Chapter 26: Magnetism: Force and Field
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1 Chapte 6: Magnetism: Foce and Field Magnets Magnetism
2 Magnetic foces Magnetism
3 Magnetic field of Eath Magnetism
4 Magnetism Magnetic monopoles? Pehaps thee exist magnetic chages, just like electic chages. Such an entity would be called a magnetic monopole (having + o magnetic chage). How can you isolate this magnetic chage? Ty cutting a ba magnet in half: S N S N S N Even an individual electon has a magnetic dipole! Many seaches fo magnetic monopoles the existence of which would explain (within famewok of QM) the quantization of electic chage (agument of Diac) No monopoles have eve been found:
5 Magnetism Souce of magnetic field What is the souce of magnetic fields, if not magnetic chage? Answe: electic chage in motion! e.g., cuent in wie suounding cylinde (solenoid) poduces vey simila field to that of ba magnet. Theefoe, undestanding souce of field geneated by ba magnet lies in undestanding cuents at atomic level within bulk matte. Obits of electons about nuclei Intinsic spin of electons (moe impotant effect)
6 Magnetism Magnetic foce: Obsevations
7 Magnetism Magnetic foce (Loentz foce)
8 Magnetic foce (cont d) Magnetism Components of the magnetic foce
9 Magnetic foce (cont d) Magnetic foce Magnetism B x x x x x x x x x x x x v x x x x x x q F B v q F B v q F = 0
10 Magnetic foce (cont d) Units of magnetic field Magnetism
11 Magnetism Magnetic foce vs. electic foce
12 Magnetic field lines Magnetic Field Lines and Flux
13 Magnetic field lines Magnetic Field Lines and Flux S N
14 Magnetic Field Lines and Flux Magnetic field lines (cont d)
15 Magnetic Field Lines and Flux Magnetic field lines (cont d) Electic Field Lines of an Electic Dipole Magnetic Field Lines of a ba magnet S N
16 Magnetic Field Lines and Flux Magnetic field lines (cont d)
17 Magnetic Field Lines and Flux Magnetic field lines (cont d)
18 Magnetic flux dφ Φ B B = = B Magnetic Field Lines and Flux da = B cosθda = B da = B nda ˆ da B da = B cosθda = B Φ = BA B Aea A B Φ B = B nda ˆ = B da = B cosθda Φ = B nˆ da B magnetic flux though a suface nˆ θ B nˆ B B // A, A = Anˆ
19 Magnetic Field Lines and Flux Magnetic flux (cont d) Units: 1webe = 1Wb = 1Tm = 1(N/A)m = 1 Nm/A Gauss s law fo magnetism No magnetic monopole has been obseved! A=C/s, T=N/[C(m/s)] -> Tm =Nm/[C/s]=Nm/A B v da v = 0 (magnetic flux though any closed suface)
20 Motion of Chaged Paticles in a Magnetic Field Case 1: Velocity pependicula to magnetic field υ pependicula to B The paticle moves at constant speed υ in a cicle in the plane pependicula to B. F/m = a povides the acceleation to the cente, so u FL q υ B qυ B υ = = a and = a = m m m R m υ hence R = v qb F x R B
21 Motion of Chaged Paticles in a Magnetic Field Case 1: Velocity pependicula to magnetic field
22 Motion of Chaged Paticles in a Magnetic Field Case 1: Velocity pependicula to magnetic field Velocity selecto
23 Motion of Chaged Paticles in a Magnetic Field Case 1: Velocity pependicula to magnetic field Mass spectomete
24 Motion of Chaged Paticles in a Magnetic Field Case 1: Velocity pependicula to magnetic field Mass spectomete
25 Motion of Chaged Paticles in a Magnetic Field Case 1: Velocity pependicula to magnetic field Mass spectomete
26 Motion of Chaged Paticles in a Magnetic Field Case 1: Velocity pependicula to magnetic field Mass spectomete
27 Motion of Chaged Paticles in a Magnetic Field Case : Geneal case (v at any angle to B) Begin by sepaating the two components of The Loentz foce: F = qv B = q [( v + v ) B] Now the coss poduct of any two paallel vectos is zeo, so The Loentz foce: [ F = q v ] B Note that F is pependicula to B and to v, so v stays constant. // // // v into v This esults in a cicula motion with the adius R = while thee is a constant v (a helical motion). // // and v mv /( qb), with espect to B.
28 Motion of Chaged Paticles in a Magnetic Field Case : Geneal case (cont d) Since the magnetic field does not exet foce on a chage that tavels in its diection, the component of velocity in the magnetic field diection does not change.
