21 MAGNETIC FORCES AND MAGNETIC FIELDS

Transcription

1 CHAPTER 1 MAGNETIC ORCES AND MAGNETIC IELDS ANSWERS TO OCUS ON CONCEPTS QUESTIONS 1. (d) Right-Hand Rule No. 1 gives the diection of the magnetic foce as x fo both dawings A and. In dawing C, the velocity is paallel to the magnetic field, so the magnetic foce is zeo.. (b) Using Right-Hand Rule No. 1 (see Section 1.), we find that the diection of the magnetic foce on a positively chaged paticle is to the west. Revesing this diection because the paticle is a negative electon, we see that the magnetic foce acting on it points to the east. 3. (a) Using Right-Hand Rule No. 1 (see Section 1.), we find that the diection of the magnetic foce on a positively chaged paticle is staight down towad the bottom of the sceen. 4. = T, south 5. (c) The electic foce points out of the sceen, in the diection of the electic field. An application of Right-Hand Rule No. 1 shows that the magnetic foce also points out of the sceen, paallel to the electic foce. When two foces have the same diection, the magnitude of thei sum has the lagest possible value. 6. (e) In this situation, the centipetal foce, c = mv / (Equation 5.3), is povided by the magnetic foce, = qv sin 9. (Equation 1.1), so mv / = qv sin 9.. Thus, q mv, and the chage magnitude q is invesely popotional to the adius. Since the / adius of cuve 1 is smalle than that of cuve, and the adius of cuve is smalle than that of cuve 3, we conclude that q 1 is lage than q, which is lage than q (a) The magnetic foce that acts on the electon in egions 1 and is always pependicula to its path, so the foce does no wok. Accoding to the wok-enegy theoem, Equation 6.3, the kinetic enegy, and hence speed, of the electon does not change when no wok is done. 8. (d) Accoding to Equation 1., the adius of the cicula path is given by mv / q. Since v, q, and ae the same fo the poton and the electon, the moe-massive poton tavels on the cicle with the geate adius. The centipetal foce c acting on the poton must point towad the cente of the cicle. In this case, the centipetal foce is povided by the magnetic foce. Accoding to Right-Hand Rule No. 1, the diection of is elated to the velocity v and the magnetic field. An application of this ule shows that the poton

2 Chapte 1 Answes to ocus on Concepts Questions 177 must tavel counteclockwise aound the cicle in ode that the magnetic foce point towad the cente of the cicle. 9. poton / electon = (c) When, fo example, a paticle moves pependicula to a magnetic field, the field exets a foce that causes the paticle to move on a cicula path. Any object moving on a cicula path expeiences a centipetal acceleation. 11. = 3. N, along the y axis 1. (e) The magnetic field is diected fom the noth pole to the south pole (Section 1.1). Accoding to Right-Hand Rule No. 1 (Section 1.5), the magnetic foce in dawing 1 points noth. 13. (c) Thee is no net foce. No foce is exeted on the top and bottom wies, because the cuent is eithe in the same o opposite diection as the magnetic field. Accoding to Right- Hand Rule No. 1 (Section 1.5), the left side of the loop expeiences a foce that is diected into the sceen, and the ight side expeiences a foce that is diected out of the sceen (towad the eade). The two foces have the same magnitude, so the net foce is zeo. The two foces on the left and ight sides, howeve, do exet a net toque on the loop with espect to the axis. 14. (d) Accoding to Right-Hand Rule No. 1 (Section 1.5), all fou sides of the loop ae subject to foces that ae diected pependiculaly towad the opposite side of the squae. In addition, the foces have the same magnitude, so the net foce is zeo. A toque consists of a foce and a leve am. o the axis of otation though the cente of the loop, the leve am fo each of the fou foces is zeo, so the net toque is also zeo. 15. N = 86 tuns 16. (a) Right-Hand Rule No. (Section 1.7) indicates that the magnetic field fom the top wie in points into the sceen and that fom the bottom wie points out of the sceen. Thus, the net magnetic field in is zeo. Also, the magnetic field fom the hoizontal wie in 4 points into the sceen and that fom the vetical wie points out of the sceen. Thus, the net magnetic field in 4 is also zeo. 17. (b) Two wies attact each othe when the cuents ae in the same diection and epel each othe when the cuents ae in the opposite diection (see Section 1.7). Wie is attacted to A and epelled by C, but the foces einfoce one anothe. Theefoe, the net foce has a magnitude of A + C, whee A and C ae the magnitudes of the foces exeted on wie by A and on wie by C. Howeve, A = C, since the wies A and C ae equidistance fom. Theefoe, the net foce on wie has a magnitude of A. The net foce exeted on wie A is less than this, because wie A is attacted to and epelled by C,

3 178 MAGNETIC ORCES AND MAGNETIC IELDS the foces patially canceling. The net foce expected on wie C is also less than that on A. It is epelled by both A and, but A is twice as fa away as. 18. (a) The magnetic field in the egion inside a solenoid is constant, both in magnitude and in diection (see Section 1.7). 19. = T, out of the sceen. (d) Accoding to Ampee s law, I is the net cuent passing though the suface bounded by the path. The net cuent is 3 A + 4 A 5 A = A.

4 Chapte 1 Poblems 179 CHAPTER 1 MAGNETIC ORCES AND MAGNETIC IELDS PROLEMS 1. SSM REASONING The electon s acceleation is elated to the net foce acting on it by Newton s second law: a = /m (Equation 4.1), whee m is the electon s mass. Since we ae ignoing the gavitational foce, the net foce is that caused by the magnetic foce, whose magnitude is expessed by Equation 1.1 as = q v sin. Thus, the magnitude of a q vsin / m. the electon s acceleation can be witten as SOLUTION We note that = 9., since the velocity of the electon is pependicula to the magnetic field. The magnitude of the electon s chage is C, and the electon s mass is kg (see the inside of the font cove), so a q v sin m C.1 1 m/s T sin m/s kg 1. REASONING The magnitude of the magnetic field is (Equation 1.1), q vsin whee is the magnitude of the magnetic foce on the chage, whose magnitude is q and whose velocity has a magnitude v and makes an angle θ with the diection of the field. oth the poton in pat a and the electon in pat b have the same chage magnitude of 19 q C. Theefoe, the magnetic field has the same magnitude in both pats of the poblem. Howeve, the diection of the field is diffeent fo the poton and the electon. This is because the poton chage is positive, wheeas the electon chage is negative. inally, we note that the magnitude of the magnetic foce is a maximum, which means that the velocity is pependicula to the magnetic field, so that 9.. SOLUTION a. Using Equation 1.1, we find that the magnitude of the magnetic field fo the poton is N q vsin C m / s sin T

5 18 MAGNETIC ORCES AND MAGNETIC IELDS Since the poton is taveling due east and the foce points due south, we find fom ight hand ule no. 1 that the magnetic field points upwad, pependicula to the eath s suface. b. o the electon, the magnitude of the field is the same as fo the poton, since the two chages have the same magnitude. Thus,.11 T. Since the electon is a negative chage, howeve, ight-hand ule no. 1 eveals that the field diection is downwad, pependicula to the eath s suface. 3. SSM REASONING Accoding to Equation 1.1, the magnitude of the magnetic foce on a moving chage is q vsin. Since the magnetic field points due noth and the poton moves eastwad, 9.uthemoe, since the magnetic foce on the moving poton balances its weight, we have mg q v sin, whee m is the mass of the poton. This expession can be solved fo the speed v. SOLUTION Solving fo the speed v, we have 7 mg ( kg)(9.8 m/s ) 3 v 4.11 m/s q sin 19 5 (1.6 1 C)(.51 T) sin REASONING The magnitude of the magnetic field is given by (Equation 1.1), and we will apply this expession diectly to obtain. q vsin SOLUTION The chage 6 q C tavels with a speed v = m/s at an angle of = 5 with espect to a magnetic field of magnitude and expeiences a foce of magnitude = N. Accoding to Equation 1.1, the field magnitude is 5.41 N T q vsin C 7.41 m/s sin 5 3 Note in paticula that it is only the magnitude q of the chage that appeas in this calculation. The algebaic sign of the chage does not affect the esult. 5. REASONING Accoding to Equation 1.1, the magnetic foce has a magnitude of = q v sin, whee q is the magnitude of the chage, is the magnitude of the magnetic field, v is the speed, and is the angle of the velocity with espect to the field. As inceases fom to 9, the foce inceases, so the angle must lie between 5 and 9.

