MAGNETIC FIELDS AND FORCES 24

Transcription

1 MAGNETIC FIELDS AND FORCES 24 Q24.1. Reason: When a ba magnet is bought nea the cente of anothe ba magnet as shown in Figue Q24.1, the foce between the ba magnets is zeo. The attactive foce between the ba magnets noth-south sides cancels the epulsive foce between the ba magnets noth-noth sides. Assess: If the ba magnet is not exactly in the cente, then it will otate until the two magnets ae paallel with noth-south and south-noth sides touching. Q24.2. Reason: By tying a sting aound the cente of the ba magnet, one can allow the ba magnet to feely otate. By suspending the magnet in the eath s atmosphee, the eath s magnetic field will inteact with the ba magnet s magnetic field. The inteaction between the fields will otate the ba magnet until it is paallel to the eath s magnetic field, as shown in the figue. Once the ba magnet is paallel to the eath s magnetic field, thee will be no net foce and the ba magnet will stop otating. The half of the ba magnet pointing to the geogaphic noth (which is located nea the eath s magnetic south pole) will be a noth pole wheeas the half of the ba magnet pointing to the geogaphic south diection will be a south pole. Assess: Anytime a feely suspended ba magnet is in the vicinity of a magnetic field, the magnet will otate until the two magnets (o magnetic fields) ae paallel with noth-south and south-noth sides touching. Q24.3. Reason: A compass points nea the geogaphic noth because the eath s magnetic field lines un nea the geogaphic South Pole (magnetic noth pole) to nea the geogaphic Noth Pole (magnetic south pole): see Figue 24.8 in textbook. The field is not dependent on the hemisphee one is standing in. Assess: In a ba magnet magnetic field lines un fom a noth pole to a south pole. To find the geogaphic Noth Pole one uses the noth pole of a magnet because the noth pole will point in the diection of the eath s magnetic south pole. Q24.4. Reason: The eath s magnetic field vectos tend to point below the hoizon in the nothen hemisphee. The angle below the hoizon is called the dip angle. But the dip angle changes with latitude. As illustated in the figue, nea o the equato, the dip angle is small and nea the noth pole, the dip angle is nealy 90. Each diffeent latitude has a cetain dip angle and the tutle is able to detemine its latitude fom the dip angle it senses. 24-1

2 24-2 Chapte 24 Assess: The way the tutle finds its latitude is simila to the way sailos can find thei latitude by looking at the position of stas. Fo example, a sailo can use the dip angle of the sun s ays at noon to find his o he latitude. Q24.5. Reason: The figue illustates the magnetic field lines fo a hoseshoe magnet. Assess: Magnetic field lines un fom a noth pole to a south pole with the aows showing the diection to and fom. The lines should always be equally spaced to show the qualitative stength of the magnetic field. Q24.6. Reason: Because the noth pole of the compass needle points counteclockwise, the magnetic field is counteclockwise. When you cul you ight hand finges counteclockwise, the thumb points out of the page. Thus, the cuent in the wie is out of the page. This is demonstated in the opposite diection in Figue of the textbook. Assess: The ight hand ule fo fields gives the elationship between a cuent and the cicling magnetic field it ceates. Q24.7. Reason: Remembe that an X epesents the magnetic field going into the page. The dot is the point of an aow and illustates the magnetic field coming out of the page. Cul you ight hand s finges aound the wie so that they come out of the page on the ight side and into the page on the left side. The diection of you thumb is the diection of the cuent. Since you thumb now points down, the cuent in the wie is down: See the figue. Assess: The ight-hand ule fo fields gives the elationship between a cuent and the cicling magnetic field it ceates. Q24.8. Reason: Conventional household wiing is composed of at least two cuent-caying conductos. One conducto caies cuent to some appliance o light, and anothe conducto seves as the electical etun. Howeve, household cuent eveses its diection about 120 times each second, o 60 cycles (evese and back) each second.

3 Magnetic Fields and Foces 24-3 Assess: It is unlikely you could detect cuent in the walls of a house with a compass fo two easons. The fist eason is if the diection of the cuent changes 120 times a second, the magnetic field suounding the cuent-caying wie changes diection 120 times each second. The mass inetia of the needle would pevent it fom flipping 120 each second. The second eason a compass would make a poo instument to find wiing is that since household wiing is composed of a pai of cuent-caying conductos close to each othe, and the two cuents will always be equal in magnitude, yet opposite in diection with each othe, thei magnetic fields will cancel. Q24.9. Reason: To find the field at any point, we need to find the field we would have if only the top wie wee pesent and the field we would have if only the bottom wie wee pesent. Let the point whee the field is given be called P and let the distance fom point P to the bottom wie be called d. At P, the field due to the lowe wie is, fom the ight-hand ule, out of the page. The field at P due to the top wie is into the page. Thus the field stength due to the top wie should be subtacted fom the field stength due to the bottom wie (the bottom wie contibutes a lage field since point P is close to the bottom wie): B B B μ I μ I μ I 0 2π d 2 π(3 d) 3πd 0 0 total = bottom top = = μ0i This field is diected out of the page. We know that at point P, B total = 2.0 mt =, fom which it follows 3 π d μ0i that 3.0 mt. 2π d = Thus μ0i Bbottom = = 3.0 mt. At point 1, the field due to the bottom wie is into the page 2π d μ0i (fom the ight-hand ule) and has magnitude = 3.0 mt. Since the top and bottom wies have the same 2π d cuent and ae equidistant to point 1, the field at point 1 due to the top wie is also 3.0 mt and fom the ighthand ule, is also into the page. Thus the total field at point 1 is 6.0 mt, into the page. Finally, at point 2, the μ0i field due to the top wie is = 3.0 mt, out of the page and the field due to the bottom wie is 2π d μ0i 3.0 mt = = 1.0mT, into the page. Since the fields ae in opposite diections, they ae subtacted, as at 2 π (3 d) 3 point P, but now the lage field is out of the page so the total field is out of the page and has magnitude: 3.0 mt 1.0 mt = 2.0 mt. Assess: You might have guessed, using symmety, that the field at point 2 would be the same as the field at point P. This is because these two points have in common that they ae a distance d fom one cuent, I, and a distance 3d fom an equal cuent flowing in the opposite diection. Q Reason: At point 1, thee ae two magnetic fields inteacting: the unifom magnetic field in the plane of the pape pointing up and the magnetic field fom the wie. The way you find the diection of the magnetic field fom the cuent-caying wie is shown in Section 24.3, using the ight-hand ule fo fields. The magnetic field fom the wie has to be pointing down with the same magnitude as the unifom magnetic field in the plane because the poblem states that the total magnetic field is zeo at point 1. When the magnetic field fom the cuent is pointing down at point 1, then the magnetic field diection is clockwise aound the wie. At point 2, the magnetic field fom the cuent-caying wie is in the ight diection (tangent to the clockwise cicula path aound the wie). The total magnetic field is theefoe the combination of the magnetic field fom the wie (to the

