Exam 2 Review W11D2 1

Transcription

1 Exam 2 Review W11D2 1

2 Exam 2 Announcements Exam Two: Thursday 20 April 7:30-9:30 pm Conflict Exam Two: If you have a regularly scheduled academic activity that conflicts with the Thursday evening exam, the conflict exam is scheduled on Friday April 21 either from 8-10 am in or 9-11 am in For conflict exam requests, please fill out the google form at 1FAIpQLScs1WmoxSutcIsxavYoFuoJm9qqzRWWE2w5oI 33GhvvVDe3mg/viewform?usp=sf_link 2

3 Exam Two Room Assignments L01 in (Walker L02 in (Walker) L03 in L04 in L05 in L06 in L07 in L08 in

4 Exam 2 Topics Current, resistance, and Ohm s Law Magnetic field Magnetic force on moving charges and current carrying wires in external magnetic fields Magnetic dipole moment vector Torque and force on a magnetic dipole in an external magnetic field (Experiment 2) Biot-Savart law Ampere s Law Faraday s Law Mutual and self-induction Energy stored in magnetic fields 4

5 What You Should Study Review Friday Problem Solving (& Solutions) Review In Class Problems (& Solutions) Review Concept Questions (& Solutions) Review Problem Sets (& Solutions) Review PowerPoint Presentations Review Relevant Parts of Course Notes (& Included Examples) Review Videos Try Practice problems on Exam 2 webpage Try Practice problems on W11D2 webpage 5

6 Current, Current Density, Resistance and Ohm s Law, Magnetic Field, Magnetic Force W05D2 6

7 CQ: Resistance 50 mm I 75 mm I 100 mm 20 mm 50 mm Two resistors are made out of the same material, but have different dimensions, as shown in the figures above. The current through these two resistors is in the directions shown. If the resistor on the left has a resistance of 1.00 ohm, the resistor on the right will have a resistance of ohm ohm ohm ohm ohm ohm 40 mm 7

8 CQ Ans.: Resistance 50 mm I 75 mm I 100 mm 20 mm Answer 1. Resistance is proportional to length and inversely proportional to area. The areas are the same but the ratio of the lengths of the resistors are 1.5, so the resistor on the right has a resistance of 1.5 ohms (Ω). R = ρ r L A 50 mm 40 mm 8

9 Current: Flow Of Charge Average current I av : Charge flowing across area A in time ΔQ Δt I av = ΔQ Δt Instantaneous current: differential limit of I av Area A I I = dq dt Units of Current: Coulomb/second = Ampere 9

10 How Big is an Ampere? Household Electronics Battery Powered Household Service Lightning Bolt To hurt you To throw you To kill you Fuse/Circuit Breaker ~1 A ~100 ma (1-10 A-Hr) 100 A 10 to 100 ka 40 (5) ma DC(AC) 60 (15) ma DC(AC) 0.5 (0.1) A DC(AC) A 10

11 Direction of the Current and Charge Flow of negative charge per sec through area A: opposite the direction of the current: A I Flow of positive charge per sec through area A: in the direction of the current A I 11

12 Why Does Current Flow? If an electric field is set up in a conductor, charge will move (making a current in direction of E) E A I E A I Note that when current is flowing, the conductor is not an equipotential surface (and E inside 0)! 12

13 Flow Tube for Charge How much charge will pass through a cross section of a wire of area A in a time interval Δt? x = v d t E I Assume each charge q is moving with drift speed v d. Consider a cylinder of length Δx = v d Δt and cross sectional area A. In a time interval Δt, all the moving charge inside that cylinder, ΔQ=nqAΔx will pass through the cross sectional area, where n is the number of moving charges per volume. Therefore the current is I = ΔQ / Δt = nqaδx / Δt = nqv d A A 13

14 Microscopic Picture: Drift Velocity + + v drift E Drift velocity is the average velocity forced by applied electric field in the presence of collisions. Magnitude is typically 4x10-5 m/sec, or 0.04 mm/ second! + + To go one meter at this speed takes about 10 hours! 14

15 Current Density Let n = number of charged objects per unit volume, q = charge of object, v q = drift velocity of object, J nqv q The current density, (current per unit area) is! J nq v! q + + v + d + + I A Generalization for many charged moving objects J i n i q i vqi 15

16 Current and Current Density! J nq v! q Current is the flow (flux) of current density through an open surface I = J ˆndA = S Special case: J S J d A v q + v q + + J + + ˆn v q + + da uniform and perpendicular to surface + S I = JA 16

17 CQ: Current Density A current I = 200 ma flows in the conductor shown below. What is the magnitude of the current density J? 1. J = 40 ma/cm 2. J = 20 ma/cm 3. J = 10 ma/cm 4. J = 1 ma/cm 2 5. J = 2 ma/cm 2 6. J = 4 ma/cm 2 5 cm 20 cm I 10 cm 17

18 CQ Ans.: Current Density Answer: 6. J = 4 ma/cm 2 I 5 cm J 20 cm 10 cm The area that matters is the cross-sectional area that the charge flows through the 50 cm 2 area. So: J = I/A = 200 ma/50 cm 2 = 4 ma/cm 2 18

19 Ohm s Law: Resistance Instead of thinking of electric field, think of potential difference across the conductor Area A I V a ΔV = V b V a = IR V b E L V = V b V a For some materials, which are called ohmic, the electric potential difference is linearly proportional to the current, the constant of proprotionality is called resistance and has SI units Ohms [Ohm] = [V A 1 ] 19

20 How Big is an Ohm? Short Copper Wire Notebook paper (thru) Typical resistors You (when dry) You (when wet) Internally (hand to foot) milliohms (m Ω ) ~1 G Ω Ω to 100 M Ω 100 k Ω 1 k Ω 500 Ω Stick your wet fingers in an electrical socket: I = V / R 120V / 1kΩ 0.1A You re dead! 20

21 Demonstrations: Temperature Effects on Resistance F4 Conducting Glass F1 Conductivity of Ionizing Water F5 21

22 Drude Model: Microscopic Model for Ohm s Law Electrons scatter on average once every τ seconds. After every collision, direction of electron is random (hard sphere model) Between collisions, electric field E gives each electron a drift momentum m e v after = m e vbefore + ( e) Eτ ( v before ) ave = 0 If we average over all the electrons, then. The initial velocities before the collision are random and add to zero so the average velocity after the collision is v drift ( v after ) ave = ( eτ ) E m e 22

23 Microscopic Ohm s Law: Conductivity Current density is proportional to the electric field. Ability of current to flow depends on density of charges & rate of scattering. V b Area A E J L V = V b V a V a! J = ne v! drift = ( ne2 τ ) E! m e!! J = σ c E σ c = ne2 τ m e σ c = conductivity 23

24 Microscopic Ohm s Law: Conductivity E = ρ J r J = σ E c and Resistivity Area A J V a V b E L ρ r 1 σ c V = V b V a The inverse of the conductivity is defined to be the resistivity ρ r. The conductivity only depends on the microscopic properties of the conductor, and not on its geometric shape. 24

