Module 19: Sections 9.1 through 9.2 Module 20: Sections 9.3 through Table of Contents

Transcription

1 Module 19: Sections 9.1 though 9. Module : Sections 9.3 though Table of Contents 9.1 Biot-Savat Law Magnetic Field of a Cuent Element Inteactive Simulation Example 9.1: Magnetic Field due to a Finite Staight Wie Example 9.: Magnetic Field due to a Cicula Cuent Loop Magnetic Field of a Moving Point Chage Magnetic Field of a Moving Chage Animation Magnetic Field of Seveal Chages Moving in a Cicle Animation Magnetic Field of a Ring of Moving Chages Animation Foce Between Two Paallel Wies Foces between Cuent-Caying Paallel Wies Animation Ampee s Law Example 9.3: Field Inside and Outside a Cuent-Caying Wie Example 9.4: Magnetic Field Due to an Infinite Cuent Sheet Solenoid Examaple 9.5: Tooid Magnetic Field of a Dipole Eath s Magnetic Field at MIT Ba Magnet in the Eath s Magnetic Field Inteactive Simulation Magnetic Mateials Magnetization Paamagnetism Diamagnetism Feomagnetism Summay These notes ae excepted Intoduction to Electicity and Magnetism by Sen-Ben Liao, Pete Doumashkin, and John Belche, Copyight 4, ISBN

2 9.8 Appendix 1: Magnetic Field off the Symmety Axis of a Cuent Loop Appendix : Helmholtz Coils Magnetic Field of the Helmholtz Coils Animation Magnetic Field of Two Coils Caying Opposite Cuents Animation Foces Between Coaxial Cuent-Caying Wies Animation Magnet Oscillating Between Two Coils Animation Magnet Suspended Between Two Coils Animation Poblem-Solving Stategies Biot-Savat Law: Ampee s law: Solved Poblems Magnetic Field of a Staight Wie Cuent-Caying Ac Rectangula Cuent Loop Haipin-Shaped Cuent-Caying Wie Two Infinitely Long Wies Non-Unifom Cuent Density Thin Stip of Metal Two Semi-Infinite Wies Conceptual Questions Additional Poblems Application of Ampee's Law Magnetic Field of a Cuent Distibution fom Ampee's Law Cylinde with a Hole The Magnetic Field Though a Solenoid Rotating Disk Fou Long Conducting Wies Magnetic Foce on a Cuent Loop Magnetic Moment of an Obital Electon Feomagnetism and Pemanent Magnets Chage in a Magnetic Field Pemanent Magnets Magnetic Field of a Solenoid Effect of Paamagnetism

3 Souces of Magnetic Fields 9.1 Biot-Savat Law Cuents which aise due to the motion of chages ae the souce of magnetic fields. When chages move in a conducting wie and poduce a cuent I, the magnetic field at any point P due to the cuent can be calculated by adding up the magnetic field contibutions, db, fom small segments of the wie d s, (Figue 9.1.1). Figue Magnetic field db at point P due to a cuent-caying element I d s. These segments can be thought of as a vecto quantity having a magnitude of the length of the segment and pointing in the diection of the cuent flow. The infinitesimal cuent souce can then be witten as I d s. Let denote as the distance fom the cuent souce to the field point P, and ˆ the coesponding unit vecto. The Biot-Savat law gives an expession fo the magnetic field contibution, db, fom the cuent souce, Id s, μ Id s ˆ db = (9.1.1) 4π whee μ is a constant called the pemeability of fee space: μ = 4π 1 7 T m/a (9.1.) Notice that the expession is emakably simila to the Coulomb s law fo the electic field due to a chage element dq: 1 dq de = ˆ 4πε (9.1.3) Adding up these contibutions to find the magnetic field at the point P equies integating ove the cuent souce, 9-3

4 μ I d s ˆ 4π B = db = (9.1.4) wie The integal is a vecto integal, which means that the expession fo B is eally thee integals, one fo each component of B. The vecto natue of this integal appeas in the coss poduct Id s ˆ. Undestanding how to evaluate this coss poduct and then pefom the integal will be the key to leaning how to use the Biot-Savat law. wie Magnetic Field of a Cuent Element Inteactive Simulation Figue 9.1. is an inteactive ShockWave display that shows the magnetic field of a cuent element fom Eq. (9.1.1). This inteactive display allows you to move the position of the obseve about the souce cuent element to see how moving that position changes the value of the magnetic field at the position of the obseve. Figue 9.1. Magnetic field of a cuent element (link) Example 9.1: Magnetic Field due to a Finite Staight Wie A thin, staight wie caying a cuent I is placed along the x-axis, as shown in Figue Evaluate the magnetic field at point P. Note that we have assumed that the leads to the ends of the wie make canceling contibutions to the net magnetic field at the point P. Figue A thin staight wie caying a cuent I. 9-4

5 Solution: This is a typical example involving the use of the Biot-Savat law. We solve the poblem using the methodology summaized in Section 9.1. (1) Souce point (coodinates denoted with a pime) Conside a diffeential element d s = dx ' î caying cuent I in the x-diection. The location of this souce is epesented by ' = x ' î. () Field point (coodinates denoted with a subscipt P ) Since the field point P is located at ( x, y ) = (, a), the position vecto descibing P is = aĵ. P (3) Relative position vecto The vecto = P ' is a elative position vecto which points fom the souce point to the field point. In this case, = a ˆ j x 'î, and the magnitude = = a + x ' is the distance fom between the souce and P. The coesponding unit vecto is given by a ˆ = = ĵ x 'î =sinθ a ĵ cos θ î + x ' (4) The coss poduct d s ˆ The coss poduct is given by d s ˆ = (dx'ˆ) i ( cos θ ˆ i + sin θ ˆj) = (dx'sin θ )kˆ (5) Wite down the contibution to the magnetic field due to Id s The expession is μ I d s ˆ μ I dx sinθ db = = kˆ 4π 4π which shows that the magnetic field at P will point in the +kˆ diection, o out of the page. (6) Simplify and cay out the integation 9-5

6 The vaiables θ, x and ae not independent of each othe. In ode to complete the integation, let us ewite the vaiables x and in tems of θ. Fom Figue 9.1.3, we have = a/sin θ = a csc θ x = a cotθ dx = a csc θ dθ Upon substituting the above expessions, the diffeential contibution to the magnetic field is obtained as μ I ( a csc θ θ μ I 4π (a csc θ ) 4π a d )sin θ db = = sinθ d θ Integating ove all angles subtended fom θ 1 to θ (a negative sign is needed fo θ 1 in ode to take into consideation the potion of the length extended in the negative x axis fom the oigin), we obtain μ B = I 4 π a θ μ sinθ d θ = I (cos θ θ ) θ + cos 1 (9.1.5) 1 4 π a The fist tem involving θ accounts fo the contibution fom the potion along the +x axis, while the second tem involving θ 1 contains the contibution fom the potion along the x axis. The two tems add! Let s examine the following cases: (i) In the symmetic case whee θ = θ 1, the field point P is located along the pependicula bisecto. If the length of the od is L, then cosθ 1 = L / L + a magnetic field is μ B = I π a cos θ = μ I L 1 π a L + a and the (9.1.6) (ii) The infinite length limit L This limit is obtained by choosing ( θ1, θ ) = (,). The magnetic field at a distance a away becomes μ I B = (9.1.7) π a 9-6

7 Note that in this limit, the system possesses cylindical symmety, and the magnetic field lines ae cicula, as shown in Figue Figue Magnetic field lines due to an infinite wie caying cuent I. In fact, the diection of the magnetic field due to a long staight wie can be detemined by the ight-hand ule (Figue 9.1.5). Figue Diection of the magnetic field due to an infinite staight wie If you diect you ight thumb along the diection of the cuent in the wie, then the finges of you ight hand cul in the diection of the magnetic field. In cylindical coodinates (, ϕ, z) whee the unit vectos ae elated by ˆ ˆφ = z ˆ, if the cuent flows in the +z-diection, then, using the Biot-Savat law, the magnetic field must point in the ϕ diection. Example 9.: Magnetic Field due to a Cicula Cuent Loop A cicula loop of adius R in the xy plane caies a steady cuent I, as shown in Figue (a) What is the magnetic field at a point P on the axis of the loop, at a distance z fom the cente? (b) If we place a magnetic dipole μ = μ z kˆ at P, find the magnetic foce expeienced by the dipole. Is the foce attactive o epulsive? What happens if the diection of the dipole is evesed, i.e., μ = μ z kˆ 9-7

8 Solution: Figue Magnetic field due to a cicula loop caying a steady cuent. (a) This is anothe example that involves the application of the Biot-Savat law. Again let s find the magnetic field by applying the same methodology used in Example 9.1. (1) Souce point In Catesian coodinates, the diffeential cuent element located at ' = R(cos φ ' î +sin φ ' ĵ) can be witten as Ids = I ( d '/ dφ ' )dφ ' = IRdφ ' ( sin φ 'î + cos φ ' ĵ). () Field point Since the field point P is on the axis of the loop at a distance z fom the cente, its position vecto is given by = zkˆ P. (3) Relative position vecto = ' The elative position vecto is given by and its magnitude P = ' = R cos φ 'î R sin φ ' ĵ + z kˆ P (9.1.8) = = ( R cos φ ') + ( R sin φ ') + z = R + z (9.1.9) is the distance between the diffeential cuent element and P. Thus, the coesponding unit vecto fom Id s to P can be witten as P ' ˆ = = P ' 9-8

9 (4) Simplifying the coss poduct The coss poduct d s ( P ') can be simplified as d s ( P ') = Rd φ '( sinφ 'î + cosφ ' ĵ) [ R cos φ 'î R sin φ 'ˆ j+ z k ˆ ] = Rdφ '[z cos φ 'î + z sin φ 'ˆ j+ R k ˆ ] (9.1.1) (5) Witing down db Using the Biot-Savat law, the contibution of the cuent element to the magnetic field at P is (6) Caying out the integation μ I d s ˆ μ I d s μ I d s ( P ') db = = = 4 π 4π 3 4π P ' 3 μ IR z cosφ 'î + z sin φ 'ˆ + R ˆ (9.1.11) j k = dφ ' 4π (R + z ) 3/ Using the esult obtained above, the magnetic field at P is μ IR π z cos φ ' î + z sin φ ' ĵ + B = R k ˆ 4π dφ ' (9.1.1) (R + z ) 3/ The x and the y components of B can be eadily shown to be zeo: μ IRz π μ IRz π B x = cos ' d ' = sin = (9.1.13) 4 (R + z ) φ φ 3/ 4π (R π + z ) φ ' 3/ μ IRz π μ IRz π B y = sin 'd ' = cos π z φ φ φ ' = (9.1.14) 4 (R + ) 3/ 4π (R + z ) 3/ On the othe hand, the z component is μ IR π μ π IR μ IR B = dφ '= z = (9.1.15) 4 π (R + z ) 3/ 4π (R + z ) 3/ ( R + z ) 3/ Thus, we see that along the symmetic axis, B z is the only non-vanishing component of the magnetic field. The conclusion can also be eached by using the symmety aguments. 9-9