4 24-4 Chapte 24 ight) and fom the unifom magnetic field (up). The total magnetic field is the esultant vecto of the two magnetic field vectos as shown. Assess: Wheneve thee ae two magnetic fields acting at the same point, you add the fields. If the magnetic fields act in 2-dimensions, then you have the add/subtact them using vecto addition. In this case, the vecto can be simply dawn as the esultant vecto shown. Q Reason: Assuming the wie with the cuent diected to the ight is the bottom wie, the bottom cuent-caying wie poduces a magnetic field pointing towad the bottom of the page at a point halfway between them. The top cuent-caying wie poduces a magnetic field pointing ight at a point halfway between them. The total magnetic field is the esultant vecto of the two magnetic fields as shown. Assess: Wheneve thee ae two magnetic fields acting at the same point, use vecto addition to add the fields and find the esultant vecto. In this case, the esultant vecto is conceptually shown in the pevious figue. Q Reason: No. Magnetic fields exet a foce on moving chages only. The equation fo magnetic foce, F = qvb sin α, shows that if the velocity (v) is zeo then the foce (F) is zeo as well. Q Reason: If a cuent is pesent in a long solenoid, the magnetic field that is ceated is unifom along its cental axis. The diection of the magnetic field is in the diection the thumb on you ight hand points when you finges ae culed in the diection of the cuent in accodance with the ight-hand ule fo fields. The magnetic field points to the ight. The magnetic field lines shown in Figue in the textbook show that the field cuves aound the outside of the solenoid in the othe diection. This means the compass needle should point left. Assess: The ight-hand ule can be used to detemine the diection of the magnetic field due to any cuent, even a cuved cuent. N Q Reason: The magnetic field inside of a solenoid is B = μ I. Note that the magnetic field is 0 L independent of the diamete of the solenoid (povided that its length, L, is lage in compaison). Theefoe, the diametes of the two concentic solenoids do not matte hee. In this case, the numbe of tuns, N, and the length, L, is the same fo both solenoids. The cuent, I, is equal, but opposite, between the two solenoids. Theefoe: N BTotal = B1 + B2 = μ0 [ I1 + ( I2)] = 0 L Assess: The field inside the two solenoids is zeo because the cuent was equal and opposite. Also, the adius of the solenoid does not affect the magnetic field inside fo sufficiently long solenoids. Q Reason: The diection of the foce detemines the diection the paticle will be deflected when it entes the magnetic field. Use the ight-hand ule fo foces to detemine the diection of the foce. (a) The velocity points to the ight and the magnetic field points into the page so the foce points towad the top of the page. (b) The velocity points upwad and the magnetic field points out of the page. But because the chaged paticle is negative, the foce points to the left.

5 Magnetic Fields and Foces 24-5 Assess: The diection of the foce on a positive paticle is govened by the ight-hand ule. Convesely, the foce on a negative paticle is in the opposite diection of the ight-hand ule. Fo example, if the ight-hand ule gives you an up diection, then, fo a negative paticle, the diection is down. Q Reason: The diection of the magnetic foce detemines the diection that the paticle will be deflected when it entes the magnetic field. Use the ight-hand ule fo foces to detemine the diection of the foce. See Figue in the textbook. (a) The velocity points to the ight and the magnetic field points towad the bottom of the page, so the foce points into the page. (b) In this case the velocity and the magnetic field ae pointing in the same diection, so thee is no foce. Assess: The eason thee is no foce fo pat (b) can be shown using the equation F = qvbsin α, whee α is the angle between the velocity and the magnetic field. So if the velocity is paallel to the magnetic field, this angle is zeo, which makes the sine and the foce equal zeo as well. Q Reason: Use the ight-hand ule fo foces to detemine the diection of the magnetic field. See Figue (a) The velocity points to the ight and the foce points down, so the magnetic field points out of the page. (b) The velocity points to the ight and the foce points into the page, so the magnetic field should point to the bottom of the page. BUT, because the chaged paticle is negative, the diection of the magnetic field is opposite. So the field points to the top of the page. Assess: The diection of the foce on a positive paticle is govened by the ight-hand ule. Convesely, the foce on a negative paticle is opposite. Q Reason: Use the ight-hand ule fo foces to detemine the diection of the magnetic field. Fo (a), the component of the magnetic field affecting the chaged paticle points 90 clockwise fom the diection of the velocity, v is paallel to the plane of the page (note the chage is negative). Fo (b), the component of the magnetic field affecting the chaged paticle points 90 counteclockwise fom the diection of the foce vecto, F, and is paallel to the plane of the page. Assess: The diection of the magnetic field fo a paticle is govened by the ight-hand ule egadless of what diection the velocity and foce ae pointing. Q Reason: The diection of the magnetic field of a cuent-caying wie is descibed by the ighthand ule. Using this we see that above the wie the magnetic field is coming out of the page. See Figue The foce is found by the ight-hand ule. Fo an electon, the foce points up. Assess: Magnetic foce depends on chage, velocity, magnetic field, and diection. If you change the sign of the chage the diection of the foce is opposite. Q Reason: Use the ight-hand ule fo foces to detemine the diection of the magnetic field. When you use the ight-hand ule on both paticles you get the magnetic field coming out of the page (in the +x-diection). Assess: The fact that the two paticles have diffeent diections does not automatically mean they ae acted upon by a diffeent magnetic field. In this case the magnetic field is unifom. Q Reason: At the eath s magnetic noth pole, the field lines come staight up out of the gound, as in the figue. Fom the ight-hand ule, the foce on a positive chage placed at point P and moving up the page would be to the ight. So the foce on an electon would be to the left. Thus the electon is steeed to the left and moves counteclockwise, as viewed fom above. Assess: The eath s magnetic noth pole is in the southen hemisphee so the evolution of an electon as descibed in this poblem is chaacteistic of the auoa austalis, o southen lights. Q Reason: The dots tell us that the field is out of the page. Fom the ight-hand ule, the foce on the poton is to the ight, so the poton is steeed to the ight and its motion is clockwise. Assess: If the chaged paticle had been an electon athe than a poton, then the foce would have been the opposite of what is pedicted by the ight-hand ule. The foce would have been to the left and the obit would have been counteclockwise. Q Reason: Without using any algeba, we can eason that an ion with moe chage will expeience a geate foce (and theefoe be steeed aound a smalle cicle) if the same magnetic field is used. Since the two goups of ions landed at the same spot, we can conclude that the one with the chage advantage, that is, the ion