25 Ohm s Law: Calculating Resistance What is relationship between electric potential difference and current? Area A I V a ΔV = V b V a = b a! E d! s = EL V b E L V = V b V a Calculating resistance from microscope Ohm s Law: E = ρ r J ΔV l = ρ r I A ΔV = I ρ r L A = I R R = ρ L r A 25

26 CQ: Resistance in a Conductor When a current flows in a wire of length L and cross sectional area A, the resistance of the wire is 1. proportional to A; inversely proportional to L. Area A I V a 2. proportional to both A and L. 3. proportional to L; inversely proportional to A. V b E L 4. inversely proportional to both L and A. 26

27 CQ Ans.: Resistance in a Conductor Answer 3. Proportional to length L, inversely proportional to area A. The longer the wire the greater the resistance R. The larger the cross-sectional area of the wire, the more ways that current can flow through it, so the lower the resistance. Area A I V a If resistivity is ρ r, then resistance is given by V b E L R = ρ r L A 27

28 Group Problem: Calculating Resistance Consider a hollow cylinder of length L and inner radius a and outer radius b. The material has resistivity. ρ r Suppose a potential difference is applied between the ends of the cylinder and produces a current flowing parallel to the axis. What is the resistance measured? 28

29 Group Problem Solution: Calculating Resistance Consider a hollow cylinder of length L and inner radius a and outer radius b. The material has resistivity. ρ r When a potential difference is applied between the ends of the cylinder, current flows parallel to the axis. In this case, the crosssectional area is A = π (b 2 a 2 ) and the resistance is given by R = ρ r L A = ρ r L π (b 2 a 2 ) 29

30 Magnetic Fields 30

31 Magnetic Field of the Earth North magnetic pole located in southern hemisphere 31

32 Demonstration: Magnetic Field Lines of a Bar Magnet G2 32

33 Magnetic Force on Moving Charges Force on positive charge! F q B = q! v q! B v q v q q + F q B B F q B q B Force on negative charge Magnetic force is perpendicular to both velocity of the charge and magnetic field 33

34 Force Law: Magnetic Field Units! F B = q!! v B q q SI Units: tesla [T] [T] = [N] [C][m / s] = [N] [A][m] cgs unit: gauss [G] 1 tesla = 10 4 gauss 34

35 How Big is a Tesla? Earth s Field Brain (at scalp) Refrigerator Magnet Inside MRI Good NMR Magnet Biggest in Lab Biggest in Pulsars LHC magnets (27 km long) 5 x 10-5 T = 0.5 G ~1 ft 1 mt 3 T 18 T 150 T (pulsed) 10 8 T 8.4 T 35

36 Vector Cross Product Magnitude: equal to the area of the parallelogram defined by the two vectors A B = A B sinθ = A ( B sinθ ) = ( A sinθ ) B (0 θ π ) B A sin B B sin A A C = A B Direction: determined by the Right-Hand-Rule B A

37 Vector Product of Unit Vectors Unit vectors in Cartesian coordinates ˆk = î ĵ î ĵ = î ĵ sin(π / 2) = 1 ĵ î î = î î sin(0) = 0 î î ĵ = ˆk î î = 0 ĵ ˆk = î ĵ ĵ = 0 ˆk î = ĵ ˆk ˆk = 0

38 CQ: Cross Product and Magnetic Force An electron is traveling up in a magnetic field that points to the right. What is the direction of the force on the electron? 1. Up. 2. Down. 3. Left. 4. Right. 5. Into plane of figure. 6. Out of plane of figure. v q q = e B 38

39 CQ Ans.: Cross Product and Magnetic Force Answer: 6. Points into the plane of the figure but the charge of the electron is negative therefore! F e!! v B q B = e! v q! B F B e v q q = e B Points out of the plane of the figure. 39

40 Demonstration: Magnetic Deflection of TV Image G6 40

41 CQ: Force Direction Is this picture (deflection direction) correct? 1. Yes 2. No 3. I don t know 41

42 Answer: 1. Yes CQ Ans.: Force Direction Field from N to S, beam velocity right to left, cross product is up. But charges are negative so force is down, as pictured. 42

43 Motion of a Moving Charged Particle in a Uniform Magnetic Field Magnetic field directed into the plane of the figure. Charged particle will undergo circular motion. B v q + + q 43

44 Group Problem: Cyclotron Motion B v q + + q A positively charged particle with charge +q and mass m is moving with speed v in a uniform magnetic field of magnitude B directed into the plane of figure. Find (1) the radius R of the orbit, (2) the period T of the motion, (3) the angular frequency ω. (4) Sketch the motion of the particle. 44

45 Magnetic Fields, Magnetic Forces, and Sources of Magnetic Fields W06D1 45

46 Maxwell s Equations: Statics Maxwell s Equations for Static Fields (SI units) " S S! E d! A! B d A! = q enc ε 0 " = 0!! # E d s = 0! C! B d! s # = µ 0 I enc C Lorentz force law on moving charges in electric and magnetic field F = q( E + v B)

47 Lorentz Force Law Outline Magnetic Force on Current Carrying Wire Sources of Magnetic Fields Biot-Savart Law 47

48 Lorentz Force Law Force on charged particles in electric and magnetic fields F = elec q E F = mag q v B Electric Force F = ( ) q E + v B Magnetic Force 48

49 CQ: Velocity Selector + C q = v e e A B E + Electrons with charge e and mass m are emitted from the cathode C and accelerated toward slit A with different speeds in the direction shown. The electrons enter a region with a downward pointing electric field (magnitude E) and a magnetic field (magnitude E) that points into the plane of the figure. The electrons that travel on a straight trajectory through the plates have speed 1. v = B/E 2. v = (1/2)(eE/m)t 2 3. v = 1/eB 4. v = E/B 49

50 CQ Ans.: Velocity Selector + C q = v e e A B E + Answer 4. Electrons that travel in a straight line have zero force exerted by the fields and therefore! F e = e(! E +! v e! B) = 0 v = E / B 50

51 Magnetic Force on Current-Carrying Wire Current source infinitesimal element dq! v dq = dq d! s dt = dq dt d! s = I d! s d! s Direction of is the direction of I. Force on source element in external magnetic field df mag dq v dq B ext d! F mag = dq! v dq! B ext = Id! s! B ext Force on a current carrying wire in an external magnetic field! F mag = wire I d! s! B ext df mag I d s B ext 51

52 Magnetic Force on Current-Carrying Wire If the wire is in a uniform magnetic field then! F mag = wire I d! s B! ext where the direction of the vector is the direction of the current d! s F mag I If the wire is also straight then! F mag = I(! L! B ext ) where the direction of the vector! is the direction of the current L L B ext 52

53 Magnetic Force on Current-Carrying Wire Current is moving charges, and we know that moving charges feel a force in a magnetic field with direction given by F mag = I( L B)! L where the direction of the vector is the direction of the current 53

54 Demonstration: Wire in a Magnetic Field (Jumping Wire) G8 d F mag = Id s B 54

55 Demonstration: Series or Parallel Current Carrying Wires G9 You tube video of experiment showing parallel wires attracting or repelling: 55