10 The behavio of B z / B whee B = μ I /R is the magnetic field stength at z =, as a function of z/ R is shown in Figue 9.1.7: Figue The atio of the magnetic field, B / B z, as a function of z/ R (b) If we place a magnetic dipole μ = μ z kˆ at the point P, as discussed in Chapte 8, due to the non-unifomity of the magnetic field, the dipole will expeience a foce given by F = (μ B B) = (μ z B z ) = μ z db z kˆ (9.1.16) dz Upon diffeentiating Eq. (9.1.15) and substituting into Eq. (9.1.16), we obtain μμ IR z F = 3 z B ( R + z ) 5/ kˆ (9.1.17) Thus, the dipole is attacted towad the cuent-caying ing. On the othe hand, if the diection of the dipole is evesed, μ = μ z kˆ, the esulting foce will be epulsive Magnetic Field of a Moving Point Chage Suppose we have an infinitesimal cuent element in the fom of a cylinde of cosssectional aea A and length ds consisting of n chage caies pe unit volume, all moving at a common velocity v along the axis of the cylinde. Let I be the cuent in the element, which we define as the amount of chage passing though any coss-section of the cylinde pe unit time. Fom Chapte 6, we see that the cuent I can be witten as = I (9.1.18) naq v The total numbe of chage caies in the cuent element is simply dn = n Ads, so that using Eq. (9.1.1), the magnetic field db due to the dn chage caies is given by 9-1

11 db = (naq ) d μ v s ˆ μ = ( n A ds ) q v ˆ μ = (dn) q v ˆ (9.1.19) 4π 4π 4π whee is the distance between the chage and the field point P at which the field is being measued, the unit vecto ˆ = / points fom the souce of the field (the chage) to P. The diffeential length vecto d s is defined to be paallel to v. In case of a single chage, dn = 1, the above equation becomes μ B = 4π q v ˆ (9.1.) Note, howeve, that since a point chage does not constitute a steady cuent, the above equation stictly speaking only holds in the non-elativistic limit whee v c, the speed of light, so that the effect of etadation can be ignoed. The esult may be eadily extended to a collection of N point chages, each moving with a diffeent velocity. Let the ith chage q i be located at ( x i, y i, z i ) and moving with velocity v i. Using the supeposition pinciple, the magnetic field at P can be obtained as: N μ B = q v (x x i )î + ( y y )ĵ+ (z z )kˆ i i i i (9.1.1) 3/ i= 1 4π (x x i ) + ( y y i ) + (z z i ) Magnetic Field of a Moving Chage Animation Figue shows one fame of the animations of the magnetic field of a moving positive and negative point chage, assuming the speed of the chage is small compaed to the speed of light. Figue The magnetic field of (a) a moving positive chage (link), and (b) a moving negative chage (link) when the speed of the chage is small compaed to c. 9-11

12 9.1.4 Magnetic Field of Seveal Chages Moving in a Cicle Animation Suppose we want to calculate the magnetic fields of a numbe of chages moving on the cicumfeence of a cicle with equal spacing between the chages. To calculate this field we have to add up vectoially the magnetic fields of each of chages using Eq. (9.1.19). Figue The magnetic field of fou chages moving in a cicle (link). We show the magnetic field vecto diections in only one plane. The bullet-like icons indicate the diection of the magnetic field at that point in the aay spanning the plane. Figue shows one fame of the animation when the numbe of moving chages is fou. Othe animations show the same situation fo N =1,, and 8. When we get to eight chages, a chaacteistic patten emeges--the magnetic dipole patten. Fa fom the ing, the shape of the field lines is the same as the shape of the field lines fo an electic dipole Magnetic Field of a Ring of Moving Chages Animation Figue shows a ShockWave display of the vectoal addition pocess fo the case whee we have 3 chages moving on a cicle. The display in Figue shows an obsevation point fixed on the axis of the ing. As the addition poceeds, we also show the esultant up to that point (lage aow in the display). Figue A ShockWave simulation of the use of the pinciple of supeposition to find the magnetic field due to 3 moving chages moving in a cicle at an obsevation point on the axis of the cicle (link) 9-1

13 Figue The magnetic field due to 3 chages moving in a cicle at a given obsevation point. The position of the obsevation point can be vaied to see how the magnetic field of the individual chages adds up to give the total field (link) In Figue , we show an inteactive ShockWave display that is simila to that in Figue 9.1.1, but now we can inteact with the display to move the position of the obseve about in space. To get a feel fo the total magnetic field, we also show a ion filings epesentation of the magnetic field due to these chages. We can move the obsevation point about in space to see how the total field at vaious points aises fom the individual contibutions of the magnetic field of to each moving chage. 9. Foce Between Two Paallel Wies We have aleady seen that a cuent-caying wie poduces a magnetic field. In addition, when placed in a magnetic field, a wie caying a cuent will expeience a net foce. Thus, we expect two cuent-caying wies to exet foce on each othe. Conside two paallel wies sepaated by a distance a and caying cuents I 1 and I in the +x-diection, as shown in Figue Figue 9..1 Foce between two paallel wies The magnetic foce, F 1, exeted on wie 1 by wie may be computed as follows: Using the esult fom the pevious example, the magnetic field lines due to I going in the +x diection ae cicles concentic with wie, with the field B pointing in the tangential 9-13

14 diection. Thus, at an abitay point P on wie 1, we have B = (μ I / π a)ĵ, which points in the diection pependicula to wie 1, as depicted in Figue Theefoe, μ I μ IIl 1 F = I l B = I l î ĵ = kˆ ( ) (9..1) π a π a Clealy F 1 points towad wie. The conclusion we can daw fom this simple calculation is that two paallel wies caying cuents in the same diection will attact each othe. On the othe hand, if the cuents flow in opposite diections, the esultant foce will be epulsive Foces between Cuent-Caying Paallel Wies Animation Figues 9.. shows paallel wies caying cuent in the same and in opposite diections. In the fist case, the magnetic field configuation is such as to poduce an attaction between the wies. In the second case the magnetic field configuation is such as to poduce a epulsion between the wies. (a) (b) Figue 9.. (a) The attaction between two wies caying cuent in the same diection. The diection of cuent flow is epesented by the motion of the oange sphees in the visualization (link). (b) The epulsion of two wies caying cuent in opposite diection (link) 9.3 Ampee s Law We have seen that moving chages o cuents ae the souce of magnetism. This can be eadily demonstated by placing compass needles nea a wie. As shown in Figue 9.3.1a, all compass needles point in the same diection in the absence of cuent. Howeve, when I, the needles will be deflected along the tangential diection of the cicula path (Figue 9.3.1b). 9-14

15 Figue Deflection of compass needles nea a cuent-caying wie Let us now divide a cicula path of adius into a lage numbe of small length vectos Δ s= Δsφˆ, that point along the tangential diection with magnitude Δs (Figue 9.3.). In the limit Δs, we obtain Figue 9.3. Ampeian loop Ñ d s = B Ñ ds = μ B I π ( π ) = μ I (9.3.1) The esult above is obtained by choosing a closed path, o an Ampeian loop that follows one paticula magnetic field line. Let s conside a slightly moe complicated Ampeian loop, as that shown in Figue Figue An Ampeian loop involving two field lines 9-15

16 The line integal of the magnetic field aound the contou abcda is Ñ B d s = B d s + B d s + B d s + B d s abcda ab bc cd cd (9.3.) = +B ( θ ) + + B 1 [ 1 ( π θ )] whee the length of ac bc is θ, and 1 ( π θ) fo ac da. The fist and the thid integals vanish since the magnetic field is pependicula to the paths of integation. With B 1 = μ I /π 1 and B = μ I /π, the above expession becomes Ñ μ I μ ( θ ) + I π π 1 π θ )] = μ I θ + μ B d s = [ ( I ( π θ ) = μ I (9.3.3) π π abcda 1 We see that the same esult is obtained whethe the closed path involves one o two magnetic field lines. As shown in Example 9.1, in cylindical coodinates (, ϕ, z) with cuent flowing in the +z-axis, the magnetic field is given by B = (μ I / π )φˆ. An abitay length element in the cylindical coodinates can be witten as which implies Ñ B d s = Ñ d s = d ˆ + d ϕ φˆ + dz ẑ (9.3.4) μ I π d ϕ = μ π I Ñ dϕ = μ closed path closed path closed path I ( π ) = μ I (9.3.5) π In othe wods, the line integal of Ñ B d s aound any closed Ampeian loop is popotional to I enc, the cuent encicled by the loop. Figue An Ampeian loop of abitay shape. 9-16

17 The genealization to any closed loop of abitay shape (see fo example, Figue 9.3.4) that involves many magnetic field lines is known as Ampee s law: Ñ B d s = μ I (9.3.6) enc Ampee s law in magnetism is analogous to Gauss s law in electostatics. In ode to apply them, the system must possess cetain symmety. In the case of an infinite wie, the system possesses cylindical symmety and Ampee s law can be eadily applied. Howeve, when the length of the wie is finite, Biot-Savat law must be used instead. Biot-Savat Law μ I d s ˆ B = 4π Ampee s law d s = μ Ienc Ñ B geneal cuent souce ex: finite wie cuent souce has cetain symmety ex: infinite wie (cylindical) Ampee s law is applicable to the following cuent configuations: 1. Infinitely long staight wies caying a steady cuent I (Example 9.3). Infinitely lage sheet of thickness b with a cuent density J (Example 9.4). 3. Infinite solenoid (Section 9.4). 4. Tooid (Example 9.5). We shall examine all fou configuations in detail. Example 9.3: Field Inside and Outside a Cuent-Caying Wie Conside a long staight wie of adius R caying a cuent I of unifom cuent density, as shown in Figue Find the magnetic field eveywhee. Figue Ampeian loops fo calculating the B field of a conducting wie of adius R. 9-17