6 Chapte 1 Poblems 181 SOLUTION Letting 1 = 5 and be the desied angle, we can apply Equation 1.1 to both situations as follows: q v sin and q v sin 1 Situation 1 Situation Dividing the equation fo situation by the equation fo situation 1 gives q v sin q v sin 1 o sin =sin =sin sin REASONING A moving chage expeiences no magnetic foce when its velocity points in the diection of the magnetic field o in the diection opposite to the magnetic field. Thus, the magnetic field must point eithe in the diection of the +x axis o in the diection of the x axis. If a moving chage expeiences the maximum possible magnetic foce when moving in a magnetic field, then the velocity must be pependicula to the field. In othe wods, the angle that the chage s velocity makes with espect to the magnetic field is = 9. SOLUTION The magnitude of the magnetic field can be detemined using Equation 1.1:.48 N qvsin 8.1 C 5.1 m/s sin T In this calculation we use = 9, because the.48-n foce is the maximum possible foce. Since the paticle expeiences no magnetic foce when it moves along the +x axis, we can conclude that the magnetic field points eithe in the diection of the +x axis o in the diection of the x axis. 7. REASONING The magnetic field applies the maximum magnetic foce to the moving chage, because the motion is pependicula to the field. This foce is pependicula to both the field and the velocity. The electic field applies an electic foce to the chage that is in the same diection as the field, since the chage is positive. These two foces ae shown in the dawing, and they ae pependicula to one anothe. Theefoe, the magnitude of the net field can be obtained using the Pythagoean theoem. magnetic SOLUTION Accoding to Equation 1.1, the magnetic foce has a magnitude of magnetic = q v sin, whee q is the magnitude of the v electic E

7 18 MAGNETIC ORCES AND MAGNETIC IELDS chage, is the magnitude of the magnetic field, v is the speed, and = 9 is the angle of the velocity with espect to the field. Thus, magnetic = q v. Accoding to Equation 18., the electic foce has a magnitude of electic = q E. Using the Pythagoean theoem, we find the magnitude of the net foce to be magnetic electic q v q E q v E C 3.11 m/s 1.1 T 4.61 N/C 1.11 N 8. REASONING Accoding to Equation 1.1, the magnetic foce has a magnitude of = q v sin. The field and the diectional angle ae the same fo each paticle. Paticle 1, howeve, tavels faste than paticle. y itself, a faste speed v would lead to a geate foce magnitude. ut the foce on each paticle is the same. Theefoe, paticle 1 must have a smalle chage to counteact the effect of its geate speed. SOLUTION Applying Equation 1.1 to each paticle, we have q v sin and q v sin 1 1 Paticle 1 Paticle Dividing the equation fo paticle 1 by the equation fo paticle and emembeing that v 1 = 3v gives q1 v1 sin q1 v1 q1 v v 1 o 1 o q v sin q v q v 3v REASONING The positive plate has a chage q and is moving downwad with a speed v at ight angles to a magnetic field of magnitude. The magnitude of the magnetic foce exeted on the positive plate is = q v sin 9.. The chage on the positive plate is elated to the magnitude E of the electic field that exists between the plates by (see Equation 18.4) q = AE, whee A is the aea of the positive plate. Substituting this expession fo q into = q v sin 9.gives the answe in tems of known quantities. SOLUTION AE v C /(N m ) m 17 N/C 3 m/s 3.6 T N

8 Chapte 1 Poblems 183 An application of Right-Hand Rule No. 1 shows that the magnetic foce is pependicula to the plane of the page and diectedoutof thepage, towad the eade. 1. REASONING The dawing on the left shows the diections of the two magnetic fields, as well as the velocity v of the paticle. Each component of the magnetic field is pependicula to the velocity, so each exets a magnetic foce on the paticle. The magnitude of the foce is = q v sin (Equation 1.1), and the diection can be detemined by using Right-Hand Rule No. 1 (RHR-1). The magnitude and diection of the net foce can be found by using tigonomety. +z +z +y y =.65 T v +y +x 1 +x x =.48 T SOLUTION a. The magnitude 1 of the magnetic foce due to the.48-t magnetic field is q v x sin 9..1 C 4.1 m/s.48 T 4. 1 N The magnitude of the magnetic foce due to the.65-t magnetic field is q v y sin 9..1 C 4.1 m/s.65 T N The diections of the foces ae found using RHR-1, and they ae indicated in the dawing on the ight. Also shown is the net foce, as well as the angle that it makes with espect to the +x axis. Since the foces ae at ight angles to each othe, we can use the Pythagoean theoem to find the magnitude of the net foce: N 5.51 N N b. The angle can be detemined by using the invese tangent function: N N tan tan 36

9 184 MAGNETIC ORCES AND MAGNETIC IELDS 11. SSM REASONING The diection in which the electons ae deflected can be detemined using Right-Hand Rule No. 1 and evesing the diection of the foce (RHR-1 applies to positive chages, and electons ae negatively chaged). Each electon expeiences an acceleation a given by Newton s second law of motion, a = /m, whee is the net foce and m is the mass of the electon. The only foce acting on the electon is the magnetic foce, = q v sin, so it is the net foce. The speed v of the electon is elated to its kinetic enegy KE by the elation KE = 1 mv. Thus, we have enough infomation to find the acceleation. SOLUTION a. Accoding to RHR-1, if you extend you ight hand so that you finges point along the diection of the magnetic field and you thumb points in the diection of the velocity v of a positive chage, you palm will face in the diection of the foce on the positive chage. o the electon in question, the finges of the ight hand should be oiented downwad (diection of ) with the thumb pointing to the east (diection of v). The palm of the ight hand points due noth (the diection of on a positive chage). Since the electon is negatively chaged, it will be deflected due south. b. The acceleation of an electon is given by Newton s second law, whee the net foce is the magnetic foce. Thus, q v sin a m m Since the kinetic enegy is KE 1 mv, the speed of the electon is v KE / m. Thus, the acceleation of the electon is a q v sin m q (KE) m m sin J C. 1 T sin kg.55 1 m/s kg REASONING Since e C, we need to detemine whethe the chage has a magnitude of q C o q 3. 1 C. We can do this by using mv (Equation 1.), which gives the adius of the cicula path in tems of the mass q

10 Chapte 1 Poblems 185 m of the chaged paticle, the paticle s speed v, and the magnitude of the magnetic field. This is possible since values ae available fo, m, v, and in Equation 1.. SOLUTION Solving Equation 1. fo q, we find that q kg4.4 1 m/s mv C 19 This chage is e C helium atom is.75 T.1 m. We can see, then, that the chage of the ionized e. 13. SSM REASONING The adius of the cicula path is given by mv (Equation q 1.), whee m and v ae the mass and speed of the paticle, espectively, q is the magnitude of the chage, and is the magnitude of the magnetic field. This expession can be solved diectly fo, since, m, and v ae given and q = +e, whee e = C. SOLUTION Solving Equation 1. fo gives kg7.1 m/s mv q C.1 m.14 T 14. REASONING The time t that it takes the paticle to complete one evolution is the time to tavel a distance d = equal to the cicumfeence of a cicle of adius at a speed v. om Equation.1, we know that speed is the atio of distance to elapsed time v d / t, so the elapsed time is the atio of distance to speed: d t (1) v v ecause the paticle follows a cicula path that is pependicula to the extenal magnetic mv field of magnitude, the adius of the path is given by (Equation 1.), whee m is q the mass and q is the magnitude of the chage of the paticle. We will use Equation 1. to detemine the speed of the paticle, and then Equation (1) to find the time fo one complete evolution. SOLUTION Solving mv (Equation 1.) fo v yields q q q v m m ()

11 186 MAGNETIC ORCES AND MAGNETIC IELDS In the last step of Equation (), we have expessed the speed v explicitly in tems of the chage-to-mass atio q /m of the paticle. Substituting Equation () into Equation (1), we obtain t v q m s q C/kg.7 T m 15. REASONING a. The dawing shows the velocity v of the paticle at the top of its path. The magnetic foce, which povides the centipetal foce, must be diected towad the cente of the cicula path. Since the diections of v,, and ae known, we can use Right-Hand Rule No. 1 (RHR-1) to detemine if the chage is positive o negative. v (out of pape) b. The adius of the cicula path followed by a chaged paticle is given by Equation 1. as mv / q. The mass m of the paticle can be obtained diectly fom this elation, since all othe vaiables ae known. SOLUTION a. If the paticle wee positively chaged, an application of RHR-1 would show that the foce would be diected staight up, opposite to that shown in the dawing. Thus, the chage on the paticle must be negative. b. Solving Equation 1. fo the mass of the paticle gives C.48 T96 m q 3 m.7 1 kg v 14 m/s 16. REASONING Equation 1. gives the adius of the cicula path as = mv/( q ), whee m, v, and q ae, espectively, the mass, speed, and chage magnitude of the paticle, and is the magnitude of the magnetic field. We wish the adius to be the same fo both the poton and the electon. The speed v and the chage magnitude q ae the same fo the poton and the electon, but the mass of the electon is kg, while that of the poton is kg. Theefoe, to offset the effect of the smalle electon mass m in Equation 1., the magnitude of the field must be educed fo the electon.