56 Group Problem: Current Loop l + current source B g A conducting rod of uniform mass per length λ and length l is suspended by two flexible wires in a uniform magnetic field of magnitude B which points out of the page and a uniform gravitational field of magnitude g pointing down. If the tension in the wires is zero, what is the direction and magnitude of the current? 56

57 Sources of Magnetic Fields 57

58 What Creates Magnetic Fields: Moving Charges 58

59 Electric Field of Point Charge An electric charge produces an electric field: E = 1 4πε o q ˆr r 2 ˆr : unit vector directed from the charged object to the field point P 59

60 v Magnetic Field of Moving Charge Moving charge with velocity field at the point P: r q > 0 P. ˆr B v ˆr : produces magnetic B = µ 0 4π q v ˆr r 2 unit vector directed from charged object to P µ 0 = 4π 10 7 T m A 1 permeability of free space 60

61 Current is a Source of Magnetic Field 61

62 Demonstration: Field Generated by Wire G

63 Continuous Moving Charge Distributions: Currents & Biot-Savart 63

64 From Charges to Currents? d! B dq! v dq ˆr = [coulomb] [meter] [sec] db dq P ˆr I d s = [coulomb] [sec] [meter] I v dq dq! v dq = dq d! s dt = dq dt d! s = I d! s 64

65 I Biot-Savart Law Current element d s of length ds pointing in direction of current I produces a magnetic field at point P: db dq v dq P ˆr I d s d B = µ 0 4π I d s ˆr r

66 The Right-Hand Rule #2 ˆ ˆr B P ˆk I current directed out of plane of figure dir d! s = ˆk dir! B(P) = ˆk ˆr = ˆθ 66

67 CQ: Biot-Savart The magnetic field at P points towards the 1. +x direction 2. +y direction 3. +z direction 4. -x direction 5. -y direction 6. -z direction 7. Field is zero ˆk ĵ I + z î + y P I +x 67

68 CQ Ans.r: Biot-Savart Answer: 6. dir B(P) direction (into page) is in the negative z The vertical line segment contributes nothing to the field at P (it is parallel to the displacement). The horizontal segment makes a field pointing into the page by the right-hand rule # 2 (right thumb in direction of current, fingers curl into page at P. ˆk ĵ I + z î + y P I +x 68

69 Repeat Demonstration: Parallel & Anti-Parallel Currents G9 69

70 CQ: Parallel Wires Consider two parallel current carrying wires. With the currents running in the opposite direction, the wires are 1. attracted (opposites attract?) 2. repelled (opposites repel?) 3. pushed another direction 4. not pushed no net force 5. I don t know I 1 I 2 wire 1 wire 2 70

71 CQ Ans.: Parallel Wires Answer 2. The wires are repelled I 1 creates a magnetic field into the page at wire 2. That makes a force on wire 2 to the right. I 2 creates a magnetic field into the page at wire 1. That makes a force on wire 1 to the left. B 1 F 12 I 1 I 2 wire 1 wire 2! F 12 = I 2! L2! B 1 B F ( ) I 1 I 2 wire 1 wire 2! F 21 = I 1! L1! B 2 ( ) 71

72 Can we understand why? Whether they attract or repel can be seen in the shape of the created B field You tube videos of field line animations showing why showing parallel wires attract: or repel: 72

73 CQ: Current Carrying Coils The above coils have 1. parallel currents that attract 2. parallel currents that repel 3. opposite currents that attract 4. opposite currents that repel 73

74 CQ Ans.: Current Carrying Coils Answer: 4. Opposite currents that repel Look at the field lines at the edge between the coils. They are jammed in, want to push out. Also, must be in opposite directions 74

75 Magnetic Field Generated by a Current Loop 75

76 Worked Example: Ring of Radius R Consider a ring with radius R and current I. Find the direction and magnitude of magnetic field B at the center (P) 1) Think about it: I Legs contribute nothing. P I parallel to r R Ring makes field into page I 2) Choose a small current element Ids 3) Pick your coordinates and integration variables 4) Apply Biot-Savart Law 76

77 Worked Example: Ring of Radius R In the circular part of the coil d! s ˆr d! s ˆr = ds d s d. P ˆr I Biot-Savart: R I db = µ 0 I 4π d! s ˆr r 2 = µ 0 I 4π ds r 2 = µ 0 I 4π R dθ R 2 = µ 0 I 4π dθ R 77

78 Worked Example: Ring of Radius R Consider a ring with radius R and current I d s d. P ˆr R I I db = µ 0 I 4π B = db = 2π 0 dθ R µ 0 I 4π dθ R B = µ 0 I 2R = µ 0 I 4π R 2π 0 direction into page dθ = µ 0 I 4π R 2π ( ) 78

79 Group Ring of Radius R Consider a ring with radius R and carrying a current I What is a vector expression for the magnetic field at point P? z ˆk ĵ R I y î I x. P 79

80 Sources of Magnetic Fields: Ampere s Law W07D1 80

81 Lorentz Force Law: Overview! F q = q(! E ext +! v q! B ext ) Newton s Second Law: q(! E ext +! v q! B ext ) = m! a q For sources with enough symmetry: Gauss s Law: Ampère s Law: closed surface! B d!! " s = µ 0 J ˆnda closed path C! E ˆn out da " = 1 ρ dv ε 0 open surface S (C ) volume enclosed 81

82 Review: Biot-Savart Law Current element of length d s carrying current I produces a magnetic field at the point P: d B(P) = µ 0 4π I d s ˆr SP r SP 2 B( r) = µ 0 4π wire Id s ( r r ) r r 3 82

83 Biot-Savart Law: Finite Wire + y P (x, y) r O r L Id s ĵ r r ˆk +x î! B(x, y)=! B(x, y) = x =+L 2 x = L 2 x =+L 2 x = L 2! B(x, y) = µ 0 I 4π y µ 0 Id x î ((x x )î + yĵ) 4π ((x x ) + y 2 ) 3/2 µ 0 4π Iyd x ˆk ((x x ) 2 + y 2 ) 3 2 = µ 0 I 4π y (x x ) (x L / 2) (x L / 2) 2 + y 2 (x + L / 2) (x + L / 2) 2 + y 2 L/2 (x x ) 2 + y 2 L/2 ˆk ˆk 83

84 Biot-Savart Law: Infinite Wire In the limit as the length L goes to infinity, the magnetic field! B( y) = lim L/2 µ 0 I 4π y! B( y) = µ 0 I 2π y ˆk (x L / 2) (x L / 2) 2 + y 2 (x + L / 2) (x + L / 2) 2 + y 2 ˆk More generally: B = µ 0 I 2πr ˆθ 84

85 The integral expression CQ: Line Integral 1. is equal to the magnetic work done around a closed path. 2. is an infinite sum of the product of the tangent component of the magnetic field along a small element of the closed path with a small element of the path up to a choice of plus or minus sign. 3. is always zero. closed path B d s 4. is equal to the magnetic potential energy between two points. 5. None of the above. 85