18 Solution: (i) Outside the wie whee R, the Ampeian loop (cicle 1) completely encicles the cuent, i.e., I enc = I. Applying Ampee s law yields which implies Ñ B d s = B Ñ ds =B( π ) = μ I μ I B = π (ii) Inside the wie whee < R, the amount of cuent encicled by the Ampeian loop (cicle ) is popotional to the aea enclosed, i.e., π I enc = π R I Thus, we have Ñ B d s =B ( π ) = μ I π π R B = μ I π R We see that the magnetic field is zeo at the cente of the wie and inceases linealy with until =R. Outside the wie, the field falls off as 1/. The qualitative behavio of the field is depicted in Figue below: Figue Magnetic field of a conducting wie of adius R caying a steady cuent I. Example 9.4: Magnetic Field Due to an Infinite Cuent Sheet Conside an infinitely lage sheet of thickness b lying in the xy plane with a unifom cuent density J = J î. Find the magnetic field eveywhee. 9-18

19 Figue An infinite sheet with cuent density J = J î. Solution: We may think of the cuent sheet as a set of paallel wies caying cuents in the +xdiection. Fom Figue 9.3.8, we see that magnetic field at a point P above the plane points in the y-diection. The z-component vanishes afte adding up the contibutions fom all wies. Similaly, we may show that the magnetic field at a point below the plane points in the +y-diection. Figue Magnetic field of a cuent sheet We may now apply Ampee s law to find the magnetic field due to the cuent sheet. The Ampeian loops ae shown in Figue Figue Ampeian loops fo the cuent sheets Fo the field outside, we integate along path C 1. The amount of cuent enclosed by C 1 is 9-19

20 Applying Ampee s law leads to Ñ B I = J d A = J ( ) bl (9.3.7) enc d s = B( ) = μ I = μ ( l l J b ) (9.3.8) enc o B = μ Jb/. Note that the magnetic field outside the sheet is constant, independent of the distance fom the sheet. Next we find the magnetic field inside the sheet. The amount of cuent enclosed by path C is Applying Ampee s law, we obtain I enc = J d A = J ( z l ) (9.3.9) Ñ B d s = B( ) l = μ I enc = μ J ( z l ) (9.3.1) o B = μ J z. At z =, the magnetic field vanishes, as equied by symmety. The esults can be summaized using the unit-vecto notation as μ Jb ĵ, z > b/ B = μ Jzĵ, b / < z < b / (9.3.11) Jb μ ĵ, z < b / Let s now conside the limit whee the sheet is infinitesimally thin, with b. In this case, instead of cuent density J = J î, we have suface cuent K = K î, whee K = J. Note that the dimension of K is cuent/length. In this limit, the magnetic field becomes μ K ĵ, z > B = (9.3.1) μ K ĵ, z < 9.4 Solenoid A solenoid is a long coil of wie tightly wound in the helical fom. Figue shows the magnetic field lines of a solenoid caying a steady cuent I. We see that if the tuns ae closely spaced, the esulting magnetic field inside the solenoid becomes faily unifom, 9-

21 povided that the length of the solenoid is much geate than its diamete. Fo an ideal solenoid, which is infinitely long with tuns tightly packed, the magnetic field inside the solenoid is unifom and paallel to the axis, and vanishes outside the solenoid. Figue Magnetic field lines of a solenoid We can use Ampee s law to calculate the magnetic field stength inside an ideal solenoid. The coss-sectional view of an ideal solenoid is shown in Figue To compute B, we conside a ectangula path of length l and width w and tavese the path in a counteclockwise manne. The line integal of B along this loop is Ñ B d s = B d s + B d s + B d s + B d s (9.4.1) = + + Bl + Figue 9.4. Ampeian loop fo calculating the magnetic field of an ideal solenoid. In the above, the contibutions along sides and 4 ae zeo because B is pependicula to d s. In addition, B= along side 1 because the magnetic field is non-zeo only inside the solenoid. On the othe hand, the total cuent enclosed by the Ampeian loop is I = NI, whee N is the total numbe of tuns. Applying Ampee s law yields o enc Ñ B d s = Bl = μ NI (9.4.) 9-1

22 B = μ NI = μ ni (9.4.3) l whee n= N / l epesents the numbe of tuns pe unit length., In tems of the suface cuent, o cuent pe unit length K = ni, the magnetic field can also be witten as, B = μ K (9.4.4) What happens if the length of the solenoid is finite? To find the magnetic field due to a finite solenoid, we shall appoximate the solenoid as consisting of a lage numbe of cicula loops stacking togethe. Using the esult obtained in Example 9., the magnetic field at a point P on the z axis may be calculated as follows: Take a coss section of tightly packed loops located at z with a thickness dz ', as shown in Figue The amount of cuent flowing though is popotional to the thickness of the coss section and is given by di = I ( ndz ') = I( N / l) dz ', whee n= N / l is the numbe of tuns pe unit length. Figue Finite Solenoid The contibution to the magnetic field at P due to this subset of loops is μ R μ R db z = di = (nidz ') (9.4.5) [( z z ') + R ] 3/ [( z z ') + R ] 3/ Integating ove the entie length of the solenoid, we obtain μ nir l / dz ' μ nir z ' z B z = = l / [( z z ') + R ] 3/ R (z z ') + R μ ni ( l / ) z ( l / ) + z + (z l / ) + R (z + l / ) + R = l / l / (9.4.6) 9-

23 A plot of B z / B, whee B = μ ni is the magnetic field of an infinite solenoid, as a function of / l = 1 l = z R is shown in Figue fo R and R. Figue Magnetic field of a finite solenoid fo (a) l = 1R, and (b) l = R. Notice that the value of the magnetic field in the egion z < l / is nealy unifom and appoximately equal to B. Examaple 9.5: Tooid Conside a tooid which consists of N tuns, as shown in Figue Find the magnetic field eveywhee. Solutions: Figue A tooid with N tuns One can think of a tooid as a solenoid wapped aound with its ends connected. Thus, the magnetic field is completely confined inside the tooid and the field points in the azimuthal diection (clockwise due to the way the cuent flows, as shown in Figue ) Applying Ampee s law, we obtain 9-3

24 Ñ B d s = Ñ Bds = B Ñ ds =B(π ) = μ NI (9.4.7) o μ NI B = (9.4.8) π whee is the distance measued fom the cente of the tooid.. Unlike the magnetic field of a solenoid, the magnetic field inside the tooid is non-unifom and deceases as1/. 9.5 Magnetic Field of a Dipole Let a magnetic dipole moment vecto μ = μkˆ be placed at the oigin (e.g., cente of the Eath) in the yz plane. What is the magnetic field at a point (e.g., MIT) a distance away fom the oigin? Figue Eath s magnetic field components In Figue we show the magnetic field at MIT due to the dipole. The y- and z- components of the magnetic field ae given by B y = μ 3μ sinθ cos θ, B = μ μ (3cos 3 z θ 1) (9.5.1) 3 4π 4π Reades ae efeed to Section 9.8 fo the detail of the deivation. In spheical coodinates (,θ,φ), the adial and the pola components of the magnetic field can be witten as B = B y sinθ + B z cos θ = μ μ cos θ (9.5.) 3 4π 9-4

25 and B θ = B y cosθ B z sin θ = μ μ sin θ (9.5.3) 3 4π espectively. Thus, the magnetic field at MIT due to the dipole becomes B = B θ θˆ + B ˆ = μ μ (sin θ θˆ + cos θ ˆ) (9.5.4) 3 4π Notice the similaity between the above expession and the electic field due to an electic dipole p (see Solved Poblem.13.6): 1 E = 4πε p 3 (sin θ θˆ + cos θ ˆ) The negative sign in Eq. (9.5.4) is due to the fact that the magnetic dipole points in the zdiection. In geneal, the magnetic field due to a dipole moment μ can be witten as The atio of the adial and the pola components is given by μ 3(μ ˆˆ ) μ B = (9.5.5) 3 4π μ μ cos θ B = 4π 3 = cot θ (9.5.6) B θ μ μ sinθ 4π Eath s Magnetic Field at MIT The Eath s field behaves as if thee wee a ba magnet in it. In Figue 9.5. an imaginay magnet is dawn inside the Eath oiented to poduce a magnetic field like that of the Eath s magnetic field. Note the South pole of such a magnet in the nothen hemisphee in ode to attact the Noth pole of a compass. It is most natual to epesent the location of a point P on the suface of the Eath using the spheical coodinates (, θ, φ), whee is the distance fom the cente of the Eath, θ is the pola angle fom the z-axis, with θ π, and φ is the azimuthal angle in the xy plane, measued fom the x-axis, with φ π (See Figue ) With the distance fixed at = E, the adius of the Eath, the point P is paameteized by the two anglesθ and φ. 9-5

26 Figue 9.5. Magnetic field of the Eath In pactice, a location on Eath is descibed by two numbes latitude and longitude. How ae they elated to θ and φ? The latitude of a point, denoted as δ, is a measue of the elevation fom the plane of the equato. Thus, it is elated to θ (commonly efeed to as the colatitude) by δ = 9 θ. Using this definition, the equato has latitude, and the noth and the south poles have latitude ± 9, espectively. The longitude of a location is simply epesented by the azimuthal angle φ in the spheical coodinates. Lines of constant longitude ae geneally efeed to as meidians. The value of longitude depends on whee the counting begins. Fo histoical easons, the meidian passing though the Royal Astonomical Obsevatoy in Geenwich, UK, is chosen as the pime meidian with zeo longitude. Figue Locating a point P on the suface of the Eath using spheical coodinates. Let the z-axis be the Eath s otation axis, and the x-axis passes though the pime meidian. The coesponding magnetic dipole moment of the Eath can be witten as μ E = μ E (sin θ cos φ î + sin θ sin φ ĵ + cos θ kˆ ) = μ E (.6 î +.18 ĵ.98kˆ ) (9.5.7) 9-6