12 Chapte 1 Poblems 187 SOLUTION Applying Equation 1. to the poton and the electon, both of which cay chages of the same magnitude q = e, we obtain mv p mv e and e e Poton p e Electon Dividing the poton-equation by the electon-equation gives Solving fo e, we obtain m mv p e m o 1 mv m e p p e e e p e kg.5 T e p 4 e m 7 p kg.7 1 T 17. SSM REASONING As discussed in Section 1.4, the mass m of a singly-ionized paticle that has been acceleated though a potential diffeence V and injected into a magnetic field of magnitude is given by e m V (1) 19 whee e 1.61 C is the magnitude of the chage of an electon and is the adius of the paticle s path. If the beyllium-1 ions each the same position in the detecto as the beyllium-7 ions, both types of ions must have the same path adius. Additionally, the e acceleating potential diffeence V is kept constant, so we see that the quantity V in Equation (1) is the same fo both types of ions. SOLUTION All that diffes between the two situations ae the masses (m 7, m 1 ) of the ions and the magnitudes of the magnetic fields ( 7, 1 ). Solving Equation (1) fo the constant e quantity V, we obtain e m1 m7 V 1 7 Same fo both ions ()

13 188 MAGNETIC ORCES AND MAGNETIC IELDS Solving Equation () fo 1, we find that 7 m1 m m 7 7 m kg o.83 T.338 T kg 18. REASONING Section 1.4 discusses how a mass spectomete detemines the mass m of an ion that has a chage of +e, whee e = C. This ion acceleates though a potential diffeence V and follows a cicula path (adius ) because of a magnetic field e (magnitude ). The mass of the ion is m. If the gold ions in this poblem had a V chage of +e, we could solve this expession diectly fo the adius. Howeve, the chage of the gold ions is +e, so that befoe using the expession, we need to eplace e by e. SOLUTION Replacing e by e in the expession fom Section 1.4 gives the mass as e m. Solving this equation fo the adius, we find that V 5 3 mv kg 1. 1 V e C.5 T m 19. REASONING The speed of the -paticle can be obtained by applying the pinciple of consevation of enegy, ecognizing that the total enegy is the sum of the paticle s kinetic enegy and electic potential enegy, the gavitational potential enegy being negligible in compaison. Once the speed is known, Equation 1.1 can be used to obtain the magnitude of the magnetic foce that acts on the paticle. Lastly, the adius of its cicula path can be obtained diectly fom Equation 1.. SOLUTION a. Using A and to denote the initial positions, espectively, the pinciple of consevation of enegy can be witten as follows: 1 1 mv EPE mv A EPEA inal kinetic inal electic Initial kinetic Initial electic enegy potential enegy enegy potential enegy (1) Using Equation 19.3 to expess the electic potential enegy of the chage q as EPE = q V, whee V is the electic potential, we find fom Equation (1) that 1 1 mv qv mv A qv A () Since the paticle stats fom est, we have that v A = m/s, and Equation () indicates that

14 Chapte 1 Poblems 189 v C V q V V m kg A m/s b. Accoding to Equation 1.1, the magnitude of the magnetic foce that acts on the paticle is q v sin 1.61 C 1.81 m/s. T sin N whee = 9., since the paticle tavels pependicula to the field at all times. c. Accoding to Equation 1., the adius of the cicula path on which the paticle tavels is kg1.8 1 m/s C. T mv q.1 m. REASONING Equation 1. gives the adius of the cicula path as = mv/( q ), whee m, v, and q ae, espectively, the mass, speed, and chage magnitude of the paticle, and is the magnitude of the magnetic field. We can detemine the speed of each paticle by employing the pinciple of consevation of enegy. The electic potential enegy lost as the paticles acceleate is conveted into kinetic enegy. Equation 19.4 indicates that the electic potential enegy lost is q V, whee q is the magnitude of the chage and V is the electic potential diffeence. Since q and V ae the same fo each paticle, each loses the same amount of potential enegy. Enegy consevation, then, dictates that each gains the same amount of kinetic enegy. Since each paticle stats fom est, each entes the magnetic field with the same amount of kinetic enegy. SOLUTION Accoding to Equation 1., = mv/( q ). To detemine the speed v with which each paticle entes the field, we use Equation 19.4 and the enegy-consevation pinciple as follows: Electic potential enegy lost 1 q V mv o v Kinetic enegy gained qv Substituting this esult into Equation 1. gives the adius of the cicula motion: mv m qv 1 mv q q m q Applying this esult to each paticle, we obtain m

15 19 MAGNETIC ORCES AND MAGNETIC IELDS 1 m V 1 m V q q 1 1 and Dividing by 1 gives Paticle 1 Paticle 1 1 mv q mv q m m m 8 1 m kg 1 cm 19 cm.3 1 kg 1. REASONING The dawing shows the velocity v of the cabon atoms as they ente the magnetic field. The diamete of the cicula path followed by the cabon-1 atoms is labeled as 1, and that of the cabon-13 atoms as 13, whee denotes the adius of the path. The adius is given by Equation 1. as mv / q, whee q is the chage on the ion (q = +e). The diffeence d in the diametes is d (see the dawing) SOLUTION The spatial sepaation between the two isotopes afte they have taveled though a half-cicle is m v m v v d m m e e e m/s C.85 T kg kg m v 1 13 (out of pape). REASONING The adius of the cicula path is given by Equation 1. as = mv/( q ), whee m is the mass of the species, v is the speed, q is the magnitude of the chage, and is the magnitude of the magnetic field. To use this expession, we must know something about the speed. Infomation about the speed can be obtained by applying the consevation of enegy pinciple. The electic potential enegy lost as a chaged paticle falls fom a highe to a lowe electic potential is gained by the paticle as kinetic enegy.

16 Chapte 1 Poblems 191 SOLUTION o an electic potential diffeence V and a chage q, the electic potential enegy lost is q V, accoding to Equation The kinetic enegy gained is 1 mv. Thus, enegy consevation dictates that 1 q V mv o v qv m Substituting this esult into Equation 1. fo the adius gives mv m qv 1 mv q q m q Using e to denote the magnitude of the chage on an electon, we note that the chage fo species X + is +e, while the chage fo species X + is +e. With this in mind, we find fo the atio of the adii that 1 mv 1 e mv e 3. SSM REASONING When the poton moves in the magnetic field, its tajectoy is a cicula path. The poton will just miss the opposite plate if the distance between the plates mv / q. is equal to the adius of the path. The adius is given by Equation 1. as This elation can be used to find the magnitude of the magnetic field, since values fo all the othe vaiables ae known. SOLUTION Solving the elation mv / q fo the magnitude of the magnetic field, and ealizing that the adius is equal to the plate sepaation, we find that kg3.51 m/s mv q C.3 m.16 T The values fo the mass and the magnitude of the chage (which is the same as that of the electon) have been taken fom the inside of the font cove.