29 Magnetic Foce on a Cuent-Caying Conducto Magnetic foce on a cuent (staight wie)
30 Magnetic Foce on a Cuent-Caying Conducto Magnetic foce on a cuent (staight wie) (cont d)
31 Magnetic Foce on a Cuent-Caying Conducto Magnetic foce on a cuent (cuved wie)
32 Magnetic Foce on a Cuent-Caying Conducto Magnetic foce on a cuent: Example1
33 Magnetic Foce on a Cuent-Caying Conducto Magnetic foce on a cuent: Example1 (cont d)
34 Magnetic Foce on a Cuent-Caying Conducto Magnetic foce on a cuent: Example1 (cont d)
35 Magnetic Foce on a Cuent-Caying Conducto Magnetic foce on a cuent: Example1 (cont d)
36 Magnetic Foce on a Cuent-Caying Conducto Magnetic foce on a cuent: Example1 (cont d)
37 Magnetic Foce on a Cuent-Caying Conducto Magnetic foce on a cuent: Example
38 Magnetic Foce on a Cuent-Caying Conducto Magnetic foce on a cuent: Example (cont d)
39 Magnetic Foce on a Cuent-Caying Conducto Magnetic foce on a cuent: Example (cont d)
40 Foce and Toque on a Cuent Loop Plane of loop is paallel to the magnetic field
41 Foce and Toque on a Cuent Loop Plane of loop : geneal case if θ=90 o
42 Foce and Toque on a Cuent Loop Plane of loop and magnetic moment
43 Foce and Toque on a Cuent Loop Plane of loop : magnetic moment (cont d) The same magnetic dipole moment fomulae wok fo any shape of plana loop. Any such loop can be filled by a ectangula mesh as in the sketch. Each sub-loop is made to cay the cuent NI. You will now see that all the inteio wies have zeo cuent and ae of no consequence. Nevetheless, each sub-loop contibutes to µ in popotion to its aea.
44 Foce and Toque on a Cuent Loop Plane of loop : magnetic moment (cont d) Analogous to the electic dipole p : τ = p E
45 Foce and Toque on a Cuent Loop Potential enegy of a magnetic dipole Wok done by the toque when the magnetic moment is otated by dφ : W dw = ( U = τdφ = µ Bsinφdφ In analogy to the case of an electic dipole in Chapte, we define a potential enegy: du = dw φ = τ dφ = φ 1 φ1 U ( φ) U = µ B sinφdφ 1 ) φ ( µ Bsinφ) dφ = µ B cosφ µ B cosφ µ B cosφ = µ B Potential enegy of a magnetic dipole at angle φ to a magnetic field 1
46 Applications Galvanomete We have seen that a magnet can exet a toque on a loop of cuent aligns the loop s dipole moment with the field. In this pictue the loop (and hence the needle) wants to otate clockwise The sping poduces a toque in the opposite diection The needle will sit at its equilibium position Cuent inceased µ = I Aea inceases Toque fom B inceases Angle of needle inceases Cuent deceased µ deceases Toque fom B deceases Angle of needle deceases
47 Moto Applications Slightly tip the loop Restoing foce fom the magnetic toque Oscillations Now tun the cuent off, just as the loop s µ is aligned with B Loop coasts aound until its µ is ~antialigned with B Tun cuent back on Magnetic toque gives anothe kick to the loop Continuous otation in steady state
48 Moto (cont d) Applications Even bette Have the cuent change diections evey half otation Toque acts the entie time Two ways to change cuent in loop: 1. Use a fixed voltage, but change the cicuit (e.g., beak connection evey half cycle DC motos. Keep the cuent fixed, oscillate the souce voltage AC motos V S I t
49 Moto (cont d) Applications flip the cuent diection
50 Hall effect Applications chages accumulate (in this case electons) Measuing Hall voltage (Hall emf) In a steady state qe H =qv d B Chages move sideways until the Hall ufield E H gows u to balance the foce due to the magnetic field: E = υ B So, E = υ B and H d As peviously, I = nqυ da = nqυ dtw nq = JB E H H VH = EHw =υdbw n can be measued d and J = I / A
51 Applications Electomagnetic ail gun