86 CQ Ans.: Line Integral 2. A line integral by definition is the sum closed path B d s = lim N i=n i=1 B i Δ s i We need to make a choice of integration direction (circulation) for the line integral. The small line element Δ s i is tangent to the line and points in the direction of circulation. The dot product therefore is the product of the tangent component of the magnetic field in the direction of the line element. So the answer depends on which way we circulate around the path. 86

87 Group Problem: Line Integral of Magnetic Field B I r ˆk ˆ ˆr Calculate line integral of magnetic field about circular path of radius r, circular path B d s for very long current carrying wire using the magnetic field B = µ 0 I 2πr ˆθ 87

88 J Current Enclosed Current density I enc = open surfacce S J ˆndA Current enclosed is the flux of the current density through an open surface S bounded by the closed path. Because the unit normal to an open surface is not uniquely defined this expression is unique up to a plus or minus sign. 88

89 3 rd Maxwell Equation: Ampere s Law B d s = µ J ˆnda 0 closed path open surface 89

90 Ampere s Law: The Idea In order to have a non-zero line integral of magnetic field around a closed path, there must be current punching through any area with path as boundary 90

91 Derivation: Line Integral When Enclosed Current is Zero (a) (b) (c) D.. C r 2 Ȧ I For path shown in (a) " closed path =! B d! s r 1. θ 0 µ 0 I r 2πr 2 dθ + µ I 0 r 2 2πr 1 dθ 1 0 = 0 θ B Same argument holds For (b) more complicated path consisting of radial and tangential legs, and (c) continuous path I 91

92 Derivation: Line Integral for Arbitrary Path Enclosing Current Circulate clockwise Circulate counterclockwise I = I + I Zero enclosed current B d s = 0 closed path Circular path B d s circular path = µ 0 I Arbitrary path B d s arbitrary path = µ 0 I Superposition of two cases must add to zero 92

93 3 rd Maxwell Equation: Ampere s Law B d s = µ J ˆnda 0 closed path open surface Open surface is bounded by closed path. 93

94 Sign Conventions: Right Hand Rule for Sign of Current B d s = µ 0 J ˆnda closed path open surface Integration direction clockwise for line integral requires that unit normal points into page for open surface integral Current positive into page, negative out of page 94

95 Sign Conventions: Right Hand Rule for Sign of Current B d s = µ J ˆnda 0 closed path open surface Integration direction counterclockwise for line integral requires that unit normal points out of page for open surface integral Current positive out of page, negative into page 95

96 CQ: Ampere s Law For the loop shown in the figure to the right, choose positive unit normal directed into the plane of the figure. Each wire carries a current of magnitude I. Integrating B around the loop yields ˆn 1. a positive number 2. a negative number 3. zero 96

97 CQ Ans.: Ampere s Law ˆn Answer 1. I enc = +2I B d s = 2µ 0 I > 0 closed path Net enclosed current is into the plane of the figure, so line integral is positive 97

98 Symmetric Sources for Ampere s Law: B w J Long Circular Symmetry Current Sheet (Infinite) 98

99 Symmetric Sources for Ampere s Law: B B = 0 outside l r B = 0 B = 0 B I I Solenoid (infinite) Torus 99

100 Biot-Savart vs. Ampere Biot- Savart Law B = µ 0 I 4π d s ˆr r 2 general current source ex: finite wire wire loop Ampere s law! B! = ds µ 0I enc symmetric current source ex: infinite wire infinite current sheet 100

101 Choosing Ampèrian Closed Path Ampère s Law holds for all closed paths. However it is useful (to calculate magnetic field) for some sources with enough symmetry in which there exists a closed path such that 1. The magnetic field on some leg of the closed path is both parallel to path and has constant magnitude on that leg. Then leg of path B d s = Bl 2. The magnetic field on some leg of the closed path is perpendicular to path or is zero. Then B d s = 0 leg of path 101

102 Applying Ampère s Law: Preliminary Steps Step 1: Identify the symmetry properties of the current distribution. Step 2: Determine the direction of the magnetic field (up to a choice of sign) Step 3: Decide how many different regions of space the current distribution determines 102

103 Applying Ampère s Law: Step 4: For each region of space determined by the current distribution closed path B d s µ 0 open surface S J da Step 5: Equate the two sides of Ampère s Law in order to determine an expression for the magnetic field 103

104 Applying Ampère s Law: Determining the Direction of the Magnetic Field Step 6: Based on the sign of the magnetic field and the choice or orientation of the path, determine direction for magnetic field For a given orientation of the path, if sign of B is positive, magnetic field is in the same direction as the path orientation. If sign of B is negative, magnetic field points opposite the direction of the path orientation. Clockwise orientation Counterclockwise orientation 104

105 CQ: Bent Wire The magnetic field at P is equal to the field of: 1. a semicircle 2. a semicircle plus the field of a long straight wire 3. a semicircle minus the field of a long straight wire 4. none of the above 105

106 CQ Ans.: Bent Wire Answer: 2. Semicircle + infinite wire All of the wire makes dir B(P) into the page. The two straight parts, if put together, would make an infinite wire. The semicircle is added to this to get the magnetic field 106

107 Worked Example: Very Long Wire (r R) Region 1: Outside wire (r R) Cylindrical symmetry à B field circular à Ampèrian circle with counterclockwise orientation. Apply Ampère s Law: B = µ 0 I 2πr B = µ 0 I 2πr closed path ˆθ ; r R B d s µ 0 open surface S B2πr µ 0 I J da 107

108 Worked Example: Very Long Wire r R Apply Ampère s Law: Region 2: inside wire (r R) Cylindrical symmetry à B field circular à Ampèrian circle with counterclockwise orientation. closed path B d s µ 0 open surface S J da B2πr µ 0 I(πr 2 / π R 2 ) B = µ 0 I r 2π R 2! B = µ 0 I r 2π R 2 ˆθ ; r R 108

109 Infinite Wire: Plot of B vs. r! B(r) = µ 0 Ir ˆθ ; r R 2π R 2 µ 0 I ˆθ ; r R 2π r 109

110 Group Problem: Non-Uniform Cylindrical Wire A cylindrical conductor has radius R and a nonuniform current density with total current: J = J 0 R r ˆn Find B everywhere 110

111 Two Loops 111

112 Two Loops Moved Closer Together 112

113 Multiple Wire Loops 113

114 Magnetic Field of Solenoid Horiz. comp. cancel loosely wound tightly wound For ideal solenoid, B is uniform inside & zero outside 114

115 Magnetic Field of Ideal Solenoid B 1 B = 0 outside 2 l 3 Using Ampere s law: Think! B d s = B d s along sides 2 and 4 B = 0 along side 3 B d s + B d s + B d s + B d s 1 2 = Bl I 4 I enc = nli " n: # of turns per unit length! B d! s = Bl = µ 0 nli n = N / L : # turns/unit length B = µ 0 nli l = µ 0 ni 115

116 Demonstration: Long Solenoid 116

117 CQ: Co-axial Solenoids Two co-axial very long solenoids are centered on the z-axis. Both solenoids have n turns per unit length. A current runs through the inner turns in a clockwise direction when viewed from above. A current of the same magnitude runs through the outer turns in a counter-clockwise direction when viewed from above. The magnetic field in the region between the coils 1. is zero. 2. points in the positive z direction. 3. points in the negative z direction. 4. Not enough information is given to determine direction of magnetic field. 117