17 19 MAGNETIC ORCES AND MAGNETIC IELDS 4. REASONING The magnitude of the magnetic foce acting on the paticle is elated to its speed v by q v sin (Equation 1.1), whee is the magnitude of the magnetic field, q is the paticle s chage, and is the angle between the magnetic field and the paticle s velocity v. As the dawing shows, the vecto v (east, to the ight) is pependicula to the vecto (south, out of the page). Theefoe, = 9, and Equation 1.1 becomes q v sin9 q v (1) In addition to the magnetic foce, thee is also an electic foce of magnitude E acting on the paticle. This foce magnitude does not depend upon the speed v of the paticle, as we see fom q E E (Equation 18.). The paticle is positively chaged, so the electic foce acting on it points upwad in the same diection as the electic field. y Right-Hand Rule No.1, the magnetic foce acting on the positively chaged paticle points down, and is theefoe opposite to the electic foce. The net foce on the paticle points upwad, so we conclude that the electic foce is geate than the magnetic foce. Thus, the magnitude of the net foce acting on the paticle is equal to the magnitude of the electic foce minus the magnitude of the magnetic foce: () E We also note that paticles taveling at a speed v = m/s expeience no net foce. Theefoe, E = fo paticles moving at the speed v. SOLUTION Substituting Equation (1) and q E (Equation 18.) into Equation () yields E q E q v (3) The magnetic field magnitude is not given, but, as noted above, fo paticles with speed v = m/s, the magnetic foce of Equation (1), q v, is equal to the electic foce q E (Equation 18.). Theefoe, we have that E q v q Substituting Equation (4) into Equation (3) yields West E E o (4) v q v E q E q q v E (5) v E Up Down (out of page) v East

18 Chapte 1 Poblems 193 Solving Equation (5) fo v, we obtain q v E v q E o 1 o v v 1 v v q E q E Substituting the given values into Equation (6), we find that (6) N 3 v v m/s m/s q 1 E 4.1 C47 N/C 5. SSM REASONING The paticle tavels in a semicicula path of adius, whee is mv given by Equation 1.. The time spent by the paticle in the magnetic field is q given by t s/v, whee s is the distance taveled by the paticle and v is its speed. The distance s is equal to one-half the cicumfeence of a cicle ( s = ). SOLUTION We find that t 8 s mv m (6.1 kg) s v v v q q 6 (7.1 C)(3. T) 6. REASONING When the electon tavels pependicula to a magnetic field, its path is a mv / q. All the vaiables cicle. The adius of the cicle is given by Equation 1. as in this elation ae known, except the speed v. Howeve, the speed is elated to the electon s kinetic enegy KE by KE = 1 mv (Equation 6.). y combining these two elations, we will be able to find the adius of the path. SOLUTION Solving the elation mv / q give KE = 1 mv fo the speed and substituting the esult into KE m mv m m KE q q q kg.1 J C5.3 1 T.71 m Values fo the mass and chage of the electon have been taken fom the inside of the font cove.

19 194 MAGNETIC ORCES AND MAGNETIC IELDS 7. REASONING a. When the paticle moves in the magnetic field, its path is cicula. To keep the paticle moving on a cicula path, it must expeience a centipetal foce, the magnitude of which is given by Equation 5.3 as mv. In the pesent situation, the magnetic foce c / funishes the centipetal foce, so c =. The mass m and speed v of the paticle ae known, but the adius of the path is not. Howeve, the paticle tavels at a constant speed, so in a time t the distance s it tavels is s = vt. ut the distance is one-quate of the cicumfeence s 1. y combining these thee elations, we can detemine the of a cicle, so 4 magnitude of the magnetic foce. b. The magnitude of the magnetic foce is given by Equation 1.1 as q vsin. Since, v,, and ae known, this elation can be used to detemine the magnitude q of the chage. SOLUTION a. The magnetic foce, which povides the centipetal foce, is mv /. Solving the s 1 fo the adius and substituting s = vt into the esult gives elation 4 s vt Using this expession fo in Equation 5.3, we find that the magnitude of the magnetic foce is kg85 m/s mv mv mv N vt t 3. 1 s b. Solving the elation q vsin fo the magnitude q of the chage and noting that = 9. (since the velocity of the paticle is pependicula to the magnetic field), we find that N 4 q C vsin m/s.31 T sin REASONING AND SOLUTION The magnitudes of the magnetic and electic foces must be equal. Theefoe, = E o q v = q E This elation can be solved to give the speed of the paticle, v = E/. We also know that when the electic field is tuned off, the paticle tavels in a cicula path of adius = mv/( q ). Substituting v = E/ into this equation and solving fo q /m gives

20 Chapte 1 Poblems m.36 T 3 q E N/C C/kg m REASONING AND SOLUTION The following dawings show the two cicula paths leading to the taget T when the poton is pojected fom the oigin O. In each case, the cente of the cicle is at C. Since the taget is located at x =.1 m and y =.1 m, the adius of each cicle is =.1 m. The speed with which the poton is pojected can be obtained fom Equation 1., if we emembe that the chage and mass of a poton ae q = C and m = kg, espectively: 19.1 m1.6 1 C.1 T q 4 v m/ s m kg 3. REASONING A magnetic field exets no foce on a cuent-caying wie that is diected along the same diection as the field. This follows diectly fom IL sin (Equation 1.3), which gives the magnitude of the magnetic foce that acts on a wie of length L that is diected at an angle with espect to a magnetic field of magnitude and caies a cuent I. When = o 18, = N. Theefoe, we need only apply Equation 1.3 to the hoizontal component of the eath s magnetic field in this poblem. The diection of the magnetic foce can be detemined with the aid of RHR-1 (finges point in diection of the field, thumb points in the diection of the cuent, palm faces in the diection of the magnetic foce). SOLUTION Accoding to Equation 1.3, the magnitude of the magnetic foce exeted on the wie by the hoizontal component of the eath s field is 5 3 IL sin 8 A 6. m 1.81 T sin N

21 196 MAGNETIC ORCES AND MAGNETIC IELDS Note that = 9. because the field component points towad the geogaphic noth and the cuent is diected pependiculaly into the gound. The application of RHR-1 (finges point due noth, thumb points pependiculaly into the gound, palm faces due east) eveals that the diection of the magnetic foce is due east 31. SSM REASONING The magnitude of the magnetic foce expeienced by the wie is given by IL sin (Equation 1.3), whee I is the cuent, L is the length of the wie, is the magnitude of the eath s magnetic field, and is the angle between the diection of the cuent and the magnetic field. Since all the vaiables ae known except, we can use this elation to find its value. SOLUTION Solving IL sin fo the magnitude of the magnetic field, we have.15 N T IL sin 75 A 45 m sin REASONING The magnitude of the foce on a cuent I is given by Equation 1.3 as IL sin (Equation 1.3), whee L is the length of the wie and is the angle between the wie and a magnetic field that has a magnitude. We can apply this equation to both situations. The key to the solution is the fact that L,, and θ, although unknown, have the same values in both cases. This fact will allow us to eliminate them algebaically fom the calculation of the unknown cuent. SOLUTION Initially we have 1.3 N and I1.7 A. In the second case, we have.47 N and I. Applying this Equation 1.3 to both situations we have I Lsin and I L sin 1 1 Dividing the ight-hand equation by the left-hand equation gives IL sin I.47 N o I I1.7 A 4. A 1 I1L sin I1 1.3 N 33. REASONING The magnitude of the extenal magnetic field is popotional to the magnitude of the magnetic foce exeted on the wie, accoding to IL sin (Equation 1.3), whee L is the length of the wie, I is the cuent it caies, and is the angle between the diections of the cuent and the magnetic field. When the wie is hoizontal, the magnetic foce is zeo, indicating that sin. The only angles fo which this holds ae = and = 18. Theefoe, the extenal magnetic field must be hoizontal, and when the wie is tilted upwads at an angle of 19, the angle between the diections of the cuent and the magnetic field must be = 19.

22 Chapte 1 Poblems 197 SOLUTION Solving IL sin (Equation 1.3) fo yields 4.41 N T IL sin 7.5 A.53 m sin REASONING We begin by noting that segments A and C ae both pependicula to the magnetic field. Theefoe, they expeience magnetic foces. Howeve, segment CD is paallel to the field. As a esult no magnetic foce acts on it. Accoding to Equation 1.3, the magnitude of the magnetic foce on a cuent I is = IL sin, whee L is the length of the wie segment and is the angle that the cuent makes with espect to the magnetic field. o both segments A and C the value of the cuent is the same and the value of is 9. The length of segment A is geate, howeve. ecause of its geate length, segment A expeiences the geate foce. SOLUTION Using = IL sin (Equation 1.3), we find that the magnitudes of the magnetic foces acting on the segments ae: Segment A Segment C Segment CD b gb gb g b gb gb g b gb gb g IL sin. 8 A 1. 1 m. 6 T sin 9. 8 N IL sin. 8 A.55 m. 6 T sin 9. 4 N IL sin. 8 A.55 m. 6 T sin N 35. SSM REASONING Accoding to Equation 1.3, the magnetic foce has a magnitude of = IL sin, whee I is the cuent, is the magnitude of the magnetic field, L is the length of the wie, and = 9 is the angle of the wie with espect to the field. SOLUTION Using Equation 1.3, we find that L I sin N A T sin58 b gc h. 7 m 36. REASONING Each wie expeiences a foce due to the magnetic field. The magnitude of the foce is given by IL sin (Equation 1.3), whee I is the cuent, L is the length of the wie, is the magnitude of the magnetic field, and is the angle between the diection of the cuent and the magnetic field. Since the cuents in the two wies ae in opposite diections, the magnetic foce acting on one wie is opposite to that acting on the othe. Thus, the net foce acting on the two-wie unit is the diffeence between the magnitudes of the foces acting on each wie. SOLUTION The length L of each wie, the magnetic field, and the angle ae the same fo both wies. Denoting the cuent in one of the wies as I 1 = 7. A and the cuent in the othe as I, the magnitude net of the net magnetic foce acting on the two-wie unit is