6 24-6 Chapte 24 with chage 2 e, must have been steeed by a weake electic field. Thus peak A coesponds to the ions with chage 2e and peak B coesponds to the ions with chage e. Anothe way to conside this question is to solve ( mv / ) Equation 24.8 fo the chage of an ion evolving in a magnetic field. Doing so, we obtain: q =. The B two goups of ions have the same mass, velocity and obital adius, so they have the same numeato. It follows that peak B, which is at a highe field will coespond to ions with a lowe chage, that is, those of chage e. Assess: The minimum chage a positive ion can have is + e, and the next smallest value is + 2, e so if it tuns out that thee ae moe than two diffeent chage values among these ions (moe than just + e and + 2e ), those ions would have geate chage and so the peaks associated with them would be to the left of peak A. Q Reason: The way you find the diection of the magnetic field fom a cuent-caying wie is shown in Section 24.3, using the ight-hand ule fo fields. The magnetic field fo a cuent coming out of the page is culing aound the wie in the counteclockwise diection. Because the magnetic field at the poton is paallel to the velocity of the poton, the foce on the poton is zeo. Assess: Wheneve the magnetic field and the velocity ae paallel to each othe the foce is always zeo. As seen by F = qvbsin α, whee α is the angle between the velocity and the magnetic field. If α is 0, then F = 0. Q Reason: The way you find the diection of the magnetic field fom a cuent caying wie is shown in Section 24.3, using the ight-hand ule fo fields. The magnetic field fo a cuent coming out of the page is culing aound the wie in the counteclockwise diection. The magnetic field vecto can be split into two components, B x and B y. The component that is paallel to the velocity ( B y ) contibutes no foce, because α is zeo in the magnetic foce equation F = qvbsin α. The component of the magnetic foce that is pependicula to the velocity ( B x ) contibutes a foce, and by the ighthand ule fo foces, the foce is pointing into the page. Assess: Even though the magnetic field was not in a convenient diection, only the component that is pependicula to the velocity exets a foce on the paticle.

7 Magnetic Fields and Foces 24-7 Q Reason: The diection of the staight wie cuent is up. At the location of the squae loop, the cuent in the staight wie geneates a magnetic field that is into the page. We can disegad the foces ceated by cuent within the top and bottom sides of the squae, since they both un pependicula to the staight wie, and ae equal distance fom the wie. The foces fom the top and bottom sides conveniently cancel each othe out. Howeve, the foces fom the two sides of the squae loop that un paallel to the staight wie (the left and ight sides) do not cancel each othe out. The magnetic foce fom the side of the squae loop that is on the left (closest to the staight wie) will be stonge than the magnetic foce felt on the ight side (futhest fom the staight wie) of the squae loop. The cuent on the left side of the squae loop is opposite fom the cuent in the wie, theefoe, the wies will epel each othe. Thee will be an attactive foce fom the ight side cuent, but this effect is less than the othe side since it is twice as fa away. The paallel sides do not cancel each othe out. Theefoe, the net foce on the squae loop is to ight, o positive x diection. Will thee be any toque on the loop? No. The cuent in the squae loop geneates a magnetic dipole, and the cuent diection in the squae loop is such that the dipole moment is aligned with the magnetic field fom the staight wie. Assess: When cuent caying wies ae paallel, they expeience an attactive foce when the cuents ae in the same diection and a epulsive foce when the cuents ae in the opposite diection. Q Reason: The high magnetic field poduced by the MRI would alte the path of the electons in the cathode ay tube. The CRT monito would only display an extemely distoted image, much like what would happen to a television in the pesence of a stong magnetic field. Assess: Magnetic fields exet a foce on moving chages. Q Reason: Once the slinky is stetched out and a cuent is passed though it, the sepaate coils act as cuent loops. The cuent flows though each loop in the same diection so they will attact each othe causing the slinky to contact. Assess: When cuent flows aound two cuent loops (o two paallel wies) in the same diection they ae attacted to each othe. This attaction causes the slinky to contact. Q Reason: The foce a magnetic field exets on a cuent-caying wie is given by Equation 24.14: Fwie = ILBsin α. In a solenoid the magnetic field inside uns paallel to the axis. If the wie uns down the middle of the solenoid then α = 0 and sin0 = 0 so thee is no foce on the wie due to the field of the solenoid. Assess: This situation could also be analyzed by looking at the moving chages in the wie and applying Equation Q Reason: To make an electomagnet, you should use wie coveed in plastic. This way, the cuent cannot leak fom one tun of the wie to an adjacent tun. To get a stong field in a solenoid, you want the cuent to move though the solenoid in a helical path. But if adjacent tuns of wie ae not insulated, then the cuent can go diectly fom one tun to the next without having to go aound the solenoid. In addition, since a nail is used, which has a lage diamete and theefoe a low esistance, much of the cuent will actually tavel though the nail, fom the point whee the solenoid stats to the point whee it ends. Assess: If we used bae wie, thee would still be an electic cuent and, consequently, a magnetic field. But insulation guides the cuent though the optimal helical path, inceasing the magnetic field stength. Q Reason: The moon contains ocks that wee magnetized, and in fact some pats of the moon s suface ae magnetized. One theoy suggests that the moon at one time had a molten coe, and when the oute suface of the Moon cooled, the magnetic field of the molten coe aligned the atoms of the metallic elements on the moon s suface. Since the suface of the moon has not been distubed o heated, the feomagnetic mateial continued to keep the alignment of the atoms, and theefoe stayed magnetized. Assess: One theoy why the eath is magnetized suggests that the eath s molten coe cycles the metal coe though convection and moves the chaged paticles with it. This cyclic motion, like any cicula electic cuent, ceates a dipole magnetic field. This eath s magnetic field ove time magnetizes any feomagnetic mateial on eath (i.e., by aligning the dipole moments in the atoms). Q Reason: Afte the bicks cool off, thei magnetic field is paallel to that of the eath. They ae then laid in the wall and ae likely eoiented in the pocess. Fo example, the bicks might be fied with thei long side lined up east-west and then placed in a wall whee thei long side is oiented noth-south. Thee is a good chance that many of the bicks used have the same angle between thei magnetic field and thei long axis. These