118 CQ Ans.: Co-axial Solenoids Answer 2. Outside the two nested solenoids the magnetic field is zero, (because that region is outside each of the two ideal solenoids.) Choose an Amperian loop with clockwise orientation as shown in figure. The enclosed current (only from the outer solenoid) is directed into the plane of the page hence positive. Therefore the magnetic field for the region R 1 < r < R 2 points in the same direction as the orientation of the Amperian loop, which is in the positive z-direction. 118

119 Ampere s Law: Infinite Current Sheet B w J J J Amperian Loops: B is Constant & Parallel OR Perpendicular OR Zero 119

120 Group Problem: Current Sheet A sheet of current (infinite in the y & z directions, of thickness d in the x direction) carries a uniform current density: J = J ˆk Find the direction and magnitude of B as a function of x. 120

121 Surface Current Density A very thin sheet of current of width w carrying a current I in the positive z-direction has a surface current density K = K ˆk K = I / w For sheet of thickness d, width w, and current I I = Jdw = Kw J = K / d 121

122 Solenoid is Two Current Sheets Consider two sheets each of thickness d with current density J. Then surface current per unit length K = Jd = ni Use either Ampere s Law or superposition principle B = µ 0 K = µ 0 Jd = µ 0 ni 122

123 Experiment 2: Magnetic Fields, Force and Torque on a Magnetic Dipole W07D2 123

124 Experiment 2: Magnetic Dipole in Helmholtz Coil The experimental apparatus consists of two coils connected to a power supply along with a tube in which a small dipole magnet hangs from a spring that can be moved by a rod. The coils can be connected to a power supply that produces a current in the coils that creates a magnetic field.! 124

125 Experiment 2 Exploration In this experiment we would like you to investigate: the magnetic field associated with a variety of coils connections; the torque on the magnetic dipole and direction of the force on the dipole when the dipole is placed along different points along the central axis for a two coil connections, the Helmholtz and Anti- Helmholtz configurations. 125

126 Experiment 2 Start-up Instructions Open Week 7 sequential and select Experiment 2. All group members enter their Kerberos ID on the login page. Move to Section 1 and start the experiment. 126

127 Animation: Magnetic Field Generated by a Current Loop 127

128 Current Loops are Magnetic Dipoles Magnetic Dipole Moment: Magnitude: product of current an area of loop Direction: Perpendicular to the plane of the loop, using right hand rule with respect to direction of current µ IA ˆn IA 128

129 Magnetic Field of Bar Magnet (1) A magnet has two poles, North (N) and South (S) (2) Magnetic field lines leave from N, end at S 129

130 Demonstration: Magnetic Field Lines from Bar Magnet G

131 Bar Magnets Are Dipoles! Bar magnets are: 1) source for dipole field 2) rotate to orient with external magnetic field Is there a magnetic monopole charge like electric charge? NO! Magnetic monopoles do not exist in isolation 131

132 Electric Dipole Magnetic Monopoles? Magnetic Dipole When cut: When cut: 2 monopoles (charges) 2 magnetic dipoles Magnetic monopoles do not exist in isolation: Our Second Maxwell s Equation! (2 of 4) S Q p + E da + Q = q in ε 0 S B da = 0 Gauss s Law Magnetic Gauss s Law 132 µ µ µ µ

133 Conservation of Magnetic Flux: S E da = q in ε 0 S B da = 0 133

134 CQ: Magnetic Field Lines The picture shows the field lines outside a permanent magnet The field lines inside the magnet point: 1. Up 2. Down 3. Left to right 4. Right to left 5. The field inside is zero 6. I don t know 134

135 CQ Ans.: Magnetic Field Lines Answer 1. They point up inside the magnet Magnetic field lines are continuous. E field lines begin and end on charges. There are no magnetic charges (monopoles) so B field lines never begin or end 135

136 Review: Magnetic Force on Current-Carrying Wire If the wire is in a uniform magnetic field then! F mag = wire I d! s B! ext where the direction of the vector is the direction of the current d! s F mag I If the wire is also straight then! F mag = I(! L! B ext ) where the direction of the vector! is the direction of the current L L B ext 136

137 Torque on a Current Loop in a Uniform Magnetic Field 137

138 Worked Example: Current Loop Place rectangular current loop in uniform B field. Net force is zero. B leg 2 I leg 1 F 2 F 4 ĵ î leg 3 leg 4 ˆk b a P leg 2 r P,4 b d l = dy ĵ dy df y 4 I leg 4 B The torque on the loop is d F! 4 = Id! l B! = Id y ĵ Bî = Id y B ˆk d! τ 4 = r! P,4 d F! 4 = (bî + y ĵ) ( Id y B ˆk) = bid y ĵ IB y d y î y =a/2! τ 4 = (bi d y y = a/2 ĵ IB y d y î) = IabBĵ 138

139 Torque on Current Loop ĵ B î I µ = Iab ˆk a Magnetic moment points out of the page! µ = I ab ˆk! τ =! µ! B! τ = µ B ĵ = I abb ĵ ˆk b! τ = µ!! B Torque tries to align the magnetic moment vector in the direction of the magnetic field 139

140 Demonstrations Deflection of a Compass Needle by a Magnet G1 page=demo.php&letnum=g%201&show=0 Galvanometer principle G10 page=demo.php&letnum=g%2010&show=0 140

141 Demonstration: Galvanometer principle G

142 CQ: Dipole in Uniform Field Starting from rest, the current ring in a uniform magnetic field will: 1. rotate clockwise, not move 2. rotate counterclockwise, not move 3. move to the right, not rotate 4. move to the left, not rotate 5. move in another direction, without rotating 6. both move and rotate 7. neither rotate nor move I µ B 142

143 CQ Ans.: Dipole in Uniform Field Answer: 1. No net force so no center of mass motion. Coil will rotate clockwise (not move) because magnetic dipole rotates to align with external field. I µ B 143

144 Magnetic Force on Current Loop In Uniform Magnetic Field If a current loop is placed in a uniform magnetic field then! F mag = " closed loop I d! s B! ext = 0! Because the vector integral of the line element around a closed path is zero I µ B " d! s = 0! closed loop 144

145 Magnetic Force on a Dipole in a Non-Uniform Magnetic Field 145

146 Force on Dipole in Non-uniform Magnetic Field Dipoles can feel magnetic force in a nonuniform magnetic field. Magnetism Bar Magnet: Like poles repel, opposite poles attract 146

147 Force on Dipole in Non-uniform Magnetic Field The forces shown d F! mag = I(d! s B)! produce a net downward force on dipole into the region of greater field strength. 147

148 Force on Magnetic Dipole µ N S N S What makes the field pictured? Bar magnet below dipole, with N pole on top. It is aligned with the dipole pictured, they attract! 148

149 Force on Magnetic Dipole B µ = I A ˆk z µ N S I Special case: Along z-axis, for current loops, the magnetic field points along z-axis and the force on a dipole situated on the z-axis is N S B F z = µ z z z Dipole is attracted to region of greater field strength 149