23 198 MAGNETIC ORCES AND MAGNETIC IELDS I Lsin IL sin I I Lsin net 1 1 Solving fo the unknown cuent I gives net 3.13 N I I1 7. A 3. A Lsin.4 m.36 T sin REASONING The magnitude of the magnetic foce exeted on a long staight wie is given by Equation 1. 3 as = IL sin. The diection of the magnetic foce is pedicted by Right-Hand Rule No. 1. The net foce on the tiangula loop is the vecto sum of the foces on the thee sides. SOLUTION a. The diection of the cuent in side A is opposite to the diection of the magnetic field, so the angle between them is = 18. The magnitude of the magnetic foce is IL sin IL sin 18 N A o the side C, the angle is = 55., and the length of the side is The magnetic foce is L. m m cos. b gb gb g IL sin 4. 7 A m 1. 8 T sin N C An application of Right-Hand No. 1 shows that the magnetic foce on side C is diected pependiculaly out of the pape, towad the eade. o the side AC, the angle is = 9.. We see that the length of the side is The magnetic foce is L = (. m) tan 55. =.86 m b gb gb g IL sin 4. 7 A. 86 m 1. 8 T sin N AC An application of Right-Hand No. 1 shows that the magnetic foce on side AC is diected pependiculaly into the pape, away fom the eade. b. The net foce is the vecto sum of the foces on the thee sides. Taking the positive diection as being out of the pape, the net foce is b g N + 4. N + 4. N N

24 Chapte 1 Poblems REASONING AND SOLUTION a. om Right-Hand Rule No. 1, if we extend the ight hand so that the finges point in the diection of the magnetic field, and the thumb points in the diection of the cuent, the palm of the hand faces the diection of the magnetic foce on the cuent. The spings will stetch when the magnetic foce exeted on the coppe od is downwad, towad the bottom of the page. Theefoe, if you extend you ight hand with you finges pointing out of the page and the palm of you hand facing the bottom of the page, you thumb points left-to-ight along the coppe od. Thus, the cuent flows left - to - ight in the coppe od. b. The downwad magnetic foce exeted on the coppe od is, accoding to Equation 1.3 IL sin (1 A)(.85 m)(.16 T) sin 9. =1.6 N Accoding to Equation 1.1, the foce Applied x Applied x equied to stetch each sping is kx, whee k is the sping constant. Since thee ae two spings, we know that the magnetic foce exeted on the cuent must equal Solving fo x, we find that Applied x, so that 1.6 N x m k (75 N/m) Applied x kx. 39. SSM REASONING Since the od does not otate about the axis at P, the net toque elative to that axis must be zeo; = (Equation 9.). Thee ae two toques that must be consideed, one due to the magnetic foce and anothe due to the weight of the od. We conside both of these to act at the od's cente of gavity, which is at the geometical cente of the od (length = L), because the od is unifom. Accoding to Right-Hand Rule No. 1, the magnetic foce acts pependicula to the od and is diected up and to the left in the dawing. Theefoe, the magnetic toque is a counteclockwise (positive) toque. Equation 1.3 gives the magnitude of the magnetic foce as = IL sin, since the cuent is pependicula to the magnetic field. The weight is mg and acts downwad, poducing a clockwise (negative) toque. The magnitude of each toque is the magnitude of the foce times the leve am (Equation 9.1). Thus, we have fo the toques: IL L mg L magnetic / and weight / cos foce leve am foce leve am Setting the sum of these toques equal to zeo will enable us to find the angle that the od makes with the gound.

25 11 MAGNETIC ORCES AND MAGNETIC IELDS SOLUTION Setting the sum of the toques equal to zeo gives = magnetic + weight =, and we have ILL / mg L / cos o IL cos mg A.45 m.36 T cos.94 kg9.8 m/s REASONING Thee ae fou foces that act on the wie: the magnetic foce (magnitude ), the weight mg of the wie, and the tension in each of the two stings (magnitude T in each sting). Since thee ae two stings, the following dawing shows the total tension as T. The magnitude of the magnetic foce is given by = I L sin (Equation 1.3), whee I is the cuent, L is the length of the wie, is the magnitude of the magnetic field, and is the angle between the wie and the magnetic field. In this poblem = 9.. The diection of the magnetic foce is given by Right-Hand Rule No. 1 (see Section 1.5). The dawing shows an end view of the wie, whee it can be seen that the magnetic foce (magnitude = ) points to the ight, in the +x diection. I (out of page) T +y +x mg In ode fo the wie to be in equilibium, the net foce x in the x-diection must be zeo, and the net foce y in the y-diection must be zeo: x = (Equation 4.9a) and y = (Equation 4.9b). These equations will allow us to detemine the angle and the tension T. SOLUTION Since the wie is in equilibium, the sum of the foces in the x diection is zeo: Tsin x Substituting in = I L sin fo the magnitude of the magnetic foce, this equation becomes

26 Chapte 1 Poblems 111 T sin I L sin 9. (1) x The sum of the foces in the y diection is also zeo: T cos mg () y Since these two equations contain two unknowns, and T, we can solve fo each of them. a. To obtain the angle, we solve Equation () fo the tension the esult into Equation (1). This gives T mg cos and substitute Thus, mg sin I L sin I Lsin 9. o cos cos mg tan 4 A. m.7 T I L tan tan 37 mg.8 kg9.8 m/s 1 1 b. The tension in each wie can be found diectly fom Equation ():.8 kg9.8 m/s mg T cos cos37.49 N 41. REASONING The following dawing shows a side view of the conducting ails and the aluminum od. Thee foces act on the od: (1) its weight mg, () the magnetic foce, and the nomal foce N. An application of Right-Hand Rule No. 1 shows that the magnetic foce is diected to the left, as shown in the dawing. Since the od slides down the ails at a constant velocity, its acceleation is zeo. If we choose the x-axis to be along the ails, Newton s second law states that the net foce along the x-diection is zeo: x = ma x =. Using the components of and mg that ae along the x-axis, Newton s second law becomes cos 3. mg sin 3. x

27 11 MAGNETIC ORCES AND MAGNETIC IELDS The magnetic foce is given by Equation 1.3 as = IL sin, whee = 9.is theangle between the magnetic field and the cuent. We can use these two elations to find the cuent in the od. Cuent (out of pape) N Aluminum od mg 3. o x Conducting ail SOLUTION Substituting the expession = IL sin into Newton s second law and solving fo the cuent I, we obtain I mg sin 3. L. kg 9. 8 m / s sin 3. sin 9. cos m. 5 T sin 9. cos 3. b g b gc h b gb g 14 A 4. REASONING Accoding to Equation 1.4, the maximum toque is max = NIA, whee N is the numbe of tuns in the coil, I is the cuent, A = is the aea of the cicula coil, and is the magnitude of the magnetic field. We can apply the maximum-toque expession to each coil, noting that max, N, and I ae the same fo each. SOLUTION Applying Equation 1.4 to each coil, we have max NI 1 1 and max NI Coil 1 Coil Dividing the expession fo coil by the expession fo coil 1 gives Solving fo, we obtain max max NI NI 1 1 o 1= T 5. cm.4 T 1 b g cm

28 Chapte 1 Poblems REASONING The magnitude of the toque that acts on a cuent-caying coil placed in a magnetic field is specified by = NIA sin (Equation 1.4), whee N is the numbe of loops in the coil, I is the cuent, A is the aea of one loop, is the magnitude of the magnetic field, and is the angle between the nomal to the coil and the magnetic field. All the vaiables in this elation ae known except fo the cuent, which can, theefoe, be obtained. SOLUTION Solving the equation = NIA sinfo the cuent I and noting that = 9. since is specified to be the maximum toque, we have 5.8 Nm I. A NA sin m. T sin REASONING The magnetic moment of a cuent-caying coil is discussed in Section 1.6, whee it is given as Magnetic moment NIA (1) In Equation (1), N is the numbe of tuns in the coil, I is the cuent it caies, and A is its aea. oth coils in this poblem ae cicula, so thei aeas ae calculated fom thei adii via A. SOLUTION ecause the magnetic moments of the two coils ae equal, Equation (1) yields Substituting A N I A N I A () into Equation () and solving fo, we obtain NI N1I 1 N I o o N I Theefoe, the adius of the second coil is 1 N I NI 144. A 179.5A NI m.53 m NI (3) 45. REASONING Accoding to Equation 1.4, the toque that the cicula coil expeiences is N I Asin, whee N is the numbe of tuns, I is the cuent, A is the aea of the cicle, is the magnetic field stength, and is the angle between the nomal to the coil and the magnetic field. To use this expession, we need the aea of the cicle, which is, whee is the adius. We do not know the adius, but we know the length L of the wie, which must equal the cicumfeence of the single tun. Thus, L, which can be solved fo the adius.