52 Execise 1 Execises If a poton moves in a cicle of adius 1 cm pependicula to a B field of 0.4 T, what is the speed of the poton and the fequency of motion? qb 1 f = πm v x x C 0.4T f = 7 π kg x x f = 10 Hz = Hz f = Hz qb v = m C 0.4T 0.1m = kg v 7 v v = m 6 m 6 m = 10 = s s s
53 Execise Execises Example of the foce on a fast moving poton due to the eath s magnetic field. (Aleady we know we can neglect gavity, but can we neglect magnetism?) Let v = 10 7 m/s moving Noth. What is the diection and magnitude of F? Take B = 0.5x10-4 T and v B to get maximum effect. F = qvb = C m 4 s T FM 17 = 8 10 N (a vey fast-moving poton) 19 FE = qe = C 100 FE = N volts mete v F N vxb is into the pape (west). Check with globe B
54 Magnetic Field of a Moving Chage Magnetic field poduced by a moving chage ˆ = µ 0 q B = 4 π υ ˆ ˆ = Note the facto of µ 0 /4π the constant of popotionality needed just as 1/(4πε 0 ) is needed in electostatics. 7 7 µ 0 = 4π 10 Tm/A = 4π 10 N/A
55 Magnetic Field of a Cuent Element Magnetic field poduced by a cuent element ds µ 0 q B = 4 π υ ˆ ˆ = db Fo an element ds of a conducto caying a cuent I thee ae n A ds chages with dift velocity υ d (using piciple of supeposition). numbe of chage q µ 0 qna ds dl υ d ˆ = 4π µ 0 I db 4 = π uu dl ds ˆ
56 Magnetic Field of a Cuent Element Biot-Savat law µ 0 I db = 4 π uu dl ˆ ds ds Note: ds is dl in the textbook. Note that ds is in the diection of I, but has a magnitude which is ds the length of wie consideed. Deduced by Biot and Savat c. 185 fom expeiments with coils
57 Magnetic Field of a Cuent Element Biot-Savat law (cont d) db P µ 0 I ds ˆ db = ; ˆ = 4π θ ds I The magnitude of the field db is: db = µ Idssinθ 0 4π Total magnetic field at P is found by summing ove all the cuent elements ds in the wie. B = db
58 Magnetic Field of a Staight Cuent Caying Conducto A staight wie of length L ˆ A thin staight wie of length L caies constant cuent I. Calculate the total B field at P. ds θ x P y db R I x db = ds ˆ always points out of the page It has magnitude µ 0 I ds ˆ 4 π dssinθ So the magnitude of db is given by: µ 0 I dssinθ µ 0 I dxsinθ db = = 4π 4π
59 Magnetic Field of a Staight Cuent Caying Conducto A staight wie of length L (cont d) ˆ ds θ x P y db R I x db z R µ 0 I ds sinθ µ 0 I dx sinθ = = 4π 4π = sin θ ; = x + R db z = µ I 0 4π Rdx ( x + R ) 3/ B z µ 0IR = 4π L / dx µ / 0IR x L µ 0I L = = / 3/ L ( x + R ) 4π R ( x + R ) 1/ L / 4πR L / 4 + R
60 Magnetic Field of a Staight Cuent Caying Conducto A staight wie of length L (cont d) θ P x ds ˆ I R y x db In the limit (L/R) + = 0 4 / 4 R L L R I B z π µ 1 4 / ) / ( / 4 / + = + R L R L R L L 0 z I B R µ π = Magnetic field by a long staight wie
61 Magnetic Field of a Staight Cuent Caying Conducto A staight wie of length L (cont d) B B I B B
62 Magnetic Field of a Staight Cuent Caying Conducto Example: A long staight wie Ion filings
63 O Example A θ A ˆ Magnetic Field of a Cuent Element ds C I C Calculate the magnetic field at point O due to the wie segment shown. The wie caies unifom cuent I, and consists of two staight segments and a cicula ac of adius R that subtends angle θ. The magnetic field due to segments A A and CC is zeo because ds is paallel to ˆ along these paths. R Along path AC, ds and ˆ ae pependicula. ds ˆ = ds µ 0 I ds ˆ µ 0I ds db = db = 4 π 4π R µ 0I µ 0I µ 0I µ 0I B= ds = Rdθ = dθ = θ 4πR 4πR 4πR 4πR Note: B field at the cente of a loop, θ=π B = µ 0I R
64 Foce Between Paallel Conductos Two paallel wies At a distance a fom the wie with cuent I 1 the magnetic field due to the wie is given by µ 0 I1 B1 = π a u u F = I L B 1 µ I µ I I π a π a F = ILB 1 = IL = L
65 Foce Between Paallel Conductos Two paallel wies (cont d) Paallel conductos caying cuent in the same diection attact each othe. Paallel conductos caying cuents in opposite diections epel each othe.