150 CQ: Dipole in Magnetic Field The current carrying coil above will feel a net force 1. upwards 2. downwards 3. of zero 4. I don t know 150

151 CQ Ans.: Force on Magnetic Dipole µ N S N S Answer 2. Dipole feels downward force. What makes the field pictured? Bar magnet below dipole, with N pole on top. It is aligned with the dipole pictured, they attract! 151

152 Work Done by Interaction to Rotate Magnetic Dipole! τ =! µ! B = ( µ cos θ ĵ+ µ sin θ î) Bî = µ cos θ B ˆk θ θ W [0,θ ] = τ z d θ = µ B cos θ d θ = µ B sin θ 0 = µ B sin θ 0 W [0,θ ] =! µ! B θ 0 152

153 Pot. Energy: Dipole in Magnetic Field ΔU = U (θ ) U (θ = 0) = W [0,θ ] = µ B Set zero reference point U (θ = 0) = 0 U (θ ) = µ B Lowest energy state (aligned): U (θ = π / 2) = µ B Highest energy state (anti-aligned): U (θ = π / 2) = µ B 153

154 Force on Magnetic Dipole µ N S U Dipole = - µ B ; U (θ = 0) = 0 F = ( µ B) F = ( µ B) Special case: Magnetic field points along z-axis N S F z = µ z B z z Dipole is attracted to region of greater field strength 154

155 Experiment 2: Concept Questions 155

156 CQ.: Magnetic Dipole in Helmholtz Coil A dipole is initially pointing along the positive x-direction and located above the two coils (z >0). A short time later, the dipole will feel: µ 1. a force but not a torque. 2. a torque but not a force. 3. both a torque and a force. 4. neither force nor torque. î! 156

157 CQ Ans.: Dipole in Helmholtz Answer: 3. A torque and a force. The dipole will rotate to align with the magnetic field and it will then feel a force that will pull the dipole into a region of greater magnetic field strength (downwards). = µ B F B î µ 157

158 CQ: Dipole in Anti-Helmholtz Coil A dipole is initially pointing along the positive z-direction and located at the center of an anti- Helmholtz coil (z = 0), with the z-component of the magnetic field shown in the figure on the lower right. A short time later the dipole will feel ˆk µ z 1. a force but not a torque. 2. a torque but not a force. 3. both a torque and a force. 4. neither force nor torque. 158

159 CQ. Ans.: Dipole in Anti-Helmholtz Coil Answer: 1. A force because there is a gradient in the magnetic field but no torque because the magnetic field at the center is zero. Because the dipole was pointing in the positive z-direction it will feel a force upwards. If dipole were pointing in the negative z-direction it would feel a force downward. ˆk µ z F B 159

160 Faraday s Law W09D1 160

161 Michael Faraday Jumping Ring 161

162 Demonstration: Jumping Ring An aluminum ring jumps into the air when the solenoid beneath it is energized 162

163 What is Going On? This is a dramatic example of Faraday s Law and Lenz s Law: When current is turned on through the solenoid the created magnetic field tries to permeate the conducting aluminum ring, currents are induced in the ring to try to keep this from happening, and the ring is repelled upwards. 163

164 Demo: Electromagnetic Induction Moving current loop through magnetic field Move a loop 164

165 Faraday s Law Applet: Move Loop Keep Magnet Fixed 165

166 Magnetic Flux Thru Wire Loop (1) Uniform B A B Φ B = B A = B Ac o sθ Φ B = B A Product of perpendicular component of magnetic field and area (2) Non-Uniform B Φ B = S B da 166

167 Electromotive Force (emf) in a current loop When a wire loop is moved with a velocity through a magnetic field. There is a magnetic force per charge equal to! F mag / q = v! B! The line integral of the magnetic force around the loop at a fixed instant in time is defined to be the electromotive force (emf). ε = 1 q " closd path! F mag d! s! v 167

168 Electromotive Force (emf) in a Current loop Emf generalizes to any force per charge integrated around a closed path. ε = 1 q " closd path! F d! s Emf has the same dimensions as electric potential rather than force, so the SI unit is the volt. If the closed path is a circuit with resistance R then the electromotive force will cause a current to flow according to ε = IR 168

169 Motional emf: w y O F B u v F P B q x ĵ î ˆk B = 0 Pull a loop with speed v through uniform magnetic field (shaded area). Magnetic force on moving charge is! F mag / q = (vî + uĵ) B( ˆk)= vb ĵ ubî At one instant in time, electromotive force is (integrating clockwise) ε = 1! F q mag d! y=w " s = (vb ĵ ubî) dy ĵ = vbw y=0 169

170 Changing Magnetic Flux: Motional emf When the loop moves a distance Δs = vδt, magnetic flux through loop is decreasing by ΔΦ B = BwΔs y s ĵ î w B v ˆk O x B = 0 ΔΦ B Δt = BwΔs Δt = BwvΔt Δt = vbw 170

171 Electromotive Force and Changing Magnetic Flux Electromotive force is equal to the negative rate of change of the magnetic flux ε = dφ B dt 171

172 CQ: Moving Loop in Uniform Field I In the figure to the right, the shaded region represents nonzero magnetic field directed out of the plane of the figure. A rectangular wire loop is pulled upward with speed v. At the instant shown in the figure, there is B = 0 B v B 1. no current in the loop. 2. an induced current in the loop. 172

173 CQ Ans.: Moving Loop in Uniform Field I Answer: 2. The magnetic flux through the loop is decreasing, therefore there is an induced current in the loop v B = 0 B B 173

174 Group Problem: Changing Magnetic Flux Conducting rod pulled along two conducting rails in a uniform magnetic field of magnitude B at constant speed v. Assume the resistance has value R. a) Calculate the magnitude of the derivative of the magnetic flux with respect to time. b) What is the magnitude of the induced current? 174

175 Electromotive Force (emf) and work Because the electromotive force is defined as an integral at one instant in time it is not magnetic work per charge. Recall that magnetic fields do no y work! ĵ w O B F P x v ˆk î B = 0 The work done per charge by the pulling force on the loop is equal to the electromotive force. 1 q f i! F P d! s = vbw = ε 175

176 CQ: Moving Loop in Uniform Field II In the figure to the right, the shaded region represents nonzero magnetic field directed out of the plane of the figure. A rectangular wire loop is pulled to the left with speed v. At the instant shown in the figure, there is v B B = 0 B 1. no current in the loop. 2. a current in the loop. 176

177 CQ Ans.: Moving Loop in Uniform Field II Answer: 1. The motion does not change the magnetic flux through the loop. The magnetic flux is constant in time. Therefore there is no induced current through the loop. The force per charge on the leg in the magnetic field is upward, and hence no pulling force is need to keep the loop moving at a constant speed. v B = 0 B B 177