27 whee μ E = A m, and we have used (θ, φ ) = (169,19 ). The expession shows that μ E has non-vanishing components in all thee diections in the Catesian coodinates. On the othe hand, the location of MIT is 4 N fo the latitude and 71 W fo the longitude ( 4 noth of the equato, and 71 west of the pime meidian), which means that θ m = 9 4 = 48, and φ m = = 89. Thus, the position of MIT can be descibed by the vecto = (sin θ cos φ î + sin θ sinφ ĵ + cos θ kˆ MIT E m m m m m ) = E (.4 î.7 ĵ +.67 kˆ ) The angle between μ E and MIT is given by (9.5.8) μ θ ME = cos 1 MIT E = cos 1 (.8) = 37 (9.5.9) μ MIT E Note that the pola angle θ is defined as θ = cos 1 (k ˆ ˆ ), the invese of cosine of the dot poduct between a unit vecto ˆ fo the position, and a unit vecto +kˆ in the positive z- diection, as indicated in Figue Thus, if we measue the atio of the adial to the pola component of the Eath s magnetic field at MIT, the esult would be B = cot (9.5.1) B θ Note that the positive adial (vetical) diection is chosen to point outwad and the positive pola (hoizontal) diection points towads the equato Ba Magnet in the Eath s Magnetic Field Inteactive Simulation Figue shows a ba magnet and compass placed on a table. The inteaction between the magnetic field of the ba magnet and the magnetic field of the eath is illustated by the field lines that extend out fom the ba magnet. Field lines that emege towads the edges of the magnet geneally econnect to the magnet nea the opposite pole. Howeve, field lines that emege nea the poles tend to wande off and econnect to the magnetic field of the eath, which, in this case, is appoximately a constant field coming at 6 degees fom the hoizontal. Looking at the compass, one can see that a compass needle will always align itself in the diection of the local field. In this case, the local field is dominated by the ba magnet. Click and dag the mouse to otate the scene. Contol-click and dag to zoom in and out. 9-7

28 Figue A ba magnet in Eath s magnetic field (link) 9.6 Magnetic Mateials The intoduction of mateial media into the study of magnetism has vey diffeent consequences as compaed to the intoduction of mateial media into the study of electostatics. When we dealt with dielectic mateials in electostatics, thei effect was always to educe E below what it would othewise be, fo a given amount of fee electic chage. In contast, when we deal with magnetic mateials, thei effect can be one of the following: (i) educe B below what it would othewise be, fo the same amount of "fee" electic cuent (diamagnetic mateials); (ii) incease B a little above what it would othewise be (paamagnetic mateials); (iii) incease B a lot above what it would othewise be (feomagnetic mateials). Below we discuss how these effects aise Magnetization Magnetic mateials consist of many pemanent o induced magnetic dipoles. One of the concepts cucial to the undestanding of magnetic mateials is the aveage magnetic field poduced by many magnetic dipoles which ae all aligned. Suppose we have a piece of mateial in the fom of a long cylinde with aea A and height L, and that it consists of N magnetic dipoles, each with magnetic dipole moment μ, spead unifomly thoughout the volume of the cylinde, as shown in Figue

29 Figue A cylinde with N magnetic dipole moments We also assume that all of the magnetic dipole moments μ ae aligned with the axis of the cylinde. In the absence of any extenal magnetic field, what is the aveage magnetic field due to these dipoles alone? To answe this question, we note that each magnetic dipole has its own magnetic field associated with it. Let s define the magnetization vecto M to be the net magnetic dipole moment vecto pe unit volume: 1 M = μ i (9.6.1) V i whee V is the volume. In the case of ou cylinde, whee all the dipoles ae aligned, the magnitude of M is simply M = Nμ / AL. Now, what is the aveage magnetic field poduced by all the dipoles in the cylinde? Figue 9.6. (a) Top view of the cylinde containing magnetic dipole moments. (b) The equivalent cuent. Figue 9.6.(a) depicts the small cuent loops associated with the dipole moments and the diection of the cuents, as seen fom above. We see that in the inteio, cuents flow in a given diection will be cancelled out by cuents flowing in the opposite diection in neighboing loops. The only place whee cancellation does not take place is nea the edge of the cylinde whee thee ae no adjacent loops futhe out. Thus, the aveage cuent in the inteio of the cylinde vanishes, wheeas the sides of the cylinde appea to cay a net cuent. The equivalent situation is shown in Figue 9.6.(b), whee thee is an equivalent cuent I eq on the sides. 9-9

30 The functional fom of I eq may be deduced by equiing that the magnetic dipole moment poduced by I eq be the same as total magnetic dipole moment of the system. The condition gives I eq A= Nμ (9.6.) o Nμ I eq = (9.6.3) A Next, let s calculate the magnetic field poduced by I eq. With I eq unning on the sides, the equivalent configuation is identical to a solenoid caying a suface cuent (o cuent pe unit length) K. The two quantities ae elated by I eq Nμ K = = = M (9.6.4) L AL Thus, we see that the suface cuent K is equal to the magnetization M, which is the aveage magnetic dipole moment pe unit volume. The aveage magnetic field poduced by the equivalent cuent system is given by (see Section 9.4) B M = μ K = μ M (9.6.5) Since the diection of this magnetic field is in the same diection as M, the above expession may be witten in vecto notation as B M = μ M (9.6.6) This is exactly opposite fom the situation with electic dipoles, in which the aveage electic field is anti-paallel to the diection of the electic dipoles themselves. The eason is that in the egion inteio to the cuent loop of the dipole, the magnetic field is in the same diection as the magnetic dipole vecto. Theefoe, it is not supising that afte a lage-scale aveaging, the aveage magnetic field also tuns out to be paallel to the aveage magnetic dipole moment pe unit volume. Notice that the magnetic field in Eq. (9.6.6) is the aveage field due to all the dipoles. A vey diffeent field is obseved if we go close to any one of these little dipoles. Let s now examine the popeties of diffeent magnetic mateials 9-3

31 9.6. Paamagnetism The atoms o molecules compising paamagnetic mateials have a pemanent magnetic dipole moment. Left to themselves, the pemanent magnetic dipoles in a paamagnetic mateial neve line up spontaneously. In the absence of any applied extenal magnetic field, they ae andomly aligned. Thus, M = and the aveage magnetic field BM is also zeo. Howeve, when we place a paamagnetic mateial in an extenal field B, the dipoles expeience a toque τ = μ B that tends to align μ with B, theeby poducing a net magnetization M paallel to B. Since B M is paallel to B, it will tend to enhance B. The total magnetic field B is the sum of these two fields: B= B + B = B + μ M M (9.6.7) Note how diffeent this is than in the case of dielectic mateials. In both cases, the toque on the dipoles causes alignment of the dipole vecto paallel to the extenal field. Howeve, in the paamagnetic case, that alignment enhances the extenal magnetic field, wheeas in the dielectic case it educes the extenal electic field. In most paamagnetic substances, the magnetization M is not only in the same diection as B, but also linealy popotional to B. This is plausible because without the extenal field B thee would be no alignment of dipoles and hence no magnetization M. The linea elation between M and B is expessed as B M = χ (9.6.8) m μ whee χ m is a dimensionless quantity called the magnetic susceptibility. Eq. (1.7.7) can then be witten as whee B = (1 + χ m )B = κ m B (9.6.9) κ m = 1 +χ m (9.6.1) is called the elative pemeability of the mateial. Fo paamagnetic substances, κ m >1, o equivalently, χ m >, although χ m is usually on the ode of 1 6 to 1 3. The magnetic pemeability μ m of a mateial may also be defined as μ = (1 +χ ) m m μ = κm μ (9.6.11) 9-31

32 Paamagnetic mateials have μ m > μ Diamagnetism In the case of magnetic mateials whee thee ae no pemanent magnetic dipoles, the pesence of an extenal field B will induce magnetic dipole moments in the atoms o molecules. Howeve, these induced magnetic dipoles ae anti-paallel to B, leading to a magnetization M and aveage field B M anti-paallel to B, and theefoe a eduction in the total magnetic field stength. Fo diamagnetic mateials, we can still define the magnetic pemeability, as in equation (8-5), although now κ m < 1, o χ m <, although χ m is usually on the ode of 1 5 to 1 9. Diamagnetic mateials have μ m < μ Feomagnetism In feomagnetic mateials, thee is a stong inteaction between neighboing atomic dipole moments. Feomagnetic mateials ae made up of small patches called domains, as illustated in Figue 9.6.3(a). An extenally applied field B will tend to line up those magnetic dipoles paallel to the extenal field, as shown in Figue 9.6.3(b). The stong inteaction between neighboing atomic dipole moments causes a much stonge alignment of the magnetic dipoles than in paamagnetic mateials. Figue (a) Feomagnetic domains. (b) Alignment of magnetic moments in the diection of the extenal field B. The enhancement of the applied extenal field can be consideable, with the total 3 4 magnetic field inside a feomagnet 1 o 1 times geate than the applied field. The pemeability κ m of a feomagnetic mateial is not a constant, since neithe the total field B o the magnetization M inceases linealy with B. In fact the elationship between M and B is not unique, but dependent on the pevious histoy of the mateial. The 9-3

33 phenomenon is known as hysteesis. The vaiation of M as a function of the extenally applied field B is shown in Figue The loop abcdef is a hysteesis cuve. Figue A hysteesis cuve. Moeove, in feomagnets, the stong inteaction between neighboing atomic dipole moments can keep those dipole moments aligned, even when the extenal magnet field is educed to zeo. And these aligned dipoles can thus poduce a stong magnetic field, all by themselves, without the necessity of an extenal magnetic field. This is the oigin of pemanent magnets. To see how stong such magnets can be, conside the fact that magnetic dipole moments of atoms typically have magnitudes of the ode of 1 3 A m. Typical atomic densities ae 1 9 atoms/m 3. If all these dipole moments ae aligned, then we would get a magnetization of ode M (1 A m )(1 atoms/m ) 1 6 A/m (9.6.1) The magnetization coesponds to values of B M = μ M of ode 1 tesla, o 1, Gauss, just due to the atomic cuents alone. This is how we get pemanent magnets with fields of ode Gauss. 9.7 Summay Biot-Savat law states that the magnetic field db at a point due to a length element ds caying a steady cuent I and located at away is given by whee = and μ = 4π 1 μ db = Ids ˆ 4π 7 T m/a is the pemeability of fee space. The magnitude of the magnetic field at a distance away fom an infinitely long staight wie caying a cuent I is 9-33

34 μ I B = π The magnitude of the magnetic foce F B between two staight wies of length l caying steady cuent of I 1 and I and sepaated by a distance is μ F I 1 I l B = π Ampee s law states that the line integal of B ds aound any closed loop is popotional to the total steady cuent passing though any suface that is bounded by the close loop: d = μ I Ñ B s enc The magnetic field inside a tooid which has N closely spaced of wie caying a cuent I is given by μ NI B = π whee is the distance fom the cente of the tooid. The magnetic field inside a solenoid which has N closely spaced of wie caying cuent I in a length of l is given by B = μ N I = μ ni l whee n is the numbe of numbe of tuns pe unit length. The popeties of magnetic mateials ae as follows: Mateials Magnetic susceptibility χ m Relative pemeability κ = 1+χ m m Magnetic pemeability μ = κ μ Diamagnetic κ m < 1 μ m < μ Paamagnetic κ m > 1 μ m > μ Feomagnetic χ m 1 κ m 1 μ m μ m m 9-34