8 24-8 Chapte 24 ideas could be used by an achaeologist to detect a bick stuctue. In a backgound of ock with magnetic field vectos all pointing in the same diection (this would be emnants of all the stuff in thei dwelling which was not fied), the achaeologist might detect a naow egion in which the magnetic field vectos point in diffeent diections fom the backgound field. This naow egion would be the wall. The achaeologist might also be able to detect the locations whee two walls join at a ight angle. Whee two walls join, the oientation of the bicks is shifted by 90 o. It follows that the oientation of the magnetic field vectos would shift at a joint in the building. O if a bick wee fo some eason laid on the wall backwad elative to the suounding bicks, that bick could be identified. These ideas ae illustated in the figue. Assess: Simply by measuing the magnetic field at multiple places in an achaeological site, the scientist can not only locate the pesence of man-made stuctues, but he can also identify individual bicks and featues of the building. Q Reason: While the metal sphee must be made of a magnetic mateial in ode to expeience the stong attaction to the magnet, it is likely not a magnet itself. So it would be equally attacted to the south pole of the ba magnet. The coect choice is A. Assess: A metal object made of a magnetic mateial but which is not a pemanent magnet will be attacted to both ends of a magnet. Ty this at home with a magnet and some pape clips. Q Reason: A magnetic field points counteclockwise aound the wie using the ight-hand ule fo cuent-caying wies. Viewed fom the top, the magnetic field cicles out of the page above the wie and into the page below the wie. This means above the wie the noth pole should face towad the bottom of the page. The coect choice is C. Assess: Choices B and D ae incoect because the compass points along the diection of the cuent wie and the magnetic field fom a cuent-caying wie cicles aound the wie. Choice A is incoect because the compass points noth below the wie. Choice C is coect because the compass points noth above the wie. Q Reason: Using the ight-hand ule fo fields, we conclude that the field at the point in question, due to the wie on the left is diected up the page. The field due to the wie on the ight is also up the page at the midpoint. So the total field, which is the sum of these two, will also be up the page. The answe is A.

9 Magnetic Fields and Foces 24-9 Assess: The oientation of the cuents was vey impotant hee. Fo example, if both cuents had been diected out of the page, then the field due to the wie on the left would still be up the page, but the field due to the wie on the ight would be down the page. The total field would be zeo. Q Reason: When a chaged paticle entes a unifom magnetic field it moves in unifom cicula motion. The adius of the cuvatue is given by = mv/( qb). The coect choice is D. Assess: Since path D has the lagest cicula adius, it has the lagest mass. The poblem stated that the paticles have the same speed (v), same chage (q), and ente the same unifom magnetic field (B). Q Reason: When a chaged paticle entes a unifom magnetic field it moves in unifom cicula motion. The adius of the cuvatue is detemined by = mv/( qb), so the lage the adius, the lage the velocity. Fom the choices given, D has the lagest adius, so it also has the highest velocity. The coect choice is D. Q Reason: By the ight-hand ule fo fields, if these wee potons, the magnetic field would be out of the plane of the pape. But because the paticles ae electons, the field must point into the plane of the pape to get the desied deflection. Theefoe the answe is D. Assess: The diection detemined by the ight-hand ule assumes a positive change, so fo a negative chage the esults must be evesed. Q Reason: The answe is C because magnets want to line up noth to south. In answes A and D, the same pole was facing each othe so they would epel. In answe B, the compasses ae fee to otate, and would not assume this configuation if they could move. Poblems P24.1. Pepae: Table 24.1 in the textbook shows magnetic field stengths. The equation fo a magnetic field due to a long staight wie is B = μ I 0 /(2 πr ). Solve: In this poblem R = 1.0 cm = 0.01 m, so the pevious equation can be ewitten to solve fo cuent, I = 0.02 πb/ μ0 0.02π 5 10 Ieath suface = = 2.5 A μ π 5 10 Iefigeato = = 250 A μ π (0.1 1) Ilaboatoy = = ,000 A μ0 0.02π10 Imagnet = = 500,000 A μ0 Assess: As one would expect, high cuent is needed to poduce a high magnetic field. P24.2. Pepae: Table 24.1 shows magnetic field stengths as shown in Poblem 1. The equation fo a magnetic field due to a long staight wie is B = μ0i/(2 πr). Solve: In this poblem I = 10 A so the pevious equation can be ewitten to solve fo distance: R = μ I 0 /(2 πb ). 10μ0 Reath suface = = 4.0 cm 2π 5 10 Solving, likewise, we get the distances of the efigeato magnet, the laboatoy magnet, and the supeconducting magnet to be 0.4 mm, 20 µm to 2 µm, and 0.20 µm, espectively. Assess: As we would expect, magnetic field stength deceases as we move away fom the cuent-caying wie. P24.3. Pepae: The equations fo the magnetic field at the cente of a loop and a wie ae B = μ I/(2 R) and B = μ I/(2 π R) loop cente 0 wie 0 Solve: (a) The adius of the loop is 0.5 cm and the magnetic field is 2.5 mt so the cuent is:

10 24-10 Chapte 24 B μ I 2R 0 loop cente = 2RB I = μ loop cente ( m)( T) 4 (10 T m/a) = = A π which will be epoted as 20 A. (b) Fo a long, staight wie that caies a cuent I, the magnetic field stength is μ0i 3 4 π(10 T m/a)(19.89 A) 3 Bwie = T = d = m 2πd 2πd Assess: The esult fo d obtained is consistent with the equations fo B wie and B loop cente. P24.4. Pepae: Fom Table 24.1 the eath magnetic field is equal to 5 10 T and the magnetic field at the cente of a cuent loop B = μ0ni/(2 R). Solve: Reaanging the equation fo a cuent loop the cuent would be 2RB 2(0.50m)( 5 10 T) I = = = 0.2 A μ N 4 π(10 T m/a)(200) 0 Assess: We expected a small cuent though the loop because eath s magnetic field stength is weak. P24.5. Pepae: Assume the wies ae infinitely long. Find the contibution of the magnetic field due to each wie. Solve: The magnetic field stength at point 1 is μ0i μ0i B1 = Btop + Bbottom =, out of page 2πd +, into page 2πd top bottom μ0i B1 = = (2)(10 T m/a)( 10A) 2 2 2π 2cm ( 4+ 2)cm 2 10 m 6 10 m B = ( T, out of page) 1 At points 2 and 3, μ0i μ0i 4 B2 =, into page +, into page = ( T, into page) 2 π(2 cm) 2 π(2 cm) μ0i μ0i B3 =, into page +, out of page = ( T, out of page) 2 π(6 cm) 2 π(2 cm) Assess: Each point is affected by both wies, so the contibutions must add accoding to the diection of the field points. The equation of the magnetic field does not give its diection, only its magnitude. To get the diection you must use the ight-hand ule. If the fields ae in the same diection, they add. If they ae in diffeent diections, they subtact. P24.6. Pepae: Use the equation fo the magnetic field fo a long staight wie. Solve: (a) The field of a tansmission line is aound μ0 I (2 10 T m/a)(200 A) B = = = = 2π d 20 m T 2.0 T (b) The eath s field is Beath = 5 10 T = 50 μt, so Bwie / Beath = 2.0 μt/(50 μt) = 0.04 = 4.0%. Assess: The field poduced by a tansmission line on the gound is much smalle than the eath s magnetic field. P24.7. Pepae: We need the fomula fo the magnetic field of a staight wie. Solve: (a) Let s estimate that the baby is 10 cm ( 4 inches ) fom a 1 A cuent. Hee the magnetic field is μ0 I (2 10 T m/a)(1 A) B = = = = 2π d 0.1 m μ T 2.0 T (b) The two magnetic fields poduced will be in opposite diections so they will cancel each othe out. Assess: The magnetic field poduced by an electic blanket on a baby, even if the wies ae not oppositely paied, is much smalle than the eath s magnetic field. μ

11 Magnetic Fields and Foces P24.8. Pepae: When calculating the magnetic field at a point, you need to use the ight-hand ule fo magnetic fields as well as set up a good coodinate system (oigin is between the two wies). The magnetic fields diection on the x-axis fo pat (a) and (b) will depend on the side of the wie: See figue. In pat (a), between the two wies, the magnetic field will be up fo the 5.0 A cuent and down fo the 3.0 A cuent. Outside the two wies ( x< 2, x> 2), the magnetic field will be in the same diection. In pat (b), between the two wies, the magnetic field will be in the same diection (up). Outside the two wies, the magnetic field will be opposite. The magnetic field will be zeo when the two magnetic fields cancel. Solve: (a) The point whee the two magnetic fields cancel lies on the x-axis in between the two wies. Let that point be a distance x away fom the oigin. Because the magnetic field of a long wie is B = μ I/(2 π ) 0 and the fields cancel when B = B, 5.0 A 3.0 A we have μ0(5.0 A) μ0(3.0 A) = 5(0.02 m x) = 3(0.02 m + x) x= m = 0.50 cm 2 π(0.02 m + x) 2 π(0.02 m x) (b) The magnetic fields due to the cuents in the two wies add in the egion 2.0 cm < x < 2.0 cm. Fo x < 2.0 cm, the magnetic fields subtact, but the field due to the 5.0 A cuent is always lage than the field due to the 3.0 A cuent. Howeve, fo x > 2.0 m, the two fields will cancel at a point on the x-axis. Let that point be a distance x away fom the oigin, so μ0(5.0 A) μ0(3.0 A) = 5( x 0.02 m) = 3( x m) x= 0.08 m = 8.0 cm 2 π( x m) 2 π( x 0.02 m) Assess: The values of x obtained seem easonable. P24.9. Pepae: Assume that the supeconducting niobium wie is vey long. Solve: The magnetic field of a long wie caying cuent I is B wie μ0i = 2π We e inteested in the magnetic field of the cuent ight at the suface of the wie, whee = 1.5 mm = m. The maximum field is 0.10 T, so the maximum cuent is 3 (2 πb ) wie 2 π( m)(0.10 T) μ0 4π 10 T m/a I = = = 750 A 8 2 Assess: The cuent density in this supeconducting wie is of the ode of 1 10 A/m (o cuent/aea = 750 A/( π ( ) m ) A/m ). This is a typical value fo conventional supeconducting mateials. μ0i P Pepae: We use the fomula fo the magnetic field due to a long staight wie: B =. 2π Solve: Plugging in the cuent and distance yields: 8 μ0i (4π 10 T m/a)( A) 12 B = = = T 3 2π 2 π( m) Assess: This is a vey small field smalle than eath s by seven odes of magnitude. Howeve, this esult is not supising if we compae it to poblem 1 in which a cuent of 2.5 A was needed to make a field equal to the