178 Ring Falling Through a Magnet Emf is equal to the gravitational work per charge on the falling ring 178

179 Magnet Moving Through a Fixed Loop 179

180 Magnet Falling Through Ring 180

181 Demonstration: Magnet falling through copper pipe 181

182 Demonstration: Magnet Falling Through Plastic Tube, Aluminum Tube, and Copper Tube 182

183 Demo: Electromagnetic Induction Move magnet with current loop fixed Moving magnet through fixed current loop 183

184 Faraday s Law Applet: Move Magnet Keep Loop Fixed 184

185 What is the emf? Conducting loop, with resistance R, is at rest, and magnet is moving as shown in figure, resulting in a changing magnetic flux through the loop and an induced current in the loop. What is the emf? ε = I ind R v S N I ind 185

186 Induced Electric Field and Electromotive Force (emf) An induced electric field appears in the current loop resulting in an electric force on the charges in the conductor. Hence there is an electromotive force that is equal to the change in magnetic flux ε =! " E d! s = d dt loop The induced current is then equal to ε = IR open surface with loop as boundary!! B d A 186

187 Faraday s Law of Induction If C is a stationary closed curve and S is a surface spanning C then C E d s = d dt S B da The changing magnetic flux through S induces a nonelectrostatic electric field whose line integral around C is non-zero ε = dφ B dt 187

188 Electric Guitar: Time Changing Magnetic Field Induces an Electric Field Pickups 188

189 Electric Guitar: Time Changing Magnetic Field Induces an Electric Field 189

190 Sign Conventions: Right Hand Rule! " E d! s = d dt closed path open surface! B d! A By the right hand rule (RHR) clockwise integration direction for line integral of electric field (emf) requires that unit normal points into plane of figure for magnetic flux surface integral Magnetic flux positive into page, negative out of page 190

191 Sign Conventions: Right Hand Rule closed path E d s = d dt open surface B d A By the right hand rule (RHR) counterclockwise integration direction for line integral of electric field (emf) requires that unit normal points out of plane of figure for magnetic flux surface integral Magnetic flux positive out of page, negative into page 191

192 CQ: Solenoid At time t: the figure on the right shows a side view of a section of a very long solenoid with radius R carrying current I with magnetic field pointing up. The figure below right shows a top view of the electric field inside the solenoid at a radius r and the direction of the magnetic field. In the solenoid, the current is 1. increasing in time. 2. constant. 3. decreasing in time. 4. cannot tell without more information. 192

193 CQ Ans: Solenoid Answer 3. In the figure on the right, the line integral of the electric field in the counterclockwise direction is positive,. E d s > 0 counterclockwise Choosing a counterclockwise circulation direction requires the unit normal to point out of the plane of the figure for the direction and so the magnetic flux is positive By Faraday s Law: d dt B da > 0 S B da > 0 S therefore the time derivative of the flux must be negative and so the magnetic field is decreasing in magnitude. 193

194 Group Problem: Induced Electric Field! " E d! s = d dt C S! B d! A Consider a uniform magnetic field which points into the page and is confined to a circular region with radius R. Suppose the magnitude increases with time, i.e. db/dt > 0. Find the magnitude and direction of the electric field in the regions (i) r < R, and (ii) r > R. (iii) Plot the magnitude of the electric field as a function r. 194

195 Lenz s Law Induced EMF is in direction that opposes the change in flux that caused it. Induced current, torque, or force is always directed to oppose the change that caused it v ε = dφ B dt S N I ind 195

196 For a Closed Conducting Path Induced current is the source of induced magnetic flux that opposes the change in external magnetic flux. v S N I ind I ind R = dφ B dt I ind 196

197 Direction of Induced Current For a counterclockwise integration direction for the line integral hence a upwards unit normal for the surface integral ε = dφ B > 0 dt I ind counterclockwise! ε = dφ B < 0 dt I ind is clockwise! 197

198 Ways to Induce Changing Magnetic ε = d dt Flux open surface! B ˆn da Quantities which can vary with time: Area A enclosed by the loop with non-zero B Magnitude of B Loop moving through non-uniform B Angle between B and normal vector to loop 198

199 CQ: Moving Loop in Uniform Field Lenz s Law In the figure to the right, the shaded region represents nonzero magnetic field directed out of the plane of the figure. A rectangular wire loop is pulled upward with speed v. At the instant shown in the figure, there is B = 0 B v B 1. a counterclockwise current in the loop. 2. a clockwise current in the loop. 3. There is no current in the loop. 199

200 CQ Ans.: Moving Loop in Uniform Field Lenz s Law Answer: 1. Define positive unit normal out of the plane of the figure. As more of the loop leaves the region of non-zero magnetic field, the magnetic flux through the loop is decreasing. According to Faraday s Law there is an induced counterclockwise current through the loop which will produce an induced magnetic field that will produce magnetic flux out of the plane of the figure opposing the change of flux. B = 0 B v B 200

201 CQ: Magnetic Field Changing in Time Lenz s Law The magnetic field through a wire loop is pointed upwards and increasing with time. The induced current in the coil is 1. clockwise as seen from the top. 2. counterclockwise as seen from the top. 3. zero. 201

202 CQ Ans.: Magnetic Field Changing in Time Lenz s Law Answer 1. Induced current is clockwise. This produces an induced B field pointing down over the area of the loop. The induced B field opposes the increasing flux through the loop Lenz s Law 202

203 CQ: Lenz s Law Moving Loop A circuit in the form of a rectangular piece of wire is pulled away from a long wire carrying current I in the direction shown in the sketch. The induced current in the rectangular circuit is 1. Clockwise 2. Counterclockwise 3. Neither, the current is zero 203

204 CQ. Ans.: Lenz s Law Moving Loop Answer: 1. Induced current is clockwise B due to I is into page; the flux through the circuit due to that field decreases as the circuit moves away. So the induced current is clockwise (to make a B into the page) Note: I ind dl x B force is left on the left segment and right on the right, but the force on the left is bigger. So the net force on the rectangular circuit is to the left, again trying to keep the flux from decreasing by slowing the circuit s motion 204

205 Lenz s Law, Inductance and Magnetic Field Energy W9D2 205

206 Announcements Prepset Week 9 due online Friday April 7 8:30 am Sunday Tutoring April 9 in from 1-5 pm PS 8 due Week Ten Tuesday at 9 pm in boxes outside Math Review Week Ten Tuesday from 9-10:30 pm in

207 Outline Lenz s Law Review Experiment 3: Faraday Law Eddy Currents Mutual Inductance Self Inductance Energy in Inductors 207

208 Michael Faraday Jumping Ring 208

209 For a Closed Conducting Path Induced current is the source of induced magnetic flux that opposes the change in external magnetic flux. v S N I ind I ind R = dφ B dt I ind 209

210 Induced Current Loop Acts Like Magnet v S N I ind I ind Force between magnet and induced current loop is like repulsive force between magnets S N N S 210

211 The figure on the right shows a side view of a section of a very long solenoid carrying current I that is decreasing. The magnetic field created by this current is pointing up and decreasing. Inside the solenoid there is a conducting ring. The figure below right shows a top view of the the solenoid and the direction of the magnetic field. The direction of the induced current in the conducting ring is 1. the counterclokwise direction. 2. the clockwise direction. 3. The induced current is zero. CQ: Solenoid conducting ring db / dt < 0 conducting ring db / dt < 0 B B top view of solenoid 211