35 9.8 Appendix 1: Magnetic Field off the Symmety Axis of a Cuent Loop In Example 9. we calculated the magnetic field due to a cicula loop of adius R lying in the xy plane and caying a steady cuent I, at a point P along the axis of symmety. Let s see how the same technique can be extended to calculating the field at a point off the axis of symmety in the yz plane. Figue Calculating the magnetic field off the symmety axis of a cuent loop. Again, as shown in Example 9.1, the diffeential cuent element is Id s = R dφ '( sin φ 'î + cos φ ' ĵ) and its position is descibed by ' = R(cos φ ' î + sin φ ' ĵ). On the othe hand, the field point P now lies in the yz plane with = y ĵ + z kˆ P, as shown in Figue The coesponding elative position vecto is with a magnitude = ' = R cos φ 'î + ( y R sin φ ') ĵ + zkˆ (9.8.1) P = = ( R cos φ ') + ( y R sin φ ') + z = R + y + z yr sin φ (9.8.) and the unit vecto ' ˆ = = P P ' pointing fom Id s to P. The coss poduct d s ˆ can be simplified as d s ˆ = Rd φ '( sin φ 'i ˆ + cos φ ' ˆ) j [ R cos φ 'i ˆ + ( y R sin φ ' )ĵ + z kˆ ] = Rdφ '[z cos φ 'î + z sin φ ' ĵ + (R y sin φ ') k] ˆ (9.8.3) 9-35

36 Using the Biot-Savat law, the contibution of the cuent element to the magnetic field at P is μ I s ˆ μ I d d s μ IR z cosφ 'î + z sinφ ' ĵ+ (R y sinφ ') kˆ db = = = dφ ' (9.8.4) 3 4π 4π 4π R + y + z yr sin Thus, magnetic field at P is ( φ ') 3/ μ cos φ 'î B(, y, z) = + z sin φ ' ĵ (R y sinφ ')kˆ IR π z + dφ ' (9.8.5) 4π (R + y + z yr sin φ ') 3/ The x-component of B can be eadily shown to be zeo μ IRz π cosφ 'dφ ' B x = 4π = (9.8.6) (R + y 3/ + z yr sin φ ') by making a change of vaiable w= R + y + z yr sin φ ', followed by a staightfowad integation. One may also invoke symmety aguments to veify that B x must vanish; namely, the contibution at φ ' is cancelled by the contibution at π φ'. On the othe hand, the y and the z components of B, and B = μ IRz π sinφ 'dφ ' y 4π (R + y + z yr sin φ ') 3/ (9.8.7) μ IR π (R y sin φ' )d φ ' B z = 4π (9.8.8) 3/ (R + y + z yr sin involve elliptic integals which can be evaluated numeically. In the limit y =, the field point P is located along the z-axis, and we ecove the esults obtained in Example 9.: and μ IRz π μ IRz π B y = sin 'd ' = cos = (9.8.9) 4 π (R + z ) φ φ φ ' 3/ 4π (R + z ) 3/ φ ') 9-36

37 B = μ IR π μ dφ '= π IR μ = IR z (9.8.1) 4 π (R + z ) 3/ 4 π (R + z ) 3/ ( R + z ) 3/ Now, let s conside the point-dipole limit whee R ( y + z ) 1/ =, i.e., the chaacteistic dimension of the cuent souce is much smalle compaed to the distance whee the magnetic field is to be measued. In this limit, the denominato in the integand can be expanded as (R y z yr sin φ ') 3/ = 1 R yr sin φ ' 3/ R yr sin φ ' = 3 1 +K (9.8.11) This leads to μ I Rz π 3 R yr sin φ ' B y 4π 3 1 sin φ 'dφ ' μ I 3R yz π φ φ μ I 3π R yz = 5 sin 'd ' = 5 4π 4π (9.8.1) and B μ I R π 3 R yr sin φ ' z 1 (R y sin ')d ' 4π 3 φ φ μ I R π 3R 3 1 9R 3Ry = ' ' 3 R sin φ sin φ φ ' d 4π μ I R 3R 3 3π Ry = 3 π R 4π = μ I π 3 4π R 3 y + highe ode te ms (9.8.13) The quantity I (π R ) may be identified as the magnetic dipole moment μ = IA, whee A = π R is the aea of the loop. Using spheical coodinates whee y = sinθ and z = cosθ, the above expessions may be ewitten as μ B y = (I π R ) 3( sin θ )( cos θ ) μ 5 = 3 μsin θcos θ 3 4π 4π (9.8.14) 9-37

38 and μ (I R ) π 3 sin θ μ μ μ μ B z = 3 = ( 3sin θ ) = (3cos θ 1) 3 3 4π 4π 4π (9.8.15) Thus, we see that the magnetic field at a point R due to a cuent ing of adius R may be appoximated by a small magnetic dipole moment placed at the oigin (Figue 9.8.). Figue 9.8. Magnetic dipole moment μ = μkˆ The magnetic field lines due to a cuent loop and a dipole moment (small ba magnet) ae depicted in Figue Figue Magnetic field lines due to (a) a cuent loop, and (b) a small ba magnet. The magnetic field at P can also be witten in spheical coodinates B = B ˆ + B θ ˆθ (9.8.16) The spheical components B and B θ ae elated to the Catesian components by By and B z B B = y sinθ + z B cos θ, B θ = By cos θ B z sin θ (9.8.17) In addition, we have, fo the unit vectos, ˆ = sinθ ˆ j + cos θ k ˆ, ˆθ = cos θ ˆ j sin θ ˆk (9.8.18) Using the above elations, the spheical components may be witten as 9-38

39 and B = μ IR cosθ π dφ ' (9.8.19) 4π (R + R sin θ sin φ ') 3/ μ IR π ( sinφ ' R sinθ )dφ ' Bθ (,θ ) = (9.8.) 4π (R + R sin θ sin φ ') 3/ In the limit whee R, we obtain and μ IR cos θ π μ π IR cos θ μ μ cos θ B 3 dφ ' = = (9.8.1) 3 3 4π 4π 4π B θ = μ IR π ( sinφ ' R sinθ )dφ ' 4π (R + R sin θ sin φ ') 3/ μ IR π 3R 3R 3R sin θ R sin θ 1 + sinφ '+ 3R sin θ sin φ ' dφ ' 4π 3 μ IR ( π R sin θ + 3π R sin θ ) = μ (Iπ R )sin θ 4π 3 4π 3 μ μ sinθ = 4π Appendix : Helmholtz Coils (9.8.) Conside two N-tun cicula coils of adius R, each pependicula to the axis of symmety, with thei centes located at z =±l /. Thee is a steady cuent I flowing in the same diection aound each coil, as shown in Figue Let s find the magnetic field B on the axis at a distance z fom the cente of one coil. Figue Helmholtz coils 9-39

40 Using the esult shown in Example 9. fo a single coil and applying the supeposition pinciple, the magnetic field at Pz (,)(a point at a distance z l/ away fom one cente and z+ l/ fom the othe) due to the two coils can be obtained as: B z = B top + B bottom = μ NI R 1 [(z l/ ) + R ] + 1 3/ [(z + l / ) + R ] 3/ (9.9.1) B z B with = μ NI A plot of / B being the field stength at z = and l = R is (5 / 4) 3/ R depicted in Figue Figue 9.9. Magnetic field as a function of z/ R. Let s analyze the popeties of B z obtain in moe detail. Diffeentiating B z with espect to z, we B ( z) = db z = μ NIR 3(z l/ ) 3(z + l / ) z dz [(z l / ) + R ] 5/ [(z + l / ) + R ] 5/ (9.9.) One may eadily show that at the midpoint, z =, the deivative vanishes: db dz z= = (9.9.3) Staightfowad diffeentiation yields B ( z) = db = Nμ IR 3 15(z l / ) z + dz [(z l / ) + R ] 5/ [(z l / ) + R ] 7 / 3 15(z+ l/ ) + [(z+ l/ ) + R ] 5/ [(z + l / ) + R ] 7 / (9.9.4) 9-4

41 At the midpoint z =, the above expession simplifies to μ NI 6 15l = + dz z= [( l / ) + R ] 5/ [( l / ) + R ] 7 / (9.9.5) μ NI 6(R l ) = [( l / ) + R ] 7 / B z () = d B Thus, the condition that the second deivative of B z vanishes at z = is l = R. That is, the distance of sepaation between the two coils is equal to the adius of the coil. A configuation with l = Ris known as Helmholtz coils. Fo small z, we may make a Taylo-seies expansion of B z ( z) about z = : B ( ) = () + B z z B z z ()z + 1 B () z z +... (9.9.6)! The fact that the fist two deivatives vanish at z = indicates that the magnetic field is faily unifom in the small z egion. One may even show that the thid deivative B z () vanishes at z = as well. Recall that the foce expeienced by a dipole in a magnetic field is F = (μ B B ). If we place a magnetic dipole μ = μ z kˆ at z =, the magnetic foce acting on the dipole is F = (μ B ) = μ B z z z db z kˆ (9.9.7) dz which is expected to be vey small since the magnetic field is nealy unifom thee Magnetic Field of the Helmholtz Coils Animation The animation in Figue 9.9.3(a) shows the magnetic field of the Helmholtz coils. In this configuation the cuents in the top and bottom coils flow in the same diection, with thei dipole moments aligned. The magnetic fields fom the two coils add up to ceate a net field that is nealy unifom at the cente of the coils. Since the distance between the coils is equal to the adius of the coils and emains unchanged, the foce of attaction between them ceates a tension, and is illustated by field lines stetching out to enclose both coils. When the distance between the coils is not fixed, as in the animation depicted in Figue 9.9.3(b), the two coils move towad each othe due to thei foce of attaction. In this animation, the top loop has only half the cuent as the bottom loop. The field configuation is shown using the ion filings epesentation. 9-41