12 24-12 Chapte 24 eath s at a distance of 1.0 cm. At 1.0 mm, only one quate of an amp would be needed. But the cuent in this poblem is much smalle than A. P Pepae: The magnetic field in the cente of a solenoid is B = μ0 IN/ L. This equation is independent of the adius of the solenoid, and theefoe only depends on the cuent (I), numbe of loops (N), and the length of the solenoid (L). Solve: Rewiting the equation fo a solenoid, one can calculate the cuent in the wie to be the following: μ0in BL B= I = L μ0n (1.0 T)(2 m) 3 I = = A = 1.6 ka (4π 10 T m/a)(1000) Assess: This cuent is vey high. It is about 1000 times geate than a nomal electical cuent. P Pepae: The magnetic field at the cente of the two loops is zeo, so the magnetic field of the inne loop must be equal and opposite to the magnetic field of the oute loop. Solve: μ0(12 A) 4 Binne = = T 2(0.03 m) Bcente = Boute + Binne = 0 μ0( 20A) 4 Boute = = T 2R R = 0.05 m = 5 cm Assess: Fo the magnetic field at the cente of the two loops to be zeo, the two magnetic fields must be equal and opposite. The cuent in the oute loop is assumed to be in the negative diection because it is the only vaiable that can be negative. A negative adius has no physical meaning. P Pepae: If we model the cuent as a single (N = 1) cicula cuent loop we will want to use Equation μ0i B = 2R We ae given B = T and R = m; we ae asked to find I. Solve: Solve fo I. 12 2RB 2(0.080m)( T) I = = = A μ0 4π 10 T m/a Assess: This is a small cuent, but it seems easonable fo a cuent in one s head. Afte all, the magnetic field poduced is also vey small. Notice the units; both m and T cancel to leave A. P Pepae: Fom Table 24.1, the magnetic field of the eath is 5 10 T. The magnetic field in the cente of a solenoid is B = μ0in/ L. This equation is independent of the adius of the solenoid, and theefoe only depends on the cuent (I), numbe of loops (N ), and the length of the solenoid (L). Solve: Rewiting the equation fo a solenoid, one can calculate the cuent in the wie to be the following: μ0in BL B= I = L μ0n ( T)(4.0 m) 3 I = = A = 32 ma (4π 10 T m/a)(5000) Assess: The magnetic field of the eath can be cancelled by using a solenoid with elatively little cuent.

13 Magnetic Fields and Foces P Pepae: This poblem can be envisioned as a supeposition of a staight cuent and a cuent loop. The magnetic field at the cente of a loop is the combination of both the magnetic field of the loop plus the magnetic field of the staight pat of the wie. Bloop = μ0i/(2 R) and Bwie = μ0i/(2 π ) B = B + B total loop wie Note that fo the magnetic field of the loop, R is the adius of the loop, wheeas, fo the magnetic field of the wie, is the distance fom the staight section of the wie. In this situation, is the distance fom the point whee the wie begins to bend into a loop. By using the ight-hand ule, we see the diections of both magnetic fields point in the same diection, down. Theefoe, the two magnetic fields add togethe. Solve: The magnetic field at the cente of the loop is: B μ I μ I μ I 1 1 2R 2π 2 R π (4π 10 T m/a)(5.0 A) 1 1 = m π (0.01m) total = + = T = Assess: Since magnetic fields ae vecto fields, we can use supeposition. P Pepae: We ae not given the length of this coil, so we assume the 100 tuns ae all close togethe and we teat it as an N-tun coil athe than a solenoid. The applicable equation is Equation We ae given N = 100, I = 1.5 A, and R = 3.5 m. Solve: (a) μ0 NI (4π 10 T m/a)(100)(1.5 A) B = = = 2R 2(3.5m) T (b) Table 24.1 indicates that the field stength at the suface of the eath is about T, and ou answe is smalle than that by about half. Assess: The shaks ae detecting a field stength that is about half as stong as the eath s field. They can, theefoe, pesumably detect the eath s field. Some othe species may also be sensitive to magnetic fields. P Pepae: The heat is compaed to a loop of cuent with adius 6 cm and magnetic field of 90 pt at its cente. Solve: The cuent needed to poduce this field can be computed fom the equation fo a magnetic field at the cente of a loop. 12 2BR 2(90 10 T)(0.06 m) μ0 (4π 10 T m/a) I = = = Assess: This is a small cuent, as we would have expected A P Pepae: The magnetic field fo a cuent loop wie is diffeent fom the magnetic field of an ideal B = μ NI/(2 R). This equation does depend on the adius of the loop (R) as well as the solenoid: cuent loop at cente 0 numbe of loops (N) and cuent in the wie (I). Since the numbe of loops was not given in the poblem, you have to figue out how many loops can be ceated based on the adius of a single loop. Solve: The equation to elate the numbe of loops to the adius is N = L/(2 π ). Fom the cuent loop equation, the magnetic field at the cente of an N-tun coil becomes B μ IN μ I L μ IL = = 2R 2R 2πR = 2πR coil cente 2 whee we used the equiement that the entie length of wie be used: i.e., N = L/(2 π ). Solving fo R, R I μ0 (1.0 m) 4 π(10 T m/a)(1.0 m)(1.0 A) = = = = 3 2 2πBcoil cente 4 π( T) The diamete of the coil is D= 2R= 2.0 cm m 1.0 cm