29 114 MAGNETIC ORCES AND MAGNETIC IELDS SOLUTION Using Equation 1.4 and the fact that the aea A of a cicle is have that A, we N I Asin N I sin (1) Since the length of the wie is the cicumfeence of the cicle, o L, it follows that the L adius of the cicle is. Substituting this esult into Equation (1) gives L N I L N I sin N I sin sin 4 The maximum toque max occus when = 9., so that max A7.1 m.5 T N I L 3 sin 9. sin N m REASONING Accoding to the discussion in Section 1.6, the magnetic moment of the cuent-caying tiangle is NIA, whee N 1 is the numbe of loops in the coil, I is the cuent in the coil, and A is the aea of the tiangle. The magnitude of the net toque NIA sin (Equation 1.4), whee is exeted on the tiangle by the magnetic field is the magnitude of the magnetic field and 9. is the angle between the magnetic field and the nomal to the plane of the tiangle. SOLUTION a. Using the fact that the aea of a tiangle is one-half the base times the height of the tiangle, we find that the magnetic moment is A 1. m. m tan A m Magnetic moment NIA Aea of the tiangle b. The magnitude of the net toque exeted on the tiangle is Magnetic moment NIA sin 13.4 A m 1.8 T sin N m (1.4) 47. REASONING The magnitude of the toque that acts on a cuent-caying coil placed in a magnetic field is given by = NIA sin (Equation 1.4), whee N is the numbe of loops in the coil (N = 1 in this poblem), I is the cuent, A is the aea of one loop, is the magnitude of the magnetic field (the same fo each coil), and is the angle (the same fo each coil) between the nomal to the coil and the magnetic field. Since we ae given that the toque fo the squae coil is the same as that fo the cicula coil, we can wite

30 Chapte 1 Poblems 115 squae 1 I A sin 1 I A sin squae squae cicle cicle This elation can be used diectly to find the atio of the cuents. SOLUTION Solving the equation above fo the atio of the cuents yields I I squae cicle A A cicle squae cicle If the length of each wie is L, the length of each side of the squae is 1 L, and the aea of 4 the squae coil is A 1 L 1 L 1 L. The aea of the cicula coil is squae Acicle, whee is the adius of the coil. Since the cicumfeence ( ) of the cicula coil is equal to the length L of the wie, we have = L, o = L /( Substituting this value fo into the expession fo the aea of the cicula coil gives A cicle L/ the cuents is L I A I A L squae cicle cicle squae Thus, the atio of 48. REASONING When the wie is used to make a single-tun squae coil, each side of the squae has a length of 1 L. When 4 the wie is used to make a two-tun coil, each side of the squae has a length of 1 L. The dawing shows these two options and 8 indicates that the total effective aea of NA is geate fo the single-tun option. Hence, moe toque is obtained by using the single-tun option. L/4 SOLUTION Accoding to Equation 1.4 the maximum toque expeienced by the coil is max = (NIA) sin 9º = NIA, whee N is the numbe of tuns, I is the cuent, A is the aea of each tun, and is the magnitude of the magnetic field. Applying this expession to each option gives Single - tun Two - tun 1 max 4 NIA NI L 1 1 max A 1. m.34 T.36 N m 4 NIA NI L A 1. m.34 T.18 N m 8 L/8

31 116 MAGNETIC ORCES AND MAGNETIC IELDS 49. SSM REASONING The toque on the loop is given by Equation 1.4, NIAsin. om the dawing in the text, we see that the angle between the nomal to the plane of the loop and the magnetic field is The aea of the loop is.7 m.5 m =.35 m. SOLUTION a. The magnitude of the net toque exeted on the loop is NIAsin (75)(4.4 A)(.35 m )(1.8 T) sin 55 = 17 Nm b. As discussed in the text, when a cuent-caying loop is placed in a magnetic field, the loop tends to otate such that its nomal becomes aligned with the magnetic field. The nomal to the loop makes an angle of 55 with espect to the magnetic field. Since this angle deceases as the loop otates, the 35 angle inceases. 5. REASONING AND SOLUTION Accoding to Equation 1.4, the maximum toque fo a single tun is max = IA. When the length L of the wie is used to make the squae, each side of the squae has a length L/4. The aea of the squae is A squae = (L/4). o the ectangle, since two sides have a length d, while the othe two sides have a length d, it follows that L = 6d, o d = L/6. The aea is A ectangle = d = (L/6). Using Equation 1.4 fo the squae and the ectangle, we obtain squae ectangle IA A squae squae L / 4 IA A L / 6 ectangle ectangle b g b g REASONING The coil in the dawing is oiented such that the nomal to the suface of the coil is pependicula to the magnetic field ( = 9). The magnetic toque is a maximum, and Equation 1.4 gives its magnitude as = NIA sin. In this expession N is the numbe of loops in the coil, I is the cuent, A is the aea of one loop, and is the magnitude of the magnetic field. The toque fom the bake balances this magnetic toque. The bake toque is bake = bake, whee bake is the bake foce, and is the adius of the shaft and also the leve am. The maximum value fo the bake foce available fom static fiction is bake = s N (Equation 4.7), whee N is the nomal foce pessing the bake shoe against the shaft. The maximum bake toque, then, is bake = s N. y setting bake = max, we will be able to detemine the magnitude of the nomal foce. SOLUTION Setting the toque poduced by the bake equal to the maximum toque poduced by the coil gives

32 Chapte 1 Poblems 117 o NIA sin bake s N N s m NIA sin 41.6 A m.3 T sin N 5. REASONING The magnetic moment of the obiting electon can be found fom the expession Magneticmoment NIA. o this situation, N 1. Thus, we need to find the cuent and the aea fo the obiting chage. SOLUTION The cuent fo the obiting chage is, by definition (see Equation.1), I q / t, whee q is the amount of chage that passes a given point duing a time inteval t. Since the chage (q = e) passes by a given point once pe evolution, we can find the cuent by dividing the total obiting chage by the peiod T of evolution. I q q T / v 19 6 (1.6 1 C)(. 1 m / s) 11 (5.31 m) A The aea of the obiting chage is Theefoe, the magnetic moment is A ( m) m 11 1 Magnetic moment NIA ( 1)( A)(8.8 1 m ) Am SSM REASONING AND SOLUTION a. In igue 1.7a the magnetic field that exists at the location of each wie points upwad. Since the cuent in each wie is the same, the fields at the locations of the wies also have the same magnitudes. Theefoe, a single extenal field that points downwad will cancel the mutual epulsion of the wies, if this extenal field has a magnitude that equals that of the field poduced by eithe wie. b. Equation 1.5 gives the magnitude of the field poduced by a long staight wie. The extenal field must have this magnitude: T m/a 5 A 4 I T.16 m