42 (a) (b) Figue (a) Magnetic field of the Helmholtz coils whee the distance between the coils is equal to the adius of the coil (link). (b) Two co-axial wie loops caying cuent in the same sense ae attacted to each othe (link) Next, let s conside the case whee the cuents in the loop flow in the opposite diections, as shown in Figue Figue Two cicula loops caying cuents in the opposite diections. Again, by supeposition pinciple, the magnetic field at a point P(,, z) with z > is B z = B 1z + B z = μ NIR 1 1 3/ 3/ (9.9.8) [( z l / ) + R ] [( z + l / ) + R ] A plot of B z / B with B = μ NI /R and l = Ris depicted in Figue

43 Figue Magnetic field as a function of z/ R. Diffeentiating B z with espect to z, we obtain B ( z) = db z = μ NIR 3(z l / ) z dz [( z l / ) + R ] 5/ At the midpoint, z =, we have 3( z + l / ) + (9.9.9) [( z + l / ) + R ] 5/ B z μ NIR z 3l () = db = dz z = [( l /) + R ] 5/ (9.9.1) Thus, a magnetic dipole μ = μ z kˆ placed at z = will expeience a net foce: Fo l db () μμ NIR z 3l F B = (μ B) = (μ z B z ) = μ z kˆ z = dz [( l /) + R ] 5/ kˆ (9.9.11) =R, the above expession simplifies to 3μ z μ NI F ˆ B = (5/ 4) 5/ R k (9.9.1) 9.9. Magnetic Field of Two Coils Caying Opposite Cuents Animation The animation depicted in Figue shows the magnetic field of two coils like the Helmholtz coils but with cuents in the top and bottom coils flowing in the opposite diections. In this configuation, the magnetic dipole moments associated with each coil ae anti-paallel. (a) (b) Figue (a) Magnetic field due to coils caying cuents in the opposite diections (link). (b) Two co-axial wie loops caying cuent in the opposite sense epel each othe. The field configuations hee ae shown using the ion filings epesentation. The bottom wie loop caies twice the amount of cuent as the top wie loop (link) 9-43

44 At the cente of the coils along the axis of symmety, the magnetic field is zeo. With the distance between the two coils fixed, the epulsive foce esults in a pessue between them. This is illustated by field lines that ae compessed along the cental hoizontal axis between the coils Foces Between Coaxial Cuent-Caying Wies Animation Figue A magnet in the TeachSpin Magnetic Foce appaatus when the cuent in the top coil is counteclockwise as seen fom the top (link) Figue shows the foce of epulsion between the magnetic field of a pemanent magnet and the field of a cuent-caying ing in the TeachSpin Magnetic Foce appaatus. The magnet is foced to have its Noth magnetic pole pointing downwad, and the cuent in the top coil of the Magnetic Foce appaatus is moving clockwise as seen fom above. The net esult is a epulsion of the magnet when the cuent in this diection is inceased. The visualization shows the stesses tansmitted by the fields to the magnet when the cuent in the uppe coil is inceased Magnet Oscillating Between Two Coils Animation Figue illustates an animation in which the magnetic field of a pemanent magnet suspended by a sping in the TeachSpinTM appaatus (see TeachSpin visualization), plus the magnetic field due to cuent in the two coils (hee we see a "cutaway" coss-section of the appaatus). Figue Magnet oscillating between two coils (link) 9-44

45 The magnet is fixed so that its noth pole points upwad, and the cuent in the two coils is sinusoidal and 18 degees out of phase. When the effective dipole moment of the top coil points upwads, the dipole moment of the bottom coil points downwads. Thus, the magnet is attacted to the uppe coil and epelled by the lowe coil, causing it to move upwads. When the conditions ae evesed duing the second half of the cycle, the magnet moves downwads. This pocess can also be descibed in tems of tension along, and pessue pependicula to, the field lines of the esulting field. When the dipole moment of one of the coils is aligned with that of the magnet, thee is a tension along the field lines as they attempt to "connect" the coil and magnet. Convesely, when thei moments ae anti-aligned, thee is a pessue pependicula to the field lines as they ty to keep the coil and magnet apat Magnet Suspended Between Two Coils Animation Figue illustates an animation in which the magnetic field of a pemanent magnet suspended by a sping in the TeachSpinTM appaatus (see TeachSpin visualization), plus the magnetic field due to cuent in the two coils (hee we see a "cutaway" coss-section of the appaatus). The magnet is fixed so that its noth pole points upwad, and the cuent in the two coils is sinusoidal and in phase. When the effective dipole moment of the top coil points upwads, the dipole moment of the bottom coil points upwads as well. Thus, the magnet the magnet is attacted to both coils, and as a esult feels no net foce (although it does feel a toque, not shown hee since the diection of the magnet is fixed to point upwads). When the dipole moments ae evesed duing the second half of the cycle, the magnet is epelled by both coils, again esulting in no net foce. This pocess can also be descibed in tems of tension along, and pessue pependicula to, the field lines of the esulting field. When the dipole moment of the coils is aligned with that of the magnet, thee is a tension along the field lines as they ae "pulled" fom both sides. Convesely, when thei moments ae anti-aligned, thee is a pessue pependicula to the field lines as they ae "squeezed" fom both sides. Figue Magnet suspended between two coils (link) 9-45

46 9.1 Poblem-Solving Stategies In this Chapte, we have seen how Biot-Savat and Ampee s laws can be used to calculate magnetic field due to a cuent souce Biot-Savat Law: The law states that the magnetic field at a point P due to a length element ds caying a steady cuent I located at away is given by μ I ds ˆ μ I d s db = = 3 4π 4π The calculation of the magnetic field may be caied out as follows: (1) Souce point: Choose an appopiate coodinate system and wite down an expession fo the diffeential cuent element I ds, and the vecto ' descibing the position of I ds. The magnitude ' = ' is the distance between I ds and the oigin. Vaiables with a pime ae used fo the souce point. () Field point: The field point P is the point in space whee the magnetic field due to the cuent distibution is to be calculated. Using the same coodinate system, wite down the position vecto fo the field point P. The quantity = P P P is the distance between the oigin and P. (3) Relative position vecto: The elative position between the souce point and the field point is chaacteized by the elative position vecto = P '. The coesponding unit vecto is ' ˆ = = P P ' whee = = ' is the distance between the souce and the field point P. P (4) Calculate the coss poduct d s ˆ o d s. The esultant vecto gives the diection of the magnetic field B, accoding to the Biot-Savat law. (5) Substitute the expessions obtained to db and simplify as much as possible. (6) Complete the integation to obtain B if possible. The size o the geomety of the system is eflected in the integation limits. Change of vaiables sometimes may help to complete the integation. 9-46

47 Below we illustate how these steps ae executed fo a cuent-caying wie of length L and a loop of adius R. Cuent distibution Finite wie of length L Cicula loop of adius R Figue ˆ ' = x ' i ' = R (cos φ ' ˆ i + sin φ ' ˆj ) (1) Souce point d s = ( d '/ dx ') dx ' = dx ' ˆ i d s = ( d '/ d φ ') d φ ' = Rd φ '( sin φ ' ˆ i + cos φ ' ˆj ) () Field point P = yˆ j (3) Relative position vecto = ' P P = yˆ j x ' ˆi P = zkˆ = = x ' + y ˆ = (4) The coss poduct d s ˆ = d s ˆ yˆ j x ' ˆi x ' + y ydx k ˆ y + x = R cos φ ' ˆ i R sin φ ' ˆ j + z k ˆ = = R + z R cos φ ' ˆ i R sin φ ' ˆ j + z k ˆ ˆ = R + z Rd φ '( z cos φ ' ˆ i + z sin φ ' ˆ j + Rk ˆ ) d s ˆ = R + z (5) Rewite db (6) Integate to get B μ I ydx k d B = 3/ 4 ( y + x ) B = B x y = μ / Iy L dx ' B z = / 3/ 4 π L ( y + x ' ) π μ I L = 4 π y y + ( L / ) ˆ μ I Rd φ '( z cos φ ' ˆ i + z sin φ ' ˆ j + Rkˆ ) d B = 3/ 4 π ( R + z ) μ IRz π B x = cos φ' dφ' = 3/ 4 π ( R + z ) μ IRz π B y = sin φ' dφ' = 3/ 4 π ( R + z ) μ IR μ IR B z = = 4 π ( R + z ) ( R + z ) π d φ ' 3/ 3/ 9.1. Ampee s law: 9-47

48 Ampee s law states that the line integal of B ds aound any closed loop is popotional to the total cuent passing though any suface that is bounded by the closed loop: Ñ B d s = μ I enc To apply Ampee s law to calculate the magnetic field, we use the following pocedue: (1) Daw an Ampeian loop using symmety aguments. () Find the cuent enclosed by the Ampeian loop. (3) Calculate the line integal Ñ B d s aound the closed loop. (4) Equate Ñ d B s with μ I enc and solve fo B. Below we summaize how the methodology can be applied to calculate the magnetic field fo an infinite wie, an ideal solenoid and a tooid. System Infinite wie Ideal solenoid Tooid Figue (1) Daw the Ampeian loop () Find the cuent enclosed by the Ampeian loop (3) Calculate B Ñ d s along the loop I enc = I I enc = NI I enc = NI B Ñ d s = B ( π ) B Ñ d s = Bl B Ñ d s = B ( π ) (4) Equate μ I enc with B Ñ d s to obtain B μ I π B = B μ NI μ NI = = μ ni B = l π 9-48

49 9.11 Solved Poblems Magnetic Field of a Staight Wie Conside a staight wie of length L caying a cuent I along the +x-diection, as shown in Figue (ignoe the etun path of the cuent o the souce fo the cuent.) What is the magnetic field at an abitay point P on the xy-plane? Solution: Figue A finite staight wie caying a cuent I. The poblem is vey simila to Example 9.1. Howeve, now the field point is an abitay point in the xy-plane. Once again we solve the poblem using the methodology outlined in Section 9.1. (1) Souce point Fom Figue 9.1.1, we see that the infinitesimal length dx descibed by the position vecto ' = x ' î constitutes a cuent souce I d s = (Idx )î. () Field point As can be seen fom Figue 9.1.1, the position vecto fo the field point P is = x î + y ĵ. (3) Relative position vecto The elative position vecto fom the souce to P is = ˆ ˆ P ' = (x x ') i + y j, with 1 = P = ' = [( x x ) + y ] being the distance. The coesponding unit vecto is ' ˆ = = P = P ' (x x ) î + y ĵ [( x x ) + y ]

50 (4) Simplifying the coss poduct The coss poduct d s can be simplified as whee we have used ˆ i ˆ i = and ˆ i ˆ j = kˆ. (5) Witing down db ( dx ' î ) [( x x ' )ˆ i + y ˆ j] = y dx 'kˆ Using the Biot-Savat law, the infinitesimal contibution due to Id s is μ I d s ˆ μ I d s μ I d db = = = x 4π 4π 3 4π [( x x y) + Thus, we see that the diection of the magnetic field is in the +kˆ diection. (6) Caying out the integation to obtain B kˆ (9.11.1) y ] 3 The total magnetic field at P can then be obtained by integating ove the entie length of the wie: B = L / μ db = Iy dx kˆ = μ I (x x ) L / 4 π[( x x ) + y ] 3 4π y wie (x x) + y = μ I (x L/ ) (x+ L / ) kˆ 4π y (x L/ ) + y (x+ L / ) + y L / kˆ L / (9.11.) Let s conside the following limits: (i) x = In this case, the field point P is at ( x, y) = (, y) on the y axis. The magnetic field becomes μ B = I 4π y L / +L / μ kˆ = I L / kˆ = μ I cos θ kˆ π y ( L / ) + y (+L / ) + y ( L / ) + y π y (9.11.3) 9-5

51 in ageement with Eq. (9.1.6). (ii) Infinite length limit Conside the limit whee L x, y. This gives back the expected infinite-length esult: μ B = I L / + L / 4 π y L / L / kˆ μ = I kˆ (9.11.4) π y If we use cylindical coodinates with the wie pointing along the +z-axis then the magnetic field is given by the expession B = μ I φˆ (9.11.5) π whee φˆ is the tangential unit vecto and the field point P is a distance away fom the wie Cuent-Caying Ac Conside the cuent-caying loop fomed of adial lines and segments of cicles whose centes ae at point P as shown below. Find the magnetic field B at P. Figue Cuent-caying ac Solution: Accoding to the Biot-Savat law, the magnitude of the magnetic field due to a diffeential cuent-caying element I d s is given by Fo the oute ac, we have μ I d s ˆ db = = μ I d θ ' μ = I dθ ' (9.11.6) 4π 4π 4π 9-51

52 μ I θ μ Iθ B oute = 4π b dθ ' = 4πb (9.11.7) The diection of B is detemined by the coss poduct d oute s ˆ which points out of the page. Similaly, fo the inne ac, we have θ μ θ B inne = μ I dθ ' = I (9.11.8) 4π a 4π a Fo B, d inne s ˆ points into the page. Thus, the total magnitude of magnetic field is B μ I θ 1 1 = B +B = inne oute 4π a b (into page) (9.11.9) Rectangula Cuent Loop Detemine the magnetic field (in tems of I, a and b) at the oigin O due to the cuent loop shown in Figue Solution: Figue Rectangula cuent loop Fo a finite wie caying a cuent I, the contibution to the magnetic field at a point P is given by Eq. (9.1.5): μ I B = (cosθ 1 + cos θ ) 4π whee θ 1 and θ ae the angles which paameteize the length of the wie. 9-5

53 To obtain the magnetic field at O, we make use of the above fomula. The contibutions can be divided into thee pats: (i) Conside the left segment of the wie which extends fom ( xy=, ) ( a, + ) to ( a, + d). The angles which paameteize this segment give cosθ 1 =1 ( θ 1 = ) and cosθ = b/ b + a. Theefoe, The diection of B is out of page, o +kˆ 1. B 1 = μ I (cos θ 1 + cos θ ) = μ I b 1 (9.11.1) 4π a 4π a b + a (ii) Next, we conside the segment which extends fom ( x, y) = ( a, + b) to ( + a, + b ). Again, the (cosine of the) angles ae given by cosθ 1 = a a + b ( ) cosθ = cos θ 1 = a a + b (9.11.1) This leads to μ I a a μ Ia B = 4π b a + b + a + b = ( ) π b a + b The diection of B is into the page, o kˆ. (iii) The thid segment of the wie uns fom ( x, y) = ( + a, + b) to ( + a, + ). One may eadily show that it gives the same contibution as the fist one: The diection of B is again out of page, o +kˆ 3. The magnetic field is B 3 = B 1 ( ) B B + B + B = + = μ I b μ 1 kˆ Ia = B B kˆ π a a + b πb a + b ( ) μ I = ˆ ( b a + b b a ) π ab a k + b 9-53

54 Note that in the limit a, the hoizontal segment is absent, and the two semi-infinite wies caying cuents in the opposite diection ovelap each othe and thei contibutions completely cancel. Thus, the magnetic field vanishes in this limit Haipin-Shaped Cuent-Caying Wie An infinitely long cuent-caying wie is bent into a haipin-like shape shown in Figue Find the magnetic field at the point P which lies at the cente of the half-cicle. Figue Haipin-shaped cuent-caying wie Solution: Again we beak the wie into thee pats: two semi-infinite plus a semi-cicula segments. (i) Let P be located at the oigin in the xy plane. The fist semi-infinite segment then extends fom ( x, y) = (, ) to (, ). The two angles which paameteize this segment ae chaacteized by cosθ 1 =1(θ 1 = ) and cosθ = ( θ = π / ). Theefoe, its contibution to the magnetic field at P is The diection of B is out of page, o +kˆ 1. B = μ I (cosθ + cos θ ) = μ I (1 + ) = μ I ( ) 1 1 4π 4π 4π (ii) Fo the semi-cicula ac of adius, we make use of the Biot-Savat law: and obtain μ I d s ˆ B = ( ) 4π B = μ I π d θ μ = I 4π 4 ( ) 9-54

55 The diection of B is out of page, o +kˆ. (iii) The thid segment of the wie uns fom ( x, y) = (, +) to (, +). One may eadily show that it gives the same contibution as the fist one: The diection of B is again out of page, o +kˆ 3. The total magnitude of the magnetic field is B 3 = B 1 = μ I ( ) 4π = + B + B = B + B = μ I kˆ + μ I kˆ = μ I B B ( + π ) k ˆ (9.11.) π 4 4π Notice that the contibution fom the two semi-infinite wies is equal to that due to an infinite wie: B + B = B = μ I kˆ (9.11.1) π Two Infinitely Long Wies Conside two infinitely long wies caying cuents ae in the x-diection. Figue Two infinitely long wies (a) Plot the magnetic field patten in the yz-plane. (b) Find the distance d along the z-axis whee the magnetic field is a maximum. Solutions: (a) The magnetic field lines ae shown in Figue Notice that the diections of both cuents ae into the page. 9-55

56 Figue Magnetic field lines of two wies caying cuent in the same diection. (b) The magnetic field at (,, z) due to wie 1 on the left is, using Ampee s law: μ I μ I B 1 = = (9.11.) π π a + z Since the cuent is flowing in the x-diection, the magnetic field points in the diection of the coss poduct Thus, we have ( ˆi) ˆ ˆ ˆ ˆ1 = ( i ) (cosθ j + sin θ k ˆ ) = sinθ j cos θ kˆ (9.11.3) B = 1 μ I (sinθ ĵ cosθ kˆ ) (9.11.4) π a + z Fo wie on the ight, the magnetic field stength is the same as the left one: B 1 = B. Howeve, its diection is given by ( î) ˆ =( î ) ( cosθ ĵ + sin θ kˆ ) = sin θ ĵ+ cos θ k ˆ (9.11.5) Adding up the contibutions fom both wies, the z-components cancel (as equied by symmety), and we aive at μ I sinθ ˆ μ Iz B=B + B = j = ˆ 1 j (9.11.6) π a + z π (a + z ) 9-56

57 Figue Supeposition of magnetic fields due to two cuent souces To locate the maximum of B, we set db / dz = and find db μ = I 1 z = μ I a z = (9.11.7) dz π a + z (a + z ) π (a + z ) which gives z = a (9.11.8) Thus, at z=a, the magnetic field stength is a maximum, with a magnitude μ I B max = (9.11.9) π a Non-Unifom Cuent Density Conside an infinitely long, cylindical conducto of adius R caying a cuent I with a non-unifom cuent density J = α (9.11.3) whee α is a constant. Find the magnetic field eveywhee. Figue Non-unifom cuent density 9-57

58 Solution: The poblem can be solved by using the Ampee s law: whee the enclosed cuent I enc is given by (a) Fo < R, the enclosed cuent is Ñ B d s = μ I enc ( ) J A ' π ' d ') (9.11.3) I enc = d = (α )( πα 3 I enc = πα' d ' = ( ) 3 Applying Ampee s law, the magnetic field at P 1 is given by o B 1 (π ) = μ πα 3 ( ) 3 B 1 = αμ ( ) 3 The diection of the magnetic field B 1 is tangential to the Ampeian loop which encloses the cuent. (b) Fo > R, the enclosed cuent is which yields R I enc = ' ' = πα R 3 πα d ( ) 3 Thus, the magnetic field at a point P outside the conducto is B (π ) = μ πα R 3 ( ) 3 B αμ R 3 = ( ) 3 A plot of B as a function of is shown in Figue : 9-58

59 Figue The magnetic field as a function of distance away fom the conducto Thin Stip of Metal Conside an infinitely long, thin stip of metal of width w lying in the xy plane. The stip caies a cuent I along the +x-diection, as shown in Figue Find the magnetic field at a point P which is in the plane of the stip and at a distance s away fom it. Solution: Figue Thin stip of metal Conside a thin stip of width d paallel to the diection of the cuent and at a distance away fom P, as shown in Figue The amount of cuent caied by this diffeential element is di = I d ( ) w Using Ampee s law, we see that the stip s contibution to the magnetic field at P is given by o db(π ) = μ I = μ (di ) (9.11.4) enc 9-59

60 db = μ di = μ I d ( ) π π w Figue A thin stip with thickness d caying a steady cuent I. Integating this expession, we obtain s w μ I d μ + B = + = I ln s w (9.11.4) s π w π w s Using the ight-hand ule, the diection of the magnetic field can be shown to point in the +z-diection, o μ I w B = ln 1 + kˆ ( ) π w s Notice that in the limit of vanishing width, w = s, ln(1 + ws / ) w/ s, and the above expession becomes B = μ I kˆ ( ) π s which is the magnetic field due to an infinitely long thin staight wie Two Semi-Infinite Wies A wie caying cuent I uns down the y axis to the oigin, thence out to infinity along the positive x axis. Show that the magnetic field in the quadant with xy>, of the xy plane is given by μ I 1 1 x y B z = ( ) 4π x y y x + y x x + y Solution: 9-6

