Electricity and Magnetism. Motional EMF Faraday s Law

Transcription

1 Electricity and Magnetism Ampère s Law Motional EMF Faraday s Law Lana heridan De Anza College Nov 19, 2015

2 Last time magnetic fields from moving charges magnetic fields around current-carrying wires forces between parallel wires Gauss s law

3 Overview Ampère s law motional emf induction Faraday s law Lenz s law

4 Gauss s Law for Magnetic Fields Gauss s Law for magnetic fields.: B da = 0 Where the integral is taken over a closed surface A. (This is like a sum over the flux through many small areas.) We can interpret it as an assertion that magnetic monopoles do not exist. The magnetic field has no sources or sinks.

5 Gauss s Law for Magnetic Fields 32-3 INDUCED MAGNETIC FIELD B da = PART 3 re complicated than does not enclose the Fig encloses no x through it is zero. nly the north pole of l. However, a south ce because magnetic like one piece of the encloses a magnetic urface II N B urface I tom faces and curved B of the uniform and and B are arbitrary of the magnetic flux

6 B-Field around a wire revisited Gauss s law will not help us find the strength of the B-field around a wire, or wires carrying current: the flux through any closed surface will be zero. Another law can: Ampère s Law.

7 Line Integrals To understand Ampère s Law, we first need to understand the basic idea of what a line integral represents. The most basic line integral is just: l = k s k = Q P ds This is just summing up the length of the line from P to Q. l is the line length. Each s k is a little line segment.

8 Line Integrals Line integrals involving fields are a little more complicated. uppose we want to evaluate the dot product between the field vector at each point along the line withe the line segment at that point. This is a measure of how much the line points along the field. E s k = k B A E ds

9 Line Integrals There are two cases that are particularly easy to calculate: 1 The field always points perpendicularly to the path: b a B ds = 0 2 The field always points parallel to the path: b a where l is the path length. B ds = Bl

10 Line Integrals There is one other special piece of notation used with some line integrals: This symbol means that the integral starts and ends at the same point. The path is a loop. B ds

11 Ampère s Law distribution of currents due to a For current-length constant currents (magnetostatics): the elements.again we However, if the distribe s law to find the mag- B ds = µ 0 I enc n be derived from the André-Marie Ampère wever, the law actually The line integral of the magnetic field around a closed loop is proportional to the current that flows through the loop. 1 (29-14) product B : ds : is to be. The current i enc is the its integral, let us first The figure shows cross i 2,and i 3 either directly p lying in the plane of The counterclockwise en direction of integra- Amperian loop i 3 i 1 i 2 Only the currents encircled by the loop are used in Ampere's law. Direction of integration Fig Ampere s law applied to an 1 That is, the current arbitrary thatamperian flows through loop that anyencircles surface two bounded by the loop. ds θ B

12 772 Ampère s Law CHAPTER 29 MAGNETIC FIELD DUE T This is how to assign a sign to a current used in Ampere's law. +i 1 i 2 Direction of integration direction of Fig , th are perpend current is in be in that pl In Fig The sca Thus, Ampe We can now Fig A right-hand rule for of the Ampe A current through Ampere s thelaw, loop to determine the general the direction signs forof your we can inte outstretched currents thumbencircled is assigned by aan plus Amperian sign, andloop. a current generally around the e in the opposite The situation direction isthat assigned of Fig. a minus sign. When w

13 Question CHECKPOINT 2 The figure here shows three equal currents shows i (two three parallel equaland currents one antiparal- i (two parallel and one The (29-19) figure here antiparallel) and lel) four and Amperian loops. Rank the loops according to the magnitude of four Amperian loops. Rank the loops s law gives us B according ds alongto each, the magnitude greatest first. of B : ds : along each, greatest first. i i a (29-20) b ortional to r, at Eqs c i d A a, b, c, d B d, b, c, a C (a and b), d, c D d, (a and c), b 1 Halliday, Resnick, Walker, page 773.

14 Question CHECKPOINT 2 The figure here shows three equal currents shows i (two three parallel equaland currents one antiparal- i (two parallel and one The (29-19) figure here antiparallel) and lel) four and Amperian loops. Rank the loops according to the magnitude of four Amperian loops. Rank the loops s law gives us B according ds alongto each, the magnitude greatest first. of B : ds : along each, greatest first. i i a (29-20) b ortional to r, at Eqs c i d A a, b, c, d B d, b, c, a C (a and b), d, c D d, (a and c), b 1 Halliday, Resnick, Walker, page 773.

15 Ampère s Law and the Magnetic Field from a Current Outside a wire All of the current is You might uppose we want encircled to know and the magnitude thus all of the magnetic fieldnitude at a B o distance r outside is used a wire. in Using Ampere's Ampère s law. Law? is that the integration of an encir Amperian We can Wire r loop surface the situatio the integra i m B 0 (i 1 i 2 ), We sh ( θ = 0) allow us to ds 3 Fig Using Ampere s law to find the magnetic field that a current i produces outside a long straight wire of circular cross Magnetic Figure 29-

16 Ampère s Law and the Magnetic Field from a Current All of the Outside current is a wire encircled and thus all is used in Ampere's law. Wire surface i r Amperian loop ( θ = 0) Fig Using Ampere s law to find the magnetic field that a current i produces outside a long straight wire of circular cross section.the Amperian loop is a concentric circle that lies outside the wire. ds B (Current i 3 is not encircled by the loop.) We can then rewrite B cos ds 0(i 1 i 2 ). You might wonder why, since current i 3 contributes to the nitude B on the left side of Eq , it is not needed on the is that the contributions of current i 3 to the magnetic field ca integration in Eq is made around the full loop. In contr of an encircled current to the magnetic field do not cancel ou We cannot solve Eq for the magnitude B of the mag Ampère s Law: the situation of Fig we do not have enough information the integral. However, we do know the outcome of the integrat m 0 (i 1 i 2 ), the value of which is set by the net current passing th We shall now apply B Ampere s ds = µ 0 law I enc to two situations in w allow us to simplify and solve the integral, hence to find the m Magnetic Field Outside a Long traight Wire with Cu To find the B-field at a distancefigure r from the shows wire s a long straight centerwire choose that carries a curren page. Equation 29-4 tells us that the magnetic field B : produc circular path of radius r. the same magnitude at all points that are the same dista By cylindrical symmetry, everywhere along the circle B ds is constant.