33 118 MAGNETIC ORCES AND MAGNETIC IELDS 54. REASONING The magnitude of the magnetic field in the inteio of a long solenoid is 7 ni (Equation 1.7), whee 4 1 T m/a is the pemeability of fee space, n is the numbe of tuns pe unit length of the solenoid, and I is the cuent. SOLUTION Using Equation 1.7, we find that 14 tuns 7 ni 4 1 T m/a 4.7 A 1.31 T.65 m 55. SSM REASONING The magnitude of the magnetic field in the inteio of a solenoid that has a length much geate than its diamete is given by ni (Equation 1.7), 7 whee 41 T m/a is the pemeability of fee space, n is the numbe of tuns pe mete of the solenoid s length, and I is the cuent in the wie of the solenoid. Since and I ae given, we can solve Equation 1.7 fo n. SOLUTION Solving Equation 1.7 fo n, we find that the numbe of tuns pe mete of length is 7. T 4 n.81 tuns/m I Tm/A.1 A 56. REASONING The toque exeted on a coil with a cuent I coil is given by NI Asin (Equation 1.4), whee N is the numbe of tuns in the coil, A is its aea, coil is the magnitude of the extenal magnetic field causing the toque, and is the angle between the nomal to the suface of the coil and the diection of the extenal magnetic field. The extenal magnetic field, in this case, is the magnetic field of the solenoid. We will use 7 ni (Equation 1.7) to detemine the magnetic field, whee 41 Tm/A is the pemeability of fee space, n is the numbe of tuns pe mete of length of the solenoid, and I is the cuent in the solenoid. The magnetic field in the inteio of a solenoid is paallel to the solenoid s axis. ecause the nomal to the suface of the coil is pependicula to the axis of the solenoid, the angle is equal to 9.. SOLUTION Substituting the expession ni (Equation 1.7) fo the magnetic field of the solenoid into NIcoil Asin (Equation 1.4), we obtain NI Asin NI A ni sin nnai I sin Theefoe, the toque exeted on the coil is coil coil coil T m/a 14 m m.5 A 3.5 A sin Nm

34 Chapte 1 Poblems SSM REASONING The magnitude of the magnetic field at the cente of a cicula loop of cuent is given by Equation 1.6 as = N I/(R), whee N is the numbe of tuns, is the pemeability of fee space, I is the cuent, and R is the adius of the loop. The field is pependicula to the plane of the loop. Magnetic fields ae vectos, and hee we have two fields, each pependicula to the plane of the loop poducing it. Theefoe, the two field vectos ae pependicula, and we must add them as vectos to get the net field. Since they ae pependicula, we can use the Pythagoean theoem to calculate the magnitude of the net field. SOLUTION Using Equation 1.6 and the Pythagoean theoem, we find that the magnitude of the net magnetic field at the common cente of the two loops is net N I N I N I G H I J R K G H I J R K G I H J R K bgc hb g T m / A 1 7 A b.4 mg.. 5 T 58. REASONING The two ods attact each othe because they each cay a cuent I in the same diection. The bottom od floats because it is in equilibium. The two foces that act on the bottom od ae the downwad foce of gavity mg and the upwad magnetic foce of attaction to the uppe od. If the two ods ae a distance s apat, the magnetic field geneated by the top od at the location of the bottom od is (see Equation 1.5) I / ( s). Accoding to Equation 1.3, the magnetic foce exeted on the bottom od is I Lsin I Lsin / ( s), whee is the angle between the magnetic field at the location of the bottom od and the diection of the cuent in the bottom od. Since the ods ae paallel, the magnetic field is pependicula to the diection of the cuent (RHR-), and 9., so that sin =1.. SOLUTION Taking upwad as the positive diection, the net foce on the bottom od is Solving fo I, we find I Lsin mg s I 3 mgs (.73 kg)(9.8 m / s )( 8. 1 m) 7 L (4 1 T m / A)(.85 m) 19 A

35 111 MAGNETIC ORCES AND MAGNETIC IELDS 59. REASONING AND SOLUTION Let the cuent in the left-hand wie be labeled I 1 and that in the ight-hand wie I. a. At point A: 1 is up and is down, so we subtact them to get the net field. We have 1 = µ I 1 /( d 1 ) = µ (8. A)/[ (.3 m)] = µ I /( d ) = µ (8. A)/[ (.15 m)] So the net field at point A is A = 1 = T b. At point : 1 and ae both down so we add the two. We have So the net field at point is 1 = (8. A)/[ (.6 m)] = (8. A)/[ (.6 m)] = 1 + = T 6. REASONING The net magnetic field is the sum of the unifom field and the field poduced by the cuent in the wie. In ode fo the net field to be zeo, the two field contibutions must have the same magnitude and opposite diections. The cuent I in the wie ceates a I field that has a magnitude that is given by (Equation 1.5), whee T m/a is the pemeability of fee space and is the pependicula distance fom the wie. We can solve this equation fo, in ode to locate the point whee the field 3 poduced by the cuent has a magnitude of 7. 1 T. Right-hand ule no. indicates that the field of the cuent has opposite diections on opposite sides of the wie. Theefoe, since the wie is pependicula to the unifom field, we can be confident that this value fo will locate a place on one side o the othe of the wie whee the net field is zeo. SOLUTION Solving Equation 1.5 fo, we find that T m/a35 A I m T 61. REASONING The magnitude i of the magnetic field at the cente of the inne coil is given by Equation 1.6 as i µ IiNi /( Ri ), whee I i, N i, and R i ae, espectively, the cuent, the numbe of tuns, and the adius of the inne coil. The magnitude o of the magnetic field at the cente of the oute coil is o µ IoNo /( Ro ). In ode that the net magnetic field at the common cente of the two coils be zeo, the individual magnetic fields must have the same

36 Chapte 1 Poblems 1111 magnitude, but opposite diections. Equating the magnitudes of the magnetic fields poduced by the inne and oute coils will allow us to find the cuent in the oute coil. SOLUTION Setting i = o gives µ I N µ I N R R i i o o Solving this expession fo the cuent in the oute coil, we have I o i R I 7. A 8.6 A N i o 14 tuns.3 m i N o R i 18 tuns.15 m In ode that the two magnetic fields have opposite diections, the cuent in the oute coil must have an opposite diection to the cuent in the inne coil. o 6. REASONING a. The compass needle lines up with the net I hoizontal magnetic field that is the vecto sum of the magnetic fields I (the field at the location E E 3. of the compass due to the cuent I in the wie) compass and E (the hoizontal component of the eath s I magnetic field): = I + E. The field lines of N the magnetic field ceated by the cuent I ae cicles centeed on the wie. ecause the wie is.8 m pependicula to the eath s suface, and the W E compass is diectly noth of the wie, the magnetic field I due to the wie must point eithe due east wie S o due west at the location of the compass, (pependicula to the page) depending on the diection of the cuent. The magnetic field I causes the compass needle to deflect east of noth, so we conclude that I points due east (see the dawing, which shows the situation as viewed fom above). We will use Right-Hand Rule No. to detemine the diection of the cuent I fom the diection of I. b. The hoizontal component E of the eath s magnetic field points noth, so it is pependicula to the magnetic field I ceated by the cuent in the wie, which points east. Theefoe, the vectos, E, and I fom a ight tiangle with seving as the hypotenuse I (see the dawing). The angle between the vectos E and, then, is given by tan (Equation 1.3). We will use I (Equation 1.5) to detemine I, whee =.8 m is E

37 111 MAGNETIC ORCES AND MAGNETIC IELDS the adial distance between the wie and the cente of the compass needle and 7 41 T m/a. Then we will use Equation 1.3 to find the magnitude E of the hoizontal component of eath s magnetic field. SOLUTION a. ecause the magnetic field I due to the cuent in the wie points east at the location of the compass, the magnetic field lines must ciculate clockwise aound the wie, as viewed fom above. Right-Hand Rule No., then, indicates that the cuent I flows into the page. Since the dawing shows a top view of the situation, the cuent flows towad the eath s suface. b. Solving tan I (Equation 1.3) fo E, we obtain E I E (1) tan Substituting I I (Equation 1.5) into Equation (1), we find that E I 4 I I 7 1 T m/a 5. A tan tan tan.8 m tan T 63. REASONING The dawing shows an end-on view of the two wies, with the cuents diected out of the plane of the pape towad you. 1 and ae the individual fields poduced by each wie. Applying RHR- (thumb points in the diection of the cuent, finges cul into the shape of a half-cicle and the 1 Wie 1 Wie 1 finge tips point in the diection of the field), we obtain the field diections shown in the dawing fo each of the thee egions mentioned in the poblem statement. Note that it is only in the egion between the wies that 1 and have opposite diections. Hence, the spot whee the net magnetic field (the vecto sum of the individual fields) is zeo must lie between the wies. At this spot the magnitudes of the fields 1 and fom the wies ae I equal and each is given by (Equation 1.5), whee μ is the pemeability of fee space, I is the cuent in the wie, and is the distance fom the wie. SOLUTION The spot we seek is located at a distance d fom wie 1 (see the dawing). Note that the wies ae sepaated by a distance of one mete. Applying Equation 1.5 to each wie, we have that 1 d