61 Let Pxy (, ) be a point in the fist quadant at a distance 1 fom a point (, y ') on the y- axis and distance fom ( x ',) on the x-axis. Figue Two semi-infinite wies Using the Biot-Savat law, the magnetic field at P is given by d d μ I s ˆ μ I s ˆ μ I d s ˆ B = db = = ( ) 4π 4π wie Let s analyze each segment sepaately. y 1 4π wie x (i) Along the y axis, conside a diffeential element ds 1 = dy ' ĵ which is located at a distance 1 = xî + ( y y ' )ˆ j fom P. This yields d s =( dy ' ˆj ) [ xi ˆ + ( y y ') ˆj ] = xdy'k ˆ ( ) 1 1 (ii) Similaly, along the x-axis, we have ds = dx 'î and = x x ' î + ĵ ( ) y which gives d s = yd x' k ˆ ( ) Thus, we see that the magnetic field at P points in the +z-diection. Using the above esults and 1 = x + ( y y ') and = x x + y, we obtain ( ) B z = μ I xdy' + μ I 3/ ydx' 4 π [x + ( y y ') ] 4π [ y + (x x ') ] 3/ ( ) The integals can be eadily evaluated using 9-61

62 bds 1 a = + [b + (a s ) ] 3/ b b a + b (9.11.5) The final expession fo the magnetic field is given by μ I 1 y 1 x B = k ˆ ( ) 4π x x x + y y y x + y We may show that the esult is consistent with Eq. (9.1.5) 9.1 Conceptual Questions 1. Compae and contast Biot-Savat law in magnetostatics with Coulomb s law in electostatics.. If a cuent is passed though a sping, does the sping stetch o compess? Explain. 3. How is the path of the integation of Ñ B d s chosen when applying Ampee s law? 4. Two concentic, coplana cicula loops of diffeent diametes cay steady cuents in the same diection. Do the loops attact o epel each othe? Explain. 5. Suppose thee infinitely long paallel wies ae aanged in such a way that when looking at the coss section, they ae at the cones of an equilateal tiangle. Can cuents be aanged (combination of flowing in o out of the page) so that all thee wies (a) attact, and (b) epel each othe? Explain Additional Poblems Application of Ampee's Law The simplest possible application of Ampee's law allows us to calculate the magnetic field in the vicinity of a single infinitely long wie. Adding moe wies with diffeing cuents will check you undestanding of Ampee's law. (a) Calculate with Ampee's law the magnetic field, B = B( ), as a function of distance fom the wie, in the vicinity of an infinitely long staight wie that caies cuent I. Show with a sketch the integation path you choose and state explicitly how you use symmety. What is the field at a distance of 1 mm fom the wie if the cuent is 1 A? 9-6

63 (b) Eight paallel wies cut the page pependiculaly at the points shown. A wie labeled with the intege k (k = 1,,..., 8) beas the cuent k times I (i.e., I k = ki ). Fo those with k = 1 to 4, the cuent flows up out of the page; fo the est, the cuent flows down into the page. Evaluate Ñ B d s along the closed path (see figue) in the diection indicated by the aowhead. (Watch you signs!) Figue Ampeian loop (c) Can you use a single application of Ampee's Law to find the field at a point in the vicinity of the 8 wies? Why? How would you poceed to find the field at an abitay point P? Magnetic Field of a Cuent Distibution fom Ampee's Law Conside the cylindical conducto with a hollow cente and coppe walls of thickness b a as shown in Figue The adii of the inne and oute walls ae a and b espectively, and the cuent I is unifomly spead ove the coss section of the coppe. (a) Calculate the magnitude of the magnetic field in the egion outside the conducto, > b. (Hint: conside the entie conducto to be a single thin wie, constuct an Ampeian loop, and apply Ampee's Law.) What is the diection of B? Figue Hollow cylinde caying a steady cuent I. 9-63

64 (b) Calculate the magnetic field inside the inne adius, < a. What is the diection of B? (c) Calculate the magnetic field within the inne conducto, a < < b. What is the diection of B? (d) Plot the behavio of the magnitude of the magnetic field B() fom = to = 4b. Is B() continuous at = a and = b? What about its slope? (e) Now suppose that a vey thin wie unning down the cente of the conducto caies the same cuent I in the opposite diection. Can you plot, oughly, the vaiation of B() without anothe detailed calculation? (Hint: emembe that the vectos db fom diffeent cuent elements can be added to obtain the total vecto magnetic field.) Cylinde with a Hole A long coppe od of adius a has an off-cente cylindical hole though its entie length, as shown in Figue The conducto caies a cuent I which is diected out of the page and is unifomly distibuted thoughout the coss section. Find the magnitude and diection of the magnetic field at the point P. Figue A cylindical conducto with a hole The Magnetic Field Though a Solenoid A solenoid has closely spaced tuns so that, fo most of its length, it may be consideed to be an ideal solenoid. It has a length of.5 m, a diamete of.1 m, and caies a cuent of.3 A. (a) Sketch the solenoid, showing clealy the otation diection of the windings, the cuent diection, and the magnetic field lines (inside and outside) with aows to show thei diection. What is the dominant diection of the magnetic field inside the solenoid? (b) Find the magnitude of the magnetic field inside the solenoid by constucting an Ampeian loop and applying Ampee's law. 9-64

65 (c) Does the magnetic field have a component in the diection of the wie in the loops making up the solenoid? If so, calculate its magnitude both inside and outside the solenoid, at adii 3 mm and 6 mm espectively, and show the diections on you sketch Rotating Disk A cicula disk of adius R with unifom chage density σ otates with an angula speed ω. Show that the magnetic field at the cente of the disk is B = 1 μ σωr Hint: Conside a cicula ing of adius and thickness d. Show that the cuent in this element is di = ( ω/ π )dq = ωσ d Fou Long Conducting Wies Fou infinitely long paallel wies caying equal cuent I ae aanged in such a way that when looking at the coss section, they ae at the cones of a squae, as shown in Figue Cuents in A and D point out of the page, and into the page at B and C. What is the magnetic field at the cente of the squae? Figue Fou paallel conducting wies Magnetic Foce on a Cuent Loop A ectangula loop of length l and width w caies a steady cuent I 1. The loop is then placed nea an finitely long wie caying a cuent I, as shown in Figue What is the magnetic foce expeienced by the loop due to the magnetic field of the wie? 9-65

66 Figue Magnetic foce on a cuent loop Magnetic Moment of an Obital Electon We want to estimate the magnetic dipole moment associated with the motion of an electon as it obits a poton. We use a semi-classical model to do this. Assume that the electon has speed v and obits a poton (assumed to be vey massive) located at the oigin. The electon is moving in a ight-handed sense with espect to the z-axis in a cicle of adius =.53 Å, as shown in Figue Note that 1 Å = 1 1 m. Figue (a) The inwad foce m v / equied to make the electon move in this cicle is e povided by the Coulomb attactive foce between the electon and poton (m e is the mass of the electon). Using this fact, and the value of we give above, find the speed of the electon in ou semi-classical model. [Ans: m/s.] (b) Given this speed, what is the obital peiod T of the electon? [Ans: s.] (c) What cuent is associated with this motion? Think of the electon as stetched out unifomly aound the cicumfeence of the cicle. In a time T, the total amount of chage q that passes an obseve at a point on the cicle is just e [Ans: 1.5 ma. Big!] (d) What is the magnetic dipole moment associated with this obital motion? Give the magnitude and diection. The magnitude of this dipole moment is one Boh magneton, μ B. [Ans: A m along the z axis.] (e) One of the easons this model is semi-classical is because classically thee is no eason fo the adius of the obit above to assume the specific value we have given. The value of is detemined fom quantum mechanical consideations, to wit that the obital 9-66

67 angula momentum of the electon can only assume integal multiples of h/π, whee h = J/s is the Planck constant. What is the obital angula momentum of the electon hee, in units of h /π? Feomagnetism and Pemanent Magnets A disk of ion has a height h = 1. mm and a adius = 1. cm. The magnetic dipole moment of an atom of ion is μ = A m. The mola mass of ion is g, and its density is 7.9 g/cm 3. Assume that all the ion atoms in the disk have thei dipole moments aligned with the axis of the disk. (a) What is the numbe density of the ion atoms? How many atoms ae in this disk? [Ans: atoms/m 3 ;.7 1 atoms.] (b) What is the magnetization M in this disk? [Ans: A/m, paallel to axis.] (c) What is the magnetic dipole moment of the disk? [Ans:.48 A m.] (d) If we wee to wap one loop of wie aound a cicle of the same adius, how much cuent would the wie have to cay to get the dipole moment in (c)? This is the equivalent suface cuent due to the atomic cuents in the inteio of the magnet. [Ans: 155 A.] Chage in a Magnetic Field A coil of adius R with its symmetic axis along the +x-diection caies a steady cuent I. A positive chage q moves with a velocity v = v ĵ when it cosses the axis at a distance x fom the cente of the coil, as shown in Figue Figue Descibe the subsequent motion of the chage. What is the instantaneous adius of cuvatue? 9-67

68 Pemanent Magnets A magnet in the shape of a cylindical od has a length of 4.8 cm and a diamete of 1.1 cm. It has a unifom magnetization M of 53 A/m, diected paallel to its axis. (a) Calculate the magnetic dipole moment of this magnet. (b) What is the axial field a distance of 1 mete fom the cente of this magnet, along its axis? [Ans: (a).4 1 A m, (b) T, o gauss.] Magnetic Field of a Solenoid (a) A 3-tun solenoid has a length of 6 cm and a diamete of 8 cm. If this solenoid caies a cuent of 5. A, find the magnitude of the magnetic field inside the solenoid by constucting an Ampeian loop and applying Ampee's Law. How does this compae to the magnetic field of the eath (.5 gauss). [Ans:.314 T, o 314 gauss, o about 6 times the magnetic field of the eath]. We make a magnetic field in the following way: We have a long cylindical shell of nonconducting mateial which caies a suface chage fixed in place (glued down) of σ C/m, as shown in Figue The cylinde is suspended in a manne such that it is fee to evolve about its axis, without fiction. Initially it is at est. We come along and spin it up until the speed of the suface of the cylinde is v. Figue (b) What is the suface cuent K on the walls of the cylinde, in A/m? [Ans: K = σ v.] (c) What is magnetic field inside the cylinde? [Ans. B = μ K = μσv, oiented along axis ight-handed with espect to spin.] (d) What is the magnetic field outside of the cylinde? Assume that the cylinde is infinitely long. [Ans: ]. 9-68