66 Foce Between Paallel Conductos Definition of ampee B 1 µ I π a u u F = I L B = F = µ II π a 0 1 L The chosen definition is that fo a = L = 1m, The ampee is made to be such that F = 10 7 N when I 1 =I =1 ampee This choice does two things (1) it makes the ampee (and also the volt) have vey convenient magnitudes fo evey day life and () it fixes the size of µ 0 = 4π Note ε 0 = 1/(µ 0 c ). All the othe units follow almost automatically.
67 Magnetic Field of a Cicula Cuent Loop Magnetic field poduced by a loop cuent Use ds B = B µ 0 Ids ˆ = 4π I ˆ ds I R µ 0 ˆ R to find B field at the cente of a loop of wie. Loop of wie lying in a plane. It has adius R and total cuent I flowing in it. Fist find d s ˆ ds ˆ is a vecto coming out of the pape at the same angle anywhee on the cicle. The angle is constant. Ids I B = µ 0 µ 0 db = ds 4π = = R 4π R Magnitude of B field at cente of loop. Diection is out of pape. R i kˆ µ 0 I 4π R v B = I R µ 0 πr kˆ
68 Magnetic Field of a Cicula Cuent Loop Example 1: Loop of wie of adius R = 5 cm and cuent I = 10 A. What is B at the cente? Magnitude and diection I µ 0I B = R B = π 10 B N A 7 4 = A (.05m) T 4 B = T = 1. Gauss Diection is out of the page.
69 Magnetic Field of a Cicula Cuent Loop Example : What is the B field at the cente of a segment o cicula ac of wie? ds I ˆ µ 0 I = ds 4π R B Total length of ac is S. θ 0 R P µ 0 I = 4π R S whee S is the ac length S =Rθ 0 θ 0 is in adians (not degees) B Why is the contibution to the B field at P equal to zeo fom the staight section of wie?
70 Ampee s Law Ampee s law : A cicula path Peviously fom the Biot-Savat s law we had On substitution fo B Conside any cicula path of adius R centeed on the wie caying cuent I. Evaluate the scala poduct B ds aound this path. Note that B and ds ae paallel at all points along the path. Also the magnitude of B is constant on this path. So the sum of all the B ds tems aound the cicle is B ds = µ 0 I B ds = B ds = B B = µ 0 I π Ampee s Law ( π )
71 Ampee s Law Ampee s law : A geneal path u uu tangential component of ds µ Idθ µ I d π π 0 0 B ds = = θ. ^ x θ θ Let us look at the integal along any shape of closed path in 3D. The most geneal ds is uu uu ds = d u + dθ τθˆ + dz k Whee unit vectos ae used fo the adial ^ ^ and the tangential diections θ and fo z ^ along the wie k. In this system we have u B = ˆ µ I 0 π z ^ θˆ τ only. ^ k kˆ µ 0 I B ds = π Fo any path which encloses the wie d θ = π Fo any path which does not enclose the wie y d θ = 0 d θ
72 Ampee s law : Ampee s Law B ds = µ 0I This law holds fo an abitay closed path that is theaded by a steady cuent. I is the total cuent that passes though a suface bounded by the closed path.
73 Ampee s Law Electic field vs. magnetic field Electic Field Geneal: Coulomb s Law High symmety: Gauss s Law Magnetic Field Geneal: Biot-Savat Law High Symmety: Ampèe s Law
74 Applications of Ampee s Law Magnetic field by a long cylindical conducto A long staight wie of adius R caies a steady cuent I that is unifomly distibuted though the coss-section of the wie. Outside R. uu B ds = B ds = B( π ) = µ 0 Itot In egion whee < R choose a cicle of adius centeed on the wie as a path of integation. Along this path, B is again constant in magnitude and is always paallel to the path. Now I tot I. Howeve, cuent is unifom ove the coss-section of the wie. Faction of the cuent I enclosed by the cicle of adius < R equals the atio of the aea of the cicle of adius and the coss section of the wie πr. I Itot = j π = π = I π R R = µ I µ I fo π = π R < 0 tot 0 B R
75 Applications of Ampee s Law Magnetic field by a long cylindical conducto µ I fo π R 0 B= < R µ 0 I B = fo R π B R
76 Applications of Ampee s Law Magnetic field by a cicula cuent Conside cicula cuent caying loop. Calculate B field at point P, a dist x fom the cente of the loop on the axis of the loop. uu µ 0 I dl ds db = 4 π uu ˆ Ids Again in this case vecto I ds is tangent to loopand pep to vecto fom cuent element to point P. db is in diection shown, pep to vectos and I ds. Magnitude db is: db µ 0 IdL ds µ 0 IdL ds = = 4π 4 π ( x + R )