17 Ampère s Law and the Magnetic Field from a Current All of the Outside current is a wire encircled and thus all is used in Ampere's law. Wire surface i r Amperian loop ( θ = 0) Fig Using Ampere s law to find the magnetic field that a current i produces outside a long straight wire of circular cross section.the Amperian loop is a concentric circle that lies outside the wire. ds B (Current i 3 is not encircled by the loop.) We can then rewrite B cos ds 0(i 1 i 2 ). You might wonder why, since current i 3 contributes to the nitude B on the left side of Eq , it is not needed on the is that the contributions of current i 3 to the magnetic field ca integration in Eq is made around the full loop. In contr of an encircled current to the magnetic field do not cancel ou We cannot solve Eq for the magnitude B of the mag Ampère s Law: the situation of Fig we do not have enough information the integral. However, we do know the outcome of the integrat m 0 (i 1 i 2 ), the value of which is set by the net current passing th We shall now apply B Ampere s ds = µ 0 law I enc to two situations in w allow us to simplify and solve the integral, hence to find the m Magnetic Field Outside a Long traight Wire with Cu To find the B-field at a distancefigure r from the shows wire s a long straight centerwire choose that carries a curren page. Equation 29-4 tells us that the magnetic field B : produc circular path of radius r. the same magnitude at all points that are the same dista By cylindrical symmetry, everywhere along the circle B ds is constant. The magnetic field lines must form a closed loop B ds = B ds.

18 Ampère s Law and All of the the current Magnetic is Field from a Current Outside encircled a wireand thus all is used in Ampere's law. Wire surface i r ds Amperian loop B ( θ = 0) You might wonder w nitude B on the left si is that the contributio integration in Eq. 29- of an encircled curren We cannot solve E the situation of Fig. 29 the integral. However, m 0 (i 1 i 2 ), the value of We shall now app allow us to simplify a And again we get Fig Using Ampere s law to find the magnetic field that a current i produces outside a long B straight ds = wire µ 0 Iof enc circular cross section. The Amperian loop is a concentric circle that lies outside the wire. B(2πr) = µ 0 I B = µ 0I 2πr Magnetic Field Ou Figure shows a page. Equation 29-4 t the same magnitude

19 Ampère s Law and the Magnetic Field from a Current Inside a wire PART 3 We can also use Ampère s Law in another context, where using the 773 Biot-avart Law is harder AMPERE LAW he wire. We can take advanmpere s law (Eqs and lar Amperian loop of radius same magnitude B at every ise, so that ds : has the direc- Eq by noting that B : is as is ds :. Thus, B : and ds : are oop, and we shall arbitrarily between ds : and B : is 0, so omes B(2 r). ment lengths ds around the e 2pr of the loop. urrent of Fig The right ve Only the current encircled by the loop is used in Ampere's law. i Fig Using Ampere s law to find the magnetic field that a current i produces inside a long straight wire of circular cross section.the current is uniformly distributed over the cross section of the wire and emerges from the page.an Amperian loop is drawn inside the wire. B r R ds Wire surface Amperian loop Now we place the Amperian loop inside the wire. We still have B ds = 2πrB, but now the current that flows through the loop is reduced.

20 Ampère s Law and the Magnetic Field from a Current Inside a wire n- d us ry c- is re ly so e We still have B ds = 2πrB, but now the current that flow through the loop is reduced. PART AMPERE LAW 773 Assuming the wire has uniform resistivity, I enc : Only the current encircled by the loop is used in Ampere's law. i Fig Using Ampere s law to find the magnetic field that a current i produces inside a long straight wire of circular cross B r R ds Wire surface Amperian loop

21 Ampère s Law and the Magnetic Field from a Current Inside a wire n- d us ry c- is re ly so e We still have B ds = 2πrB, but now the current that flow through the loop is reduced. PART AMPERE LAW 773 Assuming the wire has uniform resistivity, I enc : Only the current encircled by the loop is used in Ampere's law. i Fig Using Ampere s law to find the magnetic field that a current i produces inside a long straight wire of circular cross B r R ds Wire surface Amperian loop I enc = πr 2 πr 2 I = r 2 R 2 I Ampére s Law B ds = 2πrB = µ 0 r 2 R 2 I o, B = µ 0I r 2πR 2 = µ ( 0I r ) 2πR R

22 Ampère s Law For constant currents (magnetostatics): B ds = µ 0 I enc The line integral of the magnetic field around a closed loop is proportional to the current that flows through the loop. Later we will extend this law to deal with the situation where the fields / currents are changing.

23 olenoids Additional examples, video solenoid A helical coil of tightly wound wire that can carry a current. turn i Fig A solenoid carrying current i. i 29-5 olenoi Magnetic Field o We now turn our useful. It concerns wound helical coil that the length of t Figure sh The solenoid s ma A single complete loop of wire in a solenoid. This solenoid has 10 turns, means it has 10 complete loops.

24 uