng the rght-hand rule

12 64 CHAPTER 9 3 We note that when there s no y-component of magnetc feld from wre (whch, by the rght-hand rule, relates to when wre s at 9º = / rad), the total y-component of magnetc feld s zero (see Fg 9-58(c)) Ths means wre # s ether at +/ rad or / rad (a) We now make the assumpton that wre # must be at / rad (9º, the bottom of the cylnder) snce t would pose an obstacle for the moton of wre # (whch s needed to make these graphs) f t were anywhere n the top semcrcle (b) Lookng at the = 9º datum n Fg 9-58(b)), where there s a maxmum n net x (equal to +6 T), we are led to conclude that x 6 T T 4 T n that stuaton Usng Eq 9-4, we obtan 6 R x ( m)(4 T) 7 4 Tm/A 4 A (c) The fact that Fg 9-58(b) ncreases as progresses from to 9º mples that wre s current s out of the page, and ths s consstent wth the cancellaton of net y at 9, noted earler (wth regard to Fg 9-58(c)) (d) Referrng now to Fg 9-58(b) we note that there s no x-component of magnetc feld from wre when =, so that plot tells us that x = + T Usng Eq 9-4, we fnd the magntudes of the current to be 6 R x ( m)( T) 7 4 Tm/A A (e) We can conclude (by the rght-hand rule) that wre s current s nto the page 3 (a) Recallng the straght sectons dscusson n Sample Problem 9 Magnetc feld at the center of a crcular arc of current, we see that the current n the straght segments collnear wth P do not contrbute to the feld at that pont We use the result of Problem 9- to evaluate the contrbutons to the feld at P, notng that the nearest wre segments (each of length a) produce magnetsm nto the page at P and the further wre segments (each of length a) produce magnetsm pontng out of the page at P Thus, we fnd (nto the page) -7 4 T m A 3 A p P 8 a 8 a p p 8pa 8 47 m T T (b) The drecton of the feld s nto the page

13 65 3 Intally we have R 4r usng Eq 9-9 In the fnal stuaton we use Pythagorean theorem and wrte f z y If we square and dvde by f, we obtan R 4r [(/ R) (/ r)] f (/ R) (/ r) From the graph (see Fg 9-6(c), note the maxmum and mnmum values) we estmate / f = / =, and ths allows us to solve for r n terms of R: r = R = 3 cm or 43 cm Snce we requre r < R, then the acceptable answer s r = 3 cm 33 THINK The magnetc feld at pont P produced by the current-carryng rbbon (shown n Fg 9-6) can be calculated usng the ot-savart law EXPRESS Consder a secton of the rbbon of thckness dx located a dstance x away from pont P The current t carres s d = dx/w, and ts contrbuton to P s d P d dx x xw ANALYZE Integratng over the length of the rbbon, we obtan P 6 dwdx w (4 Tm A)(46 A) 49 dp ln ln w d x w d m 6 3 T and P ponts upward In unt-vector notaton, P (3 T) j ˆ LEARN In the lmt where d w, usng the magnetc feld becomes ln( x) x x /,

14 66 CHAPTER 9 whch s the same as that due to a thn wre P w w ln w d w d d 34 y the rght-hand rule (whch s bult-nto Eq 9-3) the feld caused by wre s current, evaluated at the coordnate orgn, s along the +y axs Its magntude s gven by Eq 9-4 The feld caused by wre s current wll generally have both an x and a y component, whch are related to ts magntude (gven by Eq 9-4), and snes and cosnes of some angle A lttle trg (and the use of the rght-hand rule) leads us to conclude that when wre s at angle (shown n Fg 9-6) then ts components are sn, cos x y The magntude-squared of ther net feld s then (by Pythagoras theorem) the sum of the square of ther net x-component and the square of ther net y-component: ( sn ) ( cos ) cos (snce sn + cos =), whch we could also have gotten drectly by usng the law of cosnes We have 6 nt, 4 nt R R Wth the requrement that the net feld have magntude = 8 nt, we fnd cos cos ( / 4) 4, where the postve value has been chosen 35 THINK The magntude of the force of wre on wre s gven by F L / r, where s the current n wre, s the current n wre, and r s the dstance between the wres EXPRESS The dstance between the wres s force s F cos,, F where cos d / d d x r d d The x component of the ANALYZE Substtutng the values gven, the x component of the force per unt length s

15 67 F d 7 3 3, x (4 T m/a)(4 A)(68 A)(5 m) L ( d d) [(4 m) (5 m) ] 884 N/m LEARN Snce the two currents flow n the opposte drectons, the force between the wres s repulsve Thus, the drecton of F s along the lne that jons the wre and s away from wre 36 We label these wres through 5, left to rght, and use Eq 9-3 Then, (a) The magnetc force on wre s l T m A 3A (m) ˆ l j ˆj ˆj d d 3d 4d 4d 4 8 m F ˆ 4 (469 N) j (b) Smlarly, for wre, we have l ˆ 5 l ˆ 4 F ˆ j j (88 N) j d 3d d (c) F 3 = (because of symmetry) (d) F (e) F 4 4 F ( 88 N)j, and 4 5 F (469 N)j ˆ ˆ 37 We use Eq 9-3 and the superposton of forces: F F F F Wth = 45, the stuaton s as shown on the rght The components of F 4 are gven by cos 45 3 F4 x F43 F4 cos pa pa 4pa and sn 45 F4 y F4 F4 sn pa pa 4pa Thus,

16 68 CHAPTER T m A 75A 4 4x 4 y F F F 4 3 N/m 4a 4a 4a 4 35m and F 4 makes an angle wth the postve x axs, where In unt-vector notaton, we have F H G I KJ F HG I K J F 4 y tan tan 6 F 3 4x -4 ˆ ˆ -4 (3 N/m)[cos6 sn6 j] ( 5 N/m) ˆ -5 F (47 N/m)j ˆ 38 (a) The fact that the curve n Fg 9-65(b) passes through zero mples that the currents n wres and 3 exert forces n opposte drectons on wre Thus, current ponts out of the page When wre 3 s a great dstance from wre, the only feld that affects wre s that caused by the current n wre ; n ths case the force s negatve accordng to Fg 9-65(b) Ths means wre s attracted to wre, whch mples (by the dscusson n Secton 9-) that wre s current s n the same drecton as wre s current: out of the page Wth wre 3 nfntely far away, the force per unt length s gven (n magntude) as 67 7 N/m We set ths equal to F /d When wre 3 s at x = 4 m the curve passes through the zero pont prevously mentoned, so the force between and 3 must equal F there Ths allows us to solve for the dstance between wre and wre : d = (4 m)(75 A)/(5 A) = m Then we solve 67 7 N/m= o /d and obtan = 5 A (b) The drecton of s out of the page 39 Usng a magnfyng glass, we see that all but are drected nto the page Wre 3 s therefore attracted to all but wre Lettng d = 5 m, we fnd the net force (per meter length) usng Eq 9-3, wth postve ndcated a rghtward force: F d d d d whch yelds 7 F / 8 N/m 4 Usng Eq 9-3, the force on, say, wre (the wre at the upper left of the fgure) s along the dagonal (pontng toward wre 3, whch s at the lower rght) Only the forces

17 69 (or ther components) along the dagonal drecton contrbute Wth = 45, we fnd the force per unt meter on wre to be 3 F F F3 F4 F cos F3 cos 45 a a a 4 Tm A5A 3 N/m 85 m The drecton of F s along rˆ ( ˆ ˆj) / In unt-vector notaton, we have -3 ( N/m) ˆ ˆ -4 ˆ -4 F ˆ ( j) (794 N/m) ( 794 N/m)j 4 The magntudes of the forces on the sdes of the rectangle that are parallel to the long straght wre (wth = 3 A) are computed usng Eq 9-3, but the force on each of the sdes lyng perpendcular to t (along our y axs, wth the orgn at the top wre and +y downward) would be fgured by ntegratng as follows: F zab sdes a y dy Fortunately, these forces on the two perpendcular sdes of length b cancel out For the remanng two (parallel) sdes of length L, we obtan L b F a a d a a b 7 4 T m/a 3A A 8cm 3 m 3 3 N, cm 8cm and F ponts toward the wre, or ĵ That s, F (3 N) j n unt-vector notaton 3 ˆ 4 The area enclosed by the loop L s A ( 4d)( 3d ) 6d Thus c ds ja T m A 5A m 6 m 45 T m 43 We use Eq 9- r / a /r for that outsde the wre (r > a) (a) At r, for the -feld nsde the wre ( r a ) and Eq 9-7

18 7 CHAPTER 9 (b) At r m, r a (m) 7 (4 Tm/A)(7A)(m) 4 85 T (c) At ra m, r a (m) 7 (4 Tm/A)(7A)(m) 3 7 T (d) At r 4m, 7 (4 Tm/A)(7A) r (4m) 4 85 T z 44 We use Ampere s law: ds, where the ntegral s around a closed loop and s the net current through the loop (a) For path, the result s ds (b) For path, we fnd 7 6 5A 3A (4 T m/a) A 5 T m ds 7 5 5A 5A 3A (4 T m/a) 3A 6 T m 45 THINK The value of the lne ntegral ds s proportonal to the net current enclosed EXPRESS y Ampere s law, we have ds, enc where enc s the current enclosed by the closed path ANALYZE (a) Two of the currents are out of the page and one s nto the page, so the net current enclosed by the path, or Amperan loop s A, out of the page Snce the path s traversed n the clockwse sense, a current nto the page s postve and a current out of the page s negatve, as ndcated by the rght-hand rule assocated wth Ampere s law Thus, ds 7 6 (4 T m/a) A 5 T m (b) The net current enclosed z by the path s zero (two currents are out of the page and two are nto the page), so ds enc LEARN The value of the Amperan loop ds depends only on the current enclosed, and not the shape of

19 7 46 A close look at the path reveals that only currents, 3, 6 and 7 are enclosed Thus, notng the dfferent current drectons descrbed n the problem, we obtan ds T m/a 45 A 83 T m 47 For r a, enc r r r Jr r J r rdr J rdr r r p p p p p a 3a (a) At r, (b) At r a/, we have r (c) At r a, 7 3 Jr (4 Tm/A)(3A/m )(3 m / ) 3 3a 3(3 m) 7 3 Ja (4 Tm/A)(3A/m )(3 m) r a T 7 4 T 48 (a) The feld at the center of the ppe (pont C) s due to the wre alone, wth a magntude of wre wre C 3R 6R For the wre we have P, wre > C, wre Thus, for P = C = C, wre, wre must be nto the page: wre P P,wre P,ppe R R Settng C = P we obtan wre = 3/8 = 3 3 3(8 A) / 8 3 A (b) The drecton s nto the page 49 (a) We use Eq 9-4 The nner radus s r = 5 cm, so the feld there s 7 4 T m/a 8 A5 N r 5 m T (b) The outer radus s r = cm The feld there s

20 7 CHAPTER Tm/A 8 A5 N r m 4 4 T 5 It s possble (though tedous) to use Eq 9-6 and evaluate the contrbutons (wth the ntent to sum them) of all loops to the feld at, say, the center of the solenod Ths would make use of all the nformaton gven n the problem statement, but ths s not the method that the student s expected to use here Instead, Eq 9-3 for the deal solenod (whch does not make use of the col radus) s the preferred method: F N n H G I K J where = 36 A, 95 m, and N = Ths yelds = 57 T 5 It s possble (though tedous) to use Eq 9-6 and evaluate the contrbutons (wth the ntent to sum them) of all loops to the feld at, say, the center of the solenod Ths would make use of all the nformaton gven n the problem statement, but ths s not the method that the student s expected to use here Instead, Eq 9-3 for the deal solenod (whch does not make use of the col dameter) s the preferred method: F N n H G I K J where = 3 A, 5 m, and N = Ths yelds 4 3 T 5 We fnd N, the number of turns of the solenod, from the magnetc feld n on / : N / Thus, the total length of wre used n makng the solenod s 3 r c 6 mhc 3 Thb3 mg rn 8 m 7 4 T m / A 8 A 53 The orbtal radus for the electron s whch we solve for : mv e nr 7A c hb g mv r mv e e n kg463 m s C4 Tm A m3 m

21 73 54 As the problem states near the end, some dealzatons are beng made here to keep the calculaton straghtforward (but are slghtly unrealstc) For crcular moton (wth speed, v, whch represents the magntude of the component of the velocty perpendcular to the magnetc feld [the feld s shown n Fg 9-]), the perod s (see Eq 8-7) T = r/v = m/e Now, the tme to travel the length of the solenod s t L / v where v s the component of the velocty n the drecton of the feld (along the col axs) and s equal to v cos where = 3º Usng Eq 9-3 ( = n) wth n = N/L, we fnd the number of revolutons made s t /T = THINK The net feld at a pont nsde the solenod s the vector sum of the felds of the solenod and that of the long straght wre along the central axs of the solenod EXPRESS The magnetc feld at a pont P s gven by s w, where s and w are the felds due to the solenod and the wre, respectvely The drecton of s s along the axs of the solenod, and the drecton of w s perpendcular to t, so the two felds are perpendcular to each other, s w For the net feld to be at 45 wth the axs, we must have s = w ANALYZE (a) Thus, n d w s w s, whch gves the separaton d to pont P on the axs: (b) The magnetc feld strength s w 6 A d 3 n A turns cm s c hb g 4 77cm s T m A A turns m 355 T LEARN In general, the angle makes wth the solenod axs s gve by w w /d w tan tan tan n d n s s s 56 We use Eq 9-6 and note that the contrbutons to P from the two cols are the same Thus,

22 74 CHAPTER 9 P 7 R N 8N 8 4 T m/a () A 3 R R 5 5R 5 5 5m P s n the postve x drecton T 57 THINK The magntude of the magnetc dpole moment s gven by = NA, where N s the number of turns, s the current, and A s the area EXPRESS The cross-sectonal area s a crcle, so A = R, where R s the radus The magnetc feld on the axs of a magnetc dpole, a dstance z away, s gven by Eq 9-7: 3 z ANALYZE (a) Substtutng the values gven, we fnd the magntude of the dpole moment to be (b) Solvng for z, we obtan z NR 3 4A 5m 4A m 7 4 Tm A36Am T 46cm LEARN Note the smlarty between, the magnetc feld of a magnetc dpole 3 z p and E, 3 z the electrc feld of an electrc dpole p (see Eq -9) 58 (a) We set z = n Eq 9-6 (whch s equvalent usng to Eq 9- multpled by the number of loops) Thus, () /R Snce case b has two loops, b Rb Ra 4 R R a a b (b) The rato of ther magnetc dpole moments s b Ab Rb a Aa Ra 5 59 THINK The magntude of the magnetc dpole moment s gven by = NA, where N s the number of turns, s the current, and A s the area

23 75 EXPRESS The cross-sectonal area s a crcle, so A = R, where R s the radus ANALYZE Wth N =, = 3 A, and R = 5 m, the magntude of the dpole moment s 3A m 47A m LEARN The drecton of s that of the normal vector n to the plane of the col, n accordance wth the rght-hand rule shown n Fg Usng Eq 9-6, we fnd that the net y-component feld s y R R ( R z ) ( R z ) 3/ 3/, where z = L (see Fg 9-74(a)) and z = y (because the central axs here s denoted y nstead of z) The fact that there s a mnus sgn between the two terms, above, s due to the observaton that the datum n Fg 9-74(b) correspondng to y = would be mpossble wthout t (physcally, ths means that one of the currents s clockwse and the other s counterclockwse) (a) As y, only the frst term contrbutes and (wth y = 7 6 T gven n ths case) we can solve for : ( R z ) R[ ( L / R) ] 3/ 3/ y R 3/ 6 (4 m)[ (3 m / 4 m) ] (7 T) 895 A 9 A 4 T m A y (b) Wth loop at y = 6 m (see Fg 9-74(b)) we are able to determne from R R ( R L ) ( R y ) 3/ 3/ We obtan = (7 3 /5 7 A 6 (a) We denote the large loop and small col wth subscrpts and, respectvely (b) The torque has magntude equal to 7 4 T m A 5A R m c hb g b g T

24 76 CHAPTER 9 sn 9 N A N r 3A 8 m 79 T 5 6 N m 6 (a) To fnd the magntude of the feld, we use Eq 9-9 for each semcrcle ( = rad), and use superposton to obtan the result: 7 (4 T m/a) 56A a 4b 4 a b 4 57m 936m T (b) y the rght-hand rule, ponts nto the paper at P (see Fg 9-7(c)) (c) The enclosed area s has magntude A a b ( ) /, whch means the magnetc dpole moment (56A) a 3 ( b ) [(57m) (936m) ] 6 A m (d) The drecton of s the same as the found n part (a): nto the paper 63 y magnng that each of the segments bg and cf (whch are shown n the fgure as havng no current) actually has a par of currents, where both currents are of the same magntude () but opposte drecton (so that the par effectvely cancels n the fnal sum), one can justfy the superposton (a) The dpole moment of path abcdefgha s ˆ ˆ ˆ j ˆ bc f gb abgha cde f c a a j ˆ 6A m j (6 Am ) ˆj (b) Snce both ponts are far from the cube we can use the dpole approxmaton For (x, y, z) = (, 5 m, ), y m) 6 (6 Tm/A)(6 m A) j (, 5 m, ) (96 T) ˆj 3 3 ˆ 64 (a) The radal segments do not contrbute to, P and the arc segments contrbute accordng to Eq 9-9 (wth angle n radans) If k^ desgnates the drecton "out of the page" then