14 24-14 Chapte 24 Assess: Using cuent loops and solenoids, you can enginee expeiments to meet you specific needs. P Pepae: Assuming the electon is a single paticle taveling in a cicula obit, the moving electon is, by definition, a cicula cuent. Any cuent moving in a cicle ceates a magnetic field at the cente of the obit. Solve: The cuent fom a single electon moving in a cicula obit is 19 6 q qv qv ( C)( m/s) I = = = = = 1.06 ma 11 t d 2π 2 π( m) With this cuent, the magnetic field can be calculated as B 3 μ0i (4π 10 T m/a)( A) = = = ( m) 12.5 T which we ll epot as 13 T. Assess: This is a vey lage magnetic field stength, but note the extemely small adius of the electon obit. P Pepae: A magnetic field exets a magnetic foce on a moving chage. The magnitude of the magnetic foce is calculated using the equation F B = q vbsin α. The ight-hand ule fo foces is used to find the diection. Solve: (a) The magnitude of the magnetic foce on the chage is F q vb α B = sin = ( C)( m/s)(0.50 T)(sin45 ) = N Using the ight-hand ule fo foces, you find that the diection of the foce is in the +y-diection o into the page. (b) Because the magnetic field is paallel to the velocity, the foce is zeo, F on q = 0 N. Assess: The diection of the foce is consistent with the ight-hand ule. P Pepae: A magnetic field exets a magnetic foce on a moving chage. The magnitude of the magnetic foce is calculated using the equation F B = qvb sin α. The ight-hand ule fo foces is used to find the diection on a positive chage. Solve: (a) The magnitude of the magnetic foce on the chage is F qvb α B = sin = ( C)( m/s)(0.50 T)(sin90 ) = N Using the ight-hand ule fo foces, you find that the diection of the foce is in the +z-diection o up. So, on the electon the foce is in the z diection, o down. (b) The magnitude of the magnetic foce on the chage is F qvb α B = sin = ( C)( m/s)(0.50 T)(sin90 ) = N Using the ight-hand ule fo foces, you find that the diection of the foce is in the +yz-diection, o 45 up and into the page. So, on the electon the foce is 45 down and out of the page. Assess: The velocity, magnetic field, and magnetic foce ae always pependicula to each othe. Using the ight-hand ule fo foces, you can detemine the diection of the magnetic foce without calculating all the individual components of velocity (v) and magnetic field (B). P Pepae: You have a magnetic field of 0.20 T applied to a negative Cl ion with a velocity of 15 cm/s. The angle between the applied magnetic field and the velocity of the ion is assumed to be 90. Solve: The magnetic foce is given by F = q vbsin θ. F = = ( C)(0.15 m/s)(0.20 T) N Assess: This is a vey small foce because the ion s speed and chage ae small. P Pepae: Chaged paticles moving pependicula to a unifom magnetic field undego unifom 2 2 cicula motion ( a = v / ) at a constant speed (v): Fnet = FB mv / = qvbo = vm/( qb). Solve: The calculations fo the adius of the cicula obit is

15 Magnetic Fields and Foces vm ( m/s)( kg) electon 19 = = = = qb ( C)(5 10 T) 4 27 vm ( m/s)( kg) poton 19 = = = 10.4 m qb ( C)(5 10 T) m 11.4 cm Assess: In a magneton, the chaged paticles have an angula velocity elated to the linea velocity by v= ω = 2 π f, whee f is the fequency and is the adius of the cicula motion. These chaged paticles ceate light that can be seen at the Noth Pole and South Pole, and the pattens ae fom these obits. P Pepae: The fequency of the cicula obit can be calculated fom angula fequency, ω = 2π f = v /, and the adius, = vm/( qb). Solve: = v/(2 π f) and = vm/( qb) v vm qb = f = 2π f qb 2πm 19 ( C)( T) 6 felecton = = Hz 31 2 π ( kg) 19 ( C)( T) fpoton = = 760 Hz 27 2 π ( kg) Assess: These ae athe high fequencies. The poton s fequency is highe because of its highe mass. P Pepae: Caefully examine Example 24.5 as a lead-in to this poblem. Instead of the time (peiod) fo one obit we want the fequency, and f = 1/ T. qb f = 2π m 31 We ae given f = 2.4 GHz, and we look up the mass of the electon: m = kg. Solve: Solve the equation fo B. 31 2πmf 2 π( kg)(2.4 GHz) B = = = T 19 q C Assess: The emitted electomagnetic waves have a wavelength of 12.5 cm and ae just ight to excite the wate molecules and cook the food. 1 Check the units. 1 C = 1 A s and 1Hz= 1s, so the units combine to give T (see Example 24.7). P Pepae: Chaged ions in a mass spectomete move in cicula obits in a unifom magnetic field. Refe to figue Equation 24.8 govens the motion of ions in a mass spectomete. Solve: The poblem states that the ion has one electon emoved, so the net chage on the ion is positive, and 19 equal to the chage of a poton, so q = C. The total mass of the ion is 85 times the mass of the poton, so m = 85( kg) = kg. Solving Equation 24.8 fo R : mv ( kg)( m/s) R = = = qb -19 ( C)(0.80 T) 1 The distance between whee the ion entes and exits the field is 2R = m m P Pepae: Detemining the mass of the paticle will be useful in detemining the type of paticle. A mass spectomete is used to detemine the mass of ions o chaged paticles. The mass of a paticle measued by a mass spectomete is given by Equation Solve: Using Equation 24.8:

16 24-16 Chapte qbr ( C)(0.05 T)(0.21m) m = = = 5 v m/s kg. This value is a few times the mass of a poton. Assuming the paticle is an ion of an element, we can detemine the numbe of poton-mass objects in the nucleus of the element by dividing by the mass of a poton: kg N = = kg A helium atom consists of two potons, two neutons, and fou electons, which is appoximately fou times the mass of a poton (neutons have appoximately the same mass of a poton, and the electons masses ae negligible). Assuming the paticle is an ion of an elemental atom, a helium atom with one electon emoved matches the popeties of the paticle. P Pepae: Using the ight-hand ule, you can detemine that the magnetic foce would point out of the page. Solve: Theefoe, thee is no foce component fom the magnetic field in the y o x diection. qvb Fz = maz = FB = qvb az = (out of the page) m Fy = may = 0 ay = 0 Fx = max = 0 ax = 0 Thus, the velocity in the y diection does not change, because acceleation is zeo and the distance the poton moves in the y diection is only dependent on the initial velocity in the y diection. vy = vsin30 y = vyt 5 6 y = ( m/s)(sin30 )(10 10 s) y = 2.8 m P Pepae: The electons ae acceleated by a potential diffeence, so we need to use the consevation of enegy to find thei final speed: mvi evi = mvf evf. Then to find the effect of the magnetic field, we 2 2 need to use magnetic foce law: Fmag = evb. Solve: (a) Electons speed up when they move to highe potential. Let us say that an electon stats at potential 0 V and eaches final speed at potential 3.0 kv. Since the electon also begins with speed 0, its initial enegy is 0. The final speed is found as follows: eVf 2( C)( V) 7 0 = mvf evf vf = = = m/s, 31 2 m ( kg) 7 which ounds to m/s, (b) Fom Newton s second law, 19 7 F evf B ( C)( m/s)(0.65 T) a = = = = m/s 31 m m kg mag 18 2

17 Magnetic Fields and Foces g 17 In tems of g, this is a= ( m/s ) = ( ) g m/s (c) If the electons wee to complete a full obit, the adius of that obit would be given by Equation 24.8, 31 7 mvf ( kg)( m/s) = = = 0.28 mm 19 qb ( C)(0.65 T) (d) Magnetic fields apply foces to chaged paticles in a diection pependicula to thei instantaneous velocity. If we conside a shot inteval of time, Δ t, duing which the velocity of the paticle is v, then the paticle s displacement is vδt while the magnetic foce on the paticle is pependicula to that displacement. When foce and displacement ae always pependicula, no wok is done. By the fist law of themodynamics, if a foce does no wok on an object, it cannot change the kinetic enegy of that object. Thus magnetic fields cannot be used to incease the speed of a chaged paticle. Assess: The adius of the electon s obit woked out in pat (c) is vey small compaed to the size of a television set. Appaently, the magnetic field expeienced by an electon is usually much less than 0.65 T, so that the field stength has a time-aveaged value much less than 0.65 T. P Pepae: Assume that the field is unifom. The wie will float in the magnetic field if the magnetic foce on the wie points upwad and has a magnitude equal to mg, allowing it to balance the downwad gavitational foce. Solve: We can use the ight-hand ule to detemine which field diection makes the wie expeience an upwad foce. The cuent being fom ight to left, the foce will be up if the magnetic field B points out of the page. The foces will balance when F = ILB= mg. 3 2 mg ( kg)(9.8 m/s ) B = = = 0.13 T IL (1.5 A)(0.10 m) Thus B = (0.13 T, out of page). P Pepae: Two paallel wies caying cuents in opposite diections exet epulsive magnetic foces on each othe. Two paallel wies caying cuents in the same diection exet attactive magnetic foces on each othe. Solve: The magnitudes of the vaious foces between the paallel wies ae μ LI I F F F F 2π d 0.02m μ0li1i3 (2 10 T m/a)(0.50 m)(10 A)(10 A) 4 F3 on 1 = = = N = F1 on 3 2π d 0.04m Now we can find the net foce each wie exets on the othe as follows: Fon 1 = F2 on 1 + F3 on 1 = ( N, up) + ( N, down) = ( N, up) 4 4 Fon 2 = F1 on 2 + F3 on 2 = ( N, up) + ( N, down) = 0 N 4 4 Fon 3 = F1 on 3 + F2 on 3 = ( N, up) + ( N, down) = ( N, down) Assess: The foces on a wie ae cumulative, so the contibution due to the both wies is added (2 10 T m/a)(0.50 m)(10 A)(10 A) 4 2 on 1 = = = N = 2 on 3 = 3 on 2 = 1 on 2 P Pepae: To find the foce on one of the 1.0 m long segments, we need the fomula fo the foce on a cuent caying wie in a magnetic field: F = BIL. The figue shows the field ceated by the wie on the left and the foce felt by the wie on the ight. We also need to know the magnetic field expeienced by one of the wies μ0i (they expeience the same field stength). This is given by B =. 2π

18 24-18 Chapte 24 Solve: Combining the fomula fo the field expeienced by a wie with the fomula fo the foce expeienced by the wie gives us the esult we want: 2 2 0I 0 I L (4 10 T m/a)(1.0 A) (1.0 m) μ μ π F = BIL = IL 2π = = = 2π 2 π(1.0 m) 2 10 N Assess: This is actually how the amp is defined. You take two long wies and un identical cuent though them. Then if you gadually vay the cuent until the foce pe unit mete on the wies is 2 10 N/m, you know that 1 amp is flowing though the wies. All the units in electicity elate back to the amp. Fo example, the Coulomb is the amount of chage which flows past a point in a wie in one second if the cuent in the wie is one amp. P Pepae: Assume that the magnetic field is unifom. Solve: The magnitude of the magnetic foce is expessed as F = ILBsin α, wheeα is the angle the wie makes with the magnetic field. F = ILBsinα F = (15 A)(3.0 m)(2.5 T) sin30 F = 56 N Using the ight-hand ule, we can detemine that the diection of the foce is into the page. Assess: Given that the cuent and the magnetic field stength ae lage, a foce of 56 N is easonable. P Pepae: Assume that the magnetic field is unifom. Solve: The magnitude of the magnetic foce is expessed as F = ILBsin α, wheeα is the angle the velocity vecto makes with the magnetic field. F = ILBsinα 100 N = (15 A)(3.0 m)(2.5 T)sinα 100 N 8 sinα = = (15 A)(3.0 m)(2.5 T) α = sin = 63 9 P Pepae: This poblem makes us eview some basic concepts in electicity. We will need to find the cuent in the coil using Ohm s law. Then to find the foce on the coil, we will need to use the fomula fo the foce on a cuent caying wie: F = BIL. Finally, to find the acceleation of the coil, we will need Newton s second law.