212 CQ Ans: Solenoid Answer 1. The induced current produces an induced magnetic flux out of the plane of the figure opposing the decreasing magnetic flux due to the current decreasing in the solenoid.. conducting ring I ind B db / dt > 0 B ind top view of solenoid 212

213 What is the emf? Conducting loop, with resistance R, is at rest, and magnet is moving as shown in figure, resulting in a changing magnetic flux through the loop and an induced current in the loop. What is the emf? ε = I ind R v S N I ind 213

214 Induced Electric Field and Electromotive Force (emf) An induced electric field appears in the current loop resulting in an electric force on the charges in the conductor. Hence there is an electromotive force that is equal to the change in magnetic flux ε =! " E d! s = d dt ring open surface! B d! A The induced current is then equal to ε = I ind R 214

215 Faraday s Law of Induction If C is a stationary closed curve and S is a surface spanning C then C E d s = d dt S B da The changing magnetic flux through S induces a nonelectrostatic electric field whose line integral around C is non-zero ε = dφ B dt 215

216 Group Problem: Induced Electric Field! " E d! s = d dt C S! B d! A Consider a uniform magnetic field which points into the page and is confined to a circular region with radius R. Suppose the magnitude increases with time, i.e. db/dt > 0. Find the magnitude and direction of the electric field in the regions (i) r < R, and (ii) r > R. (iii) Plot the magnitude of the electric field as a function r. 216

217 CQ: Lenz s Law Force and Induced Current A coil moves up from underneath a magnet with its north pole pointing upward. The current in the coil and the force on the coil: N S 1. Current clockwise; force up 2. Current counterclockwise; force up 3. Current clockwise; force down 4. Current counterclockwise; force down 217

218 CQ Ans.: Lenz s Law Force and Induced Current Answer 3. Current is clockwise; force is down The clockwise current creates a self-field downward, trying to offset the increase of magnetic flux through the coil as it moves upward into stronger fields (Lenz s Law). The I dl x B force on the coil is a force which is trying to keep the flux through the coil from increasing by slowing it down (Lenz s Law again). 218

219 Experiment 3: Faraday s Law of Induction 219

220 Experiment 3: Faraday s Law In this experiment we would like you to investigate: The induced current in a wire loop that is moving through a non-uniform magnetic field The force on an aluminum sleeve that is moving through a non-uniform magnetic field 220

221 Experiment 3 Start-up Instructions Open Week 9 sequential and select Experiment 3. All group members enter their Kerberos ID on the login page. Move to Section 1 and start the experiment. 221

222 Lenz s Law: Review Induced EMF is in direction that opposes the change in flux that caused it. Induced current, torque, or force is always directed to oppose the change that caused it v ε = dφ B dt S N I ind 222

223 Eddy Currents If a conducting material is moved in a non-uniform magnetic field then currents will be generated in the material called eddy currents. The effect of the eddy currents will be to create a drag force that opposes the motion of the conducting material. Similarly if a magnet is moved through a conducting material in such a way as to set up eddy currents the magnet will feel a drag force opposing its motion. 223

224 Copper Pendulum Between Poles of a Magnet H &show=0 224

225 Eddy Current Braking What happened to the kinetic energy of pendulum? 225

226 Eddy Current Braking The magnet induces currents in the metal that dissipate the energy through Joule heating: 1. Current is induced counter-clockwise (out from center) 2. Force is opposing motion (creates slowing torque) 226

227 Eddy Current Braking The magnet induces currents in the metal that dissipate the energy through Joule heating: 1. Current is induced clockwise (out from center) 2. Force is opposing motion (creates slowing torque) 3. EMF proportional to angular frequency 227

228 Demo: Levitating Magnet H &show=0 228

229 Demo: Mutual Inductance Two Small Coils and Radio H &show=0 229

230 Mutual Inductance Current I 2 in coil 2, induces magnetic flux Φ 12 in coil 1. Mutual inductance M 12 : Φ 12 M 12 I 2 M 12 = Φ 12 / I 2 M 12 = M 21 = M Change current in coil 2; induce emf in coil 1. ε 12 M 12 di 2 dt 230

231 Group Problem: Mutual Inductance An infinite straight wire carrying current I is placed to the left of a rectangular loop of wire with width w and length l. What is the mutual inductance of the system? 231

232 Self Inductance What if is the effect of putting current into coil 1? There is self flux : Φ B LI Faraday s Law ε = L di dt 232

233 Calculating Self Inductance Unit: Henry L = Φ self 1. Assume a current I is flowing in your device 2. Calculate the B field due to that I 3. Calculate the flux due to that B field 4. Calculate the self inductance (divide out I) I 1 H = 1 V s A 233

234 Worked Example: Self-Inductance of a Toroid Calculate the selfinductance L of a toroid with a square cross section with inner radius a, outer radius b = a+h, (height h) and N square windings. 1. Assume a current I is flowing in your device 2. Ampere s Law: Calculate the B field due B2π R = µ 0 NI B = µ NI 0 to that I 2π R 234

235 Worked Example: Self-Inductance of a 3. Calculate the flux due to that B field: Toroid Φ total = NΦ turn = N! B d! A turn b a = Nh Bdr = h µ 0 N 2 I 2π ln b a 4. Calculate the self inductance (divide out I) L = Φ total I = h µ 0N 2 2π ln b a 235

236 Group Problem: Solenoid Calculate the self-inductance L of a solenoid (n turns per meter, length l, radius R) Remember 1. Assume a current I is flowing in your device 2. Calculate the B field due to that I 3. Calculate the flux due to that B field 4. Calculate the self inductance (divide out I) L = Φ self I 236

237 Solenoid Inductance B d s = Bs = µ 0 I enc = µ 0 ns ( ) I B = µ 0 ni Φ turn = B d A = BA = µ 0 niπ R 2 L = Φ self I = NΦ turn I n = N / l = Nµ 0 nπ R 2 = µ 0 n 2 π R 2 l 237

238 CQ: Solenoid A very long solenoid consisting of N turns has radius R and length d, (d>>r). Suppose the number of turns is halved keeping all the other parameters fixed. The self inductance 1. remains the same. 2. doubles. 3. is halved. 4. is four times as large. 5. is four times as small. 6. None of the above. 238

239 CQ Ans.: Solenoid Solution 5. The self-induction of the solenoid is equal to the total flux through the object which is the product of the number of turns time the flux through each turn. The flux through each turn is proportional to the magnitude of magnetic field which is proportional to the number of turns per unit length or hence proportional to the number of turns. Hence the selfinduction of the solenoid is proportional to the square of the number of turns. If the number of turns is halved keeping all the other parameters fixed then he self inductance is four times as small. 239

240 Inductor Behavior ε = L di dt L I Inductor with constant current does nothing 240

241 Back emf ε = L di dt ε = L di dt I I di dt > 0 ε L < 0 di dt < 0 ε L > 0 241

242 Marconi Coil H &show=0 242

243 Marconi Coil: On the Titanic Titanic Movie Titanic Marconi Telegraph 243