25 77 P (7 / 4 rad) ˆ (7 / 4 rad) k kˆ 4 (4 m) 4 ( m) where = A Ths yelds = 75 8 k^ T, or = 75 8 T (b) The drecton s ˆk, or nto the page 65 Usng Eq 9-, r R, we fnd that r = 8 m gves the desred feld value 66 (a) We desgnate the wre along y = r A = m wre A and the wre along y = r = 5 m wre Usng Eq 9-4, we have A ˆ ˆ 6 ˆ net A k k ( 5 T)k pr pr A (b) Ths wll occur for some value r < y < r A such that Solvng, we fnd y = 3/6 83 m A r y y r b A g b g (c) We elmnate the y < r possblty due to wre carryng the larger current We expect a soluton n the regon y > r A where Solvng, we fnd y = 7/4 75 m A y r y r b Ag b g 67 Let the length of each sde of the square be a The center of a square s a dstance a/ from the nearest sde There are four sdes contrbutng to the feld at the center The result s center a a p a 4a 4 On the other hand, the magnetc feld at the center of a crcular wre of radus R s /R (eg, Eq 9-) Thus, the problem s equvalent to showng that a

26 78 CHAPTER 9 4 a R a R To do ths we must relate the parameters a and R If both wres have the same length L then the geometrcal relatonshps 4a = L and R = L provde the necessary connecton: R 4a R a Thus, our proof conssts of the observaton that 4 8 a R R, as one can check numercally (that 8 ) 68 We take the current ( = 5 A) to flow n the +x drecton, and the electron to be at a pont P, whch s r = 5 m above the wre (where up s the +y drecton) Thus, the feld produced by the current ponts n the +z drecton at P Then, combnng Eq 9-4 wth Eq 8-, we obtan F e e r v k (a) The electron s movng down: v b ge j v j (where v = 7 m/s s the speed) so or 6 F e 3 N 6 ev F ˆ (3 N) ˆ e, pr (b) In ths case, the electron s n the same drecton as the current: v v so or 6 F e 3 N 6 ev F ˆj (3 N) ˆ e j, r (c) Now, v vk so F e k k 69 (a) y the rght-hand rule, the magnetc feld (evaluated at a) produced by wre (the wre at bottom left) s at = 5 (measured counterclockwse from the +x axs, n the xy plane), and the feld produced by wre (the wre at bottom rght) s at = y symmetry d we observe that only the x-components survve, yeldng

27 79 ˆ 5 cos 5 ( 346 T) where = A, = m, and Eq 9-4 has been used To cancel ths, wre b must carry current nto the page (that s, the k drecton) of value r (87 m) b 4 Tm/A T 5A 7 ˆ where r 3 87 m and Eq 9-4 has agan been used (b) As stated above, to cancel ths, wre b must carry current nto the page (that s, the drecton) 7 The radal segments do not contrbute to (at the center), and the arc segments contrbute accordng to Eq 9-9 (wth angle n radans) If k^ desgnates the drecton "out of the page" then ( rad) ˆ ( / rad) ˆ ( / rad) k k kˆ 4 (4 m) 4 ( m) 4 (4 m) z where = A Ths yelds = (57 7 T) k^, or 7 57 T 7 Snce the radus s R = 3 m, then the = 5 A produces 7 (4 Tm/A)(5 A) R (3 m) 3 77 T at the edge of the wre The three equatons, Eq 9-4, Eq 9-7, and Eq 9-, agree at ths pont 7 (a) Wth cylndrcal symmetry, we have, external to the conductors, enc r whch produces enc = 5 ma from the gven nformaton Therefore, the thn wre must carry 5 ma (b) The drecton s downward, opposte to the 3 ma carred by the thn conductng surface 73 (a) The magnetc feld at a pont wthn the hole s the sum of the felds due to two current dstrbutons The frst s that of the sold cylnder obtaned by fllng the hole and