77 Applications of Ampee s Law Magnetic field by a cicula cuent (cont d) db µ 0 IdL ds µ 0 IdL ds Ids = = 4π 4 π ( x + R ) Integate aound loop, all components of db pep to axis (e.g. db y ). integate to zeo. Only db x, the components paallel to axis contibute. R µ 0 I dl ds R dbx = db sinθ = db = 4 x R x R π + + x + R Field due to entie loop obtained by integating: Bx dbx =
78 Applications of Ampee s Law Magnetic field by a cicula cuent (cont d) But I, R and x ae constant B x = db Bx = dbx µ 0 IR dl ds µ 0 IR = = 3 3 4π 4 ( x + R ) π ( x + R ) dl ds Ids x B x 0 IR ( π R) 0 R I 3 3 µ µ = = 4π ( x + R ) ( x + R ) B on the axis of a cuent loop
79 Applications of Ampee s Law Magnetic field by a cicula cuent (cont d) µ 0 RI Bx = 3 ( x + R ) Limits: x 0 x >>R µ 0I µ 0 π R I µ 0 µ B = Bx = = 3 3 R 4π x 4π x µ = IπR = IA mag moment of loop Compae case of electic field on axis of electic dipole fa fom dipole E x p = vs. 3 4πε x 0 B x 0 R I µ π µ µ = = 4π x 4π x
80 Applications of Ampee s Law Magnetic field by a solenoid When the coils of the solenoid ae closely spaced, each tun can be egaded as a cicula loop, and the net magnetic field is the vecto sum of the magnetic field fo each loop. This poduces a magnetic field that is appoximately constant inside the solenoid, and nealy zeo outside the solenoid. I
81 Applications of Ampee s Law Magnetic field by a solenoid (cont d) The ideal solenoid is appoached when the coils ae vey close togethe and the length of the solenoid is much geate than its adius. Then we can appoximate the magnetic field as constant inside and zeo outside the solenoid. I
82 Applications of Ampee s Law Magnetic field by a solenoid (cont d) Use Ampèe s Law to find B inside an ideal solenoid. uu uu uu uu uu B ds = B ds + B ds + B ds + B ds uu B ds = B L = BL = µ NI N B = µ 0I = µ 0I times numbe of tuns pe unit length L 0
83 Applications of Ampee s Law Magnetic field by a tooid A tooid can be consideed as a solenoid bent into a cicle as shown. We can apply Ampèe s law along the cicula path inside the tooid. Bds =µ 0Iencl Bds = B ds= B( π ) Iencl = NI N is the numbe of loops in the tooid, and I is the cuent in each loop B = µ NI 0 π
84 Poblem 1 Execises The wie semicicles shown in Fig. have adii a and b. Calculate the net magnetic field that the cuent in the wies poduces at point P. I Since point P is located at a symmetic position with espect to the two staight sections whee the I b cuent I moves (anti)paallel to the adial diection. So thee is no contibutions fom these segments. a The contibution fom the semicicle of adius a is P a half of that fom a complete cicle of the same adius: 1 µ 0 I B a = (out of page) a Similaly the contibution fom the semicicle of adius b is: 1 µ 0 I B b = (into page) b Fom pinciple of supeposition, the net magnetic field at point P is: B = B a B b 1 µ 0I 1 1 µ I = = 0 1 a b 4a a b (out of page)
85 Poblem Execises Long, staight conductos with squae coss sections and each caying cuent I ae laid side-by-side to fom an infinite cuent sheet. The conductos lie in the xy-plane, ae paallel to the y-axis and cay cuent in the +y diection. Thee ae n conductos pe unit length measued along the x-axis. (a) What ae the magnitude and diection of the magnetic a distance a below the cuent sheets? (b) What ae the magnitude and diection of the magnetic field a distance a above the cuent sheet? y z x
86 Poblem (cont d) Execises B a) Below sheet, all the magnetic field contibutions fom diffeent wies add up to poduce a magnetic L B field that points in the positive x-diection. Components in the z-diection cancel. Using Ampee s law, whee we use the fact that the field is antisymmetic above and below the cuent sheets, and that the legs of the path pependicula povide nothing to the integal. So, at a distance a beneath the sheet the magnetic field is: µ 0nI I encl = nli, B ds = B(L) = µ 0nLI B = in the positive x - diection. b) The field has the same magnitude above the sheet, but points in the negative x-diection.
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