38 Chapte 1 Poblems 1113 Since I 1 4I, this esult becomes 1 1 o I I d 1. m d 4I I 4 1 o o 41. m d d d 1. m d d 1. m d Solving fo d, we find that d.8 m 64. REASONING Using Right-Hand Rule No., we can see that at point A the magnetic field due to the hoizontal cuent points pependiculaly out of the plane of the pape. Similaly, the magnetic field due to the vetical cuent points pependiculaly into the plane of the pape. Point A is close to the hoizontal wie, so that the effect of the hoizontal cuent dominates and the net field is diected out of the plane of the pape. Using Right-Hand Rule No., we can see that at point the magnetic field due to the hoizontal cuent points pependiculaly into the plane of the pape. Similaly, the magnetic field due to the vetical cuent points pependiculaly out of the plane of the pape. Point is close to the hoizontal wie, so that the effect of the hoizontal cuent dominates and the net field is diected into the plane of the pape. Points A and ae the same distance of. m fom the hoizontal wie. They ae also the same distance of.4 m fom the vetical wie. Theefoe, the magnitude of the field contibution fom the hoizontal cuent is the same at both points, although the diections of the field contibutions ae opposite. Likewise, the magnitude of the field contibution fom the vetical cuent is the same at both points, although the diections of the field contibutions ae opposite. At eithe point the magnitude of the net field is the magnitude of the diffeence between the two contibutions, and this is the same at points A and. SOLUTION Accoding to Equation 1.5, the magnitude of the magnetic field fom the cuent in a long staight wie is = I/( ), whee is the pemeability of fee space, I is the cuent, and is the distance fom the wie. Applying this equation to each wie at each point, we see that the magnitude of the net field net is Point A net I I I 1 1 A, H A, V A, H A, V Hoizontal wie Vetical wie T m/a 5.6 A T. m.4 m

39 1114 MAGNETIC ORCES AND MAGNETIC IELDS Point net I I I 1 1, H, V, H, V Hoizontal wie Vetical wie T m/a 5.6 A T. m.4 m 65. SSM REASONING Accoding to Equation 1.6 the magnetic field at the cente of a cicula, cuent-caying loop of N tuns and adius is N I / ( ). The numbe of tuns N in the coil can be found by dividing the total length L of the wie by the cicumfeence afte it has been wound into a cicle. The cuent in the wie can be found by using Ohm's law, I V / R. SOLUTION The numbe of tuns in the wie is L N The cuent in the wie is V 1. V. 31 I R ( / m) L L Theefoe, the magnetic field at the cente of the coil is I L I LI NG I H J K G H I J K G H I J K 4 c T m / A h LG H 4 (. 14 m).31 L 3 A I K J 3 A T 66. REASONING In the dawing at the ight, we have labeled the cuent on the left as I 1 and its distance fom point P as 1. Similaly, we have labeled the cuent on the ight as I and its distance fom point P as. In fact, howeve, as the dawing in the text indicates, I1 I and (the dashed tiangle is an equilateal tiangle). It is 1 1 3º 3º 3º 3º 1 9º I 1 x I impotant to note that within the tiangle the angle between 1 and the y axis is 3º, as is the angle between and the y axis, because the y axis bisects the 6º apex angle of the tiangle. Moeove, it follows fom ight-hand ule no. that the magnetic field 1 (fom cuent I 1 ) is pependicula to 1 and also that the magnetic field (fom cuent I ) is pependicula to. This, in tun, means that the fields 1 and make +y P +x

40 Chapte 1 Poblems º angles with the y axis, as indicated in the dawing. We will use this geomety in dealing with the components of the two magnetic fields. It is necessay to use components to find the net field at point P, since the fields ae vectos. SOLUTION The cuent I in each wie ceates a field that has a magnitude that is given by I 7 (Equation 1.5), whee 4 1 T m/a is the pemeability of fee space and is the pependicula distance fom the wie. Using Equation 1.5 fo the magnetic fields 1 and, we list the components of the fields at point P and the net components as follows: ield x component y component 1 sin6. I sin6. 1x 1 cos6. I cos6. 1y 1 sin6. I sin6. x Net cos6. I cos6. y I cos6. The net field at point P has a zeo x component, so its magnitude is just the magnitude of the y component: 7 Magnitude I 4 1 T m/a 85. A 4 of net field cos6. cos T at point P.15 m Since the net field has only a y component that is negative, the net field points downwad and is pependicula to the dashed line that is the base of the tiangle in ou dawing. 67. REASONING AND SOLUTION The foces acting on each wie ae the magnetic foce, the gavitational foce mg, and the tension T in the stings. Each sting makes an angle of 7.5 with espect to the vetical. om the dawing below at the ight we can elate the magnetic foce to the gavitational foce. Since the wie is in equilibium, Newton s second law equies that x = and y =. These equations become T +y 7. 5 o Wie +x T sin 7.5 and T cos 7.5 mg x y mg

41 1116 MAGNETIC ORCES AND MAGNETIC IELDS Solving the fist equation fo T, and then substituting the esult into the second equation gives (afte some simplification) tan 7.5 mg (1) I L The magnetic foce exeted on one wie by the othe is, whee d is the d distance between the wies [d/ = (1. m) sin 7.5, so that d =.31 m], I is the cuent (which is the same fo each wie), and L is the length of each wie. Substituting this elation fo into Equation (1) and then solving fo the cuent, gives I m L g d tan 7. 5 H G I K J H G I K J L M. 31 m O b. 5 kg / mgc 9. 8 m / s h b g tan 7. 5 N Q P 3 A 68. REASONING AND SOLUTION Ampèe's law in the fom of Equation 1.8 indicates that I. Since the magnetic field is eveywhee pependicula to the plane of the pape, it is eveywhee pependicula to the cicula path and has no component that is paallel to the cicula path. Theefoe, Ampèe's law educes to I, so that the net cuent passing though the cicula suface is zeo. 69. REASONING Since the two wies ae next to each othe, the net magnetic field is eveywhee paallel to in igue Moeove, the net magnetic field has the same magnitude at each point along the cicula path, because each point is at the same distance fom the wies. Thus, in Ampèe's law (Equation 1.8),, I I1 I, and we have I I 1 ut is just the cicumfeence () of the cicle, so Ampèe's law becomes This expession can be solved fo. SOLUTION I I 1

42 Chapte 1 Poblems 1117 a. When the cuents ae in the same diection, we find that 7 I1 I 4 1 T m/a 8 A 1 A T.7 m b. When the cuents have opposite diections, a simila calculation shows that 7 I1 I 4 1 T m/a 8 A 1 A T.7 m 7. REASONING oth pats of this poblem can be solved using Ampèe s law with the cicula closed paths suggested in the Hint given with the poblem statement. The cicula closed paths ae used because of the symmety in the way the cuent is distibuted on the coppe cylinde. SOLUTION a. An end-on view of the coppe cylinde is a cicle, as the dawing at the ight shows. The dots aound the cicle epesent the cuent coming out of the pape towad you. The lage dashed cicle of adius is the closed path used in Ampèe's law and is centeed on the axis of the cylinde. Equation 1.8 gives Ampèe's law I. ecause of the symmety of as the aangement in the dawing, we have fo all on the cicula path, so that Ampèe's law becomes I In this esult, I is the net cuent though the cicula suface bounded by the dashed path. In othe wods, it is the cuent I in the coppe tube. uthemoe, is the cicumfeence of the cicle, so we find that I I o

43 1118 MAGNETIC ORCES AND MAGNETIC IELDS b. The setup hee is simila to that in pat a, except that the smalle dashed cicle of adius is now the closed path used in Ampèe's law (see the dawing at the ight). With this change, the deivation then poceeds exactly as in pat a. Now, howeve, thee is no cuent though the cicula suface bounded by the dashed path, because all of the cuent is outside the path. Theefoe, I = A, and I T 71. SSM REASONING AND SOLUTION The dawing at the ight shows an end-on view of the solid cylinde. The dots epesent the cuent in the cylinde coming out of the pape towad you. The dashed cicle of adius is the closed path used in Ampèe's law and is centeed on the axis of the cylinde. Equation 1.8 gives Ampèe's law as I. ecause of the symmety of the aangement in the dawing, we have fo all on the cicula path, so that Ampèe's law becomes I In this esult, =, the cicumfeence of the cicle. The cuent I is the pat of the total cuent that comes though the aea bounded by the dashed path. We can calculate this cuent by using the cuent pe unit coss-sectional aea of the solid cylinde. This cuent pe unit aea is called the cuent density. The cuent I is the cuent density times the aea : Thus, Ampèe's law becomes I I R I cuent density I R R I I o o R