28 8 CHAPTER 9 has a current densty that s the same as that n the orgnal cylnder (wth the hole) The second s the sold cylnder that flls the hole It has a current densty wth the same magntude as that of the orgnal cylnder but s n the opposte drecton If these two stuatons are superposed the total current n the regon of the hole s zero Now, a sold cylnder carryng current, whch s unformly dstrbuted over a cross secton, produces a magnetc feld wth magntude r R at a dstance r from ts axs, nsde the cylnder Here R s the radus of the cylnder For the cylnder of ths problem the current densty s J A a b c h, where A = (a b ) s the cross-sectonal area of the cylnder wth the hole The current n the cylnder wthout the hole s a I JA Ja a b and the magnetc feld t produces at a pont nsde, a dstance r from ts axs, has magntude Ir r a r a a a b a b The current n the cylnder that flls the hole s b I Jb a b c h c h and the feld t produces at a pont nsde, a dstance r from the ts axs, has magntude Ir rb r b b a b a b c h c h At the center of the hole, ths feld s zero and the feld there s exactly the same as t would be f the hole were flled Place r = d n the expresson for and obtan 7 (4 Tm/A) 55A (m) d a b [(4m) (5m) ] 5 53 T for the feld at the center of the hole The feld ponts upward n the dagram f the current s out of the page

29 8 d (b) If b = the formula for the feld becomes Ths correctly gves the feld of a a sold cylnder carryng a unform current, at a pont nsde the cylnder a dstance d from the axs If d = the formula gves = Ths s correct for the feld on the axs of a cylndrcal shell carryng a unform current Note: One may apply Ampere s law to show that the magnetc feld n the hole s unform Consder a rectangular path wth two long sdes (sde and, each wth length L) and two short sdes (each of length less than b) If sde s drectly along the axs of the hole, then sde would also be parallel to t and n the hole To ensure that the short sdes do not contrbute sgnfcantly to the ntegral n Ampere s law, we mght wsh to make L very long (perhaps longer than the length of the cylnder), or we mght appeal to an argument regardng the angle between and the short sdes (whch s 9 at the axs of the hole) In any case, the ntegral n Ampere s law reduces to z z rectangle z sde sde ds ds ds d L sde sde enclosed n hole where sde s the feld along the axs found n part (a) Ths shows that the feld at offaxs ponts (where sde s evaluated) s the same as the feld at the center of the hole; therefore, the feld n the hole s unform 74 Equaton 9-4 gves R m T m A 6 b gc Th 3 A 75 THINK In ths problem, we apply the ot-savart law to calculate the magnetc feld due to a current-carryng segment at varous locatons EXPRESS The ot-savart law can be wrtten as s ˆr s r x, y, z 3 4 r 4 r Wth s s j and r xˆ yˆ j zk, ˆ ther cross product s s r ( sˆ j) ( xˆ yˆ j zk) ˆ s( zˆ xk) ˆ

30 8 CHAPTER 9 where we have used ˆ j ˆ k, ˆ ˆ j ˆ j, and ˆ jk ˆ ˆ Thus, the ot-savart equaton becomes s zˆx kˆ x, y, z 3 4 x y z ANALYZE (a) The feld on the z axs (at x =, y =, and z = 5 m) s 7 4 T m/a A 3 m 5 m ˆ,, 5 m (4 T) 3/ 4 5 m (b) Smlarly, (, 6 m, ) =, snce x = z = (c) The feld n the xy plane, at (x, y, z) = (7 m, 7 m, ), s (4 T m/a)( A)(3 m)( 7 m)kˆ 7 m,7 m, ( 43 T)k ˆ 3/ 4 7 m 7 m (d) The feld n the xy plane, at (x, y, z) = ( 3, 4, ), s (4 T m/a)( A)(3 m)(3 m)kˆ 3 m, 4m, (4 T )k ˆ 3/ 4 m 4 m LEARN Along the x and z axes, the expressons for smplfy to s ˆ s x,, k,,, z ˆ 4 x 4 z The magnetc feld at any pont on the y axs vanshes because the current flows n the +y drecton, so ds ˆr ˆ 76 We note that the dstance from each wre to P s r d the current s = A 7m In both parts, (a) Wth the currents parallel, applcaton of the rght-hand rule (to determne each of ther contrbutons to the feld at P) reveals that the vertcal components cancel and the horzontal components add, yeldng the result: pr 4 cos 45 4 T

31 83 and drected n the x drecton In unt-vector notaton, we have ( 4 T) 4 ˆ (b) Now, wth the currents ant-parallel, applcaton of the rght-hand rule shows that the horzontal components cancel and the vertcal components add Thus, pr 4 sn 45 4 T and drected n the +y drecton In unt-vector notaton, we have (4 T)j 77 We refer to the center of the crcle (where we are evaluatng ) as C Recallng the straght sectons dscusson n Sample Problem 9 Magnetc feld at the center of a crcular arc of current, we see that the current n the straght segments that are collnear wth C do not contrbute to the feld there Eq 9-9 (wth = / rad) and the rght-hand rule ndcates that the currents n the two arcs contrbute b g R b g R to the feld at C Thus, the nonzero contrbutons come from those straght segments that are not collnear wth C There are two of these sem-nfnte segments, one a vertcal dstance R above C and the other a horzontal dstance R to the left of C oth contrbute felds pontng out of the page (see Fg 9-7(c)) Snce the magntudes of the two contrbutons (governed by Eq 9-7) add, then the result s 4 R R F H G exactly what one would expect from a sngle nfnte straght wre (see Eq 9-4) For such a wre to produce such a feld (out of the page) wth a leftward current requres that the pont of evaluatng the feld be below the wre (agan, see Fg 9-7(c)) 78 The ponts must be along a lne parallel to the wre and a dstance r from t, where r satsfes wre ext, or r r c T m AhbAg 3 4 m 3 T ext c 79 (a) The feld n ths regon s entrely due to the long wre (wth, presumably, neglgble thckness) Usng Eq 9-7, I K J h 4 ˆ

32 84 CHAPTER 9 where w = 4 A and r = m w T r (b) Now the feld conssts of two contrbutons (whch are ant-parallel) from the wre (Eq 9-7) and from a porton of the conductor (Eq 9- modfed for annular area): w enc w c r R r r r r R R where r = 3 m, R = m, R o = 4 m, and c = 4 A Thus, we fnd 4 93 T (c) Now, n the external regon, the ndvdual felds from the two conductors cancel completely (snce c = w ): 8 Usng Eq 9- and Eq 9-7, we have r pr pr 4 4 where r 4 m, 8 T, r m, and T Pont s known to be external to the wre snce From the second equaton, we fnd = A Pluggng ths nto the frst equaton yelds R = 53 3 m 8 THINK The objectve of ths problem s to calculate the magnetc feld due to an nfnte current sheet by applyng Ampere s law EXPRESS The current per unt x-length may be vewed as current densty multpled by the thckness y of the sheet; thus, = Jy Ampere s law may be (and often s) expressed n terms of the current densty vector as follows: z z ds J da where the area ntegral s over the regon enclosed by the path relevant to the lne ntegral (and J s n the +z drecton, out of the paper) Wth J unform throughout the sheet, then t s clear that the rght-hand sde of ths verson of Ampere s law should reduce, n ths problem, to JA = J yx = x ANALYZE (a) Fgure 9-84 certanly has the horzontal components of drawn correctly at ponts P and P', so the queston becomes: s t possble for to have vertcal components n the fgure?

33 85 Our focus s on pont P Suppose the magnetc feld s not parallel to the sheet, as shown n the upper left dagram If we reverse the drecton of the current, then the drecton of the feld wll also be reversed (as shown n the upper mddle dagram) Now, f we rotate the sheet by 8 about a lne that s perpendcular to the sheet, the feld wll rotate and pont n the drecton shown n the dagram on the upper rght The current dstrbuton now s exactly the same as the orgnal; however, comparng the upper left and upper rght dagrams, we see that the felds are not the same, unless the orgnal feld s parallel to the sheet and only has a horzontal component That s, the feld at P must be purely horzontal, as drawn n Fg 9-84 z (b) The path used n evaluatng ds s rectangular, of horzontal length x (the horzontal sdes passng through ponts P and P', respectvely) and vertcal sze y > y The vertcal sdes have no contrbuton to the ntegral snce s purely horzontal (so the scalar dot product produces zero for those sdes), and the horzontal sdes contrbute two equal terms, as shown next Ampere s law yelds x x LEARN In order to apply Ampere s law, the system must possess certan symmetry In the case of an nfnte current sheet, the symmetry s planar 8 Equaton 9-7 apples for each wre, wth r R d / g (by the Pythagorean theorem) The vertcal components of the felds cancel, and the two (dentcal) horzontal components add to yeld the fnal result d / d r r R d/ 6 5 T where (d/)/r s a trgonometrc factor to select the horzontal component It s clear that ths s equvalent to the expresson n the problem statement Usng the rght-hand rule, we fnd both horzontal components pont n the +x drecton Thus, n unt-vector 6 notaton, we have (5 T) ˆ 83 THINK The magnetc feld at P s the vector sum of the felds of the ndvdual wre segments b,

34 86 CHAPTER 9 EXPRESS The two small wre segments, each of length a/4, shown n Fg 9-86 nearest to pont P, are labeled and 8 n the fgure (below left) Let ˆk be a unt vector pontng nto the page We use the result of Problem 9-7: namely, the magnetc feld at P (shown n Fg 9-44 and upper rght) s L P 4 R L R Therefore, the magnetc felds due to the 8 segments are and P 8 a 4 P 8 P 8 3a 4 P4 5 P 4 a 4 P 7 b g b, a g, 6a b g b P 4 3a 4 P3 6 3a 4 g b g 3a 4 a 4 3, a a 4 a 4 3a 4 3 a b g b g b g ANALYZE Addng up all the contrbutons, the total magnetc feld at P s P 8 ˆ 3 ˆ n a 6 3 Pn( k) ( k) 4 T ( k) ˆ 4 T m A A 3 ( k) ˆ m 6 3

35 87 LEARN If pont P s located at the center of the square, then each segment would contrbute makng the total feld P P P8, 4 a 8 4 a center 8 P 84 (a) All wres carry parallel currents and attract each other; thus, the top wre s pulled downward by the other two: where L = 3 m Thus, L 5A 3A L 5A 5A F m m F 7 4 N (b) Now, the top wre s pushed upward by the center wre and pulled downward by the bottom wre: L5A3A L5A5A 5 F N m m 85 THINK The hollow conductor has cylndrcal symmetry, so Ampere s law can be appled to calculate the magnetc feld due to the current dstrbuton EXPRESS Ampere s law states that ds, enc where enc s the current enclosed by the closed path, or Amperan loop We choose the Amperan loop to be a crcle of radus r and concentrc wth the cylndrcal shell Snce the current s unformly dstrbuted throughout the cross secton of the shell, the enclosed current s r b r b enc a b a b ANALYZE (a) Thus, n the regon b < r < a, we have whch gves ca b r b enc ds r a b F hhg r b r (b) At r = a, the magnetc feld strength s I KJ

36 88 F hhg I KJ a b a b a a c CHAPTER 9 At r b, r b Fnally, for b = whch agrees wth Eq 9- r r a r a (c) The feld s zero for r < b and s equal to Eq 9-7 for r > a, so ths along wth the result of part (a) provdes a determnaton of over the full range of values The graph (wth SI unts understood) s shown below LEARN For r < b, the feld s zero, and for r > a, the feld decreases as /r In the regon b < r < a, the feld ncreases wth r as r b / r 86 We refer to the sde of length L as the long sde and that of length W as the short sde The center s a dstance W/ from the mdpont of each long sde, and s a dstance L/ from the mdpont of each short sde There are two of each type of sde, so the result of Problem 9-7 leads to L W L 4 W W L W 4 L b g b g b g b g The fnal form of ths expresson, shown n the problem statement, derves from fndng the common denomnator of the above result and addng them, whle notng that L W W L W L

37 89 87 (a) Equaton 9- apples for r < c Our sgn choce s such that s postve n the smaller cylnder and negatve n the larger one r, c r c (b) Equaton 9-7 apples n the regon between the conductors:, c r b r (c) Wthn the larger conductor we have a superposton of the feld due to the current n the nner conductor (stll obeyng Eq 9-7) plus the feld due to the (negatve) current n that part of the outer conductor at radus less than r The result s r b, r r a b b r a If desred, ths expresson can be smplfed to read r F HG a a r b (d) Outsde the coaxal cable, the net current enclosed s zero So = for r a (e) We test these expressons for one case If a and b (such that a > b) then we have the stuaton descrbed on page 696 of the textbook (f) Usng SI unts, the graph of the feld s shown to the rght I KJ 88 (a) Consder a segment of the projectle between y and y + dy We use Eq 9- to fnd the magnetc force on the segment, and Eq 9-7 for the magnetc feld of each semnfnte wre (the top ral referred to as wre and the bottom as wre ) The current n ral s n the drecton, and the current n ral s n the drecton The feld (n the regon between the wres) set up by wre s nto the paper (the k drecton) and that set up by wre s also nto the paper The force element (a functon of y) actng on the segment of the projectle (n whch the current flows n the j drecton) s gven below The coordnate orgn s at the bottom of the projectle

38 9 CHAPTER 9 df df df dy ˆj dy ˆj ˆ dy î dy 4 R w y 4y Thus, the force on the projectle s Rw ˆ w F df dy ln ˆ 4 R R w y y R (b) Usng the work-energy theorem, we have Thus, the fnal speed of the projectle s v K mv W z F ds FL f ext F W w m m R L f H G / / I K J L F I NM HG K J O ext ln QP L c T m / Ahc Ah ln b cm / 67cmgb mgo NM 3 c kgh QP 3 3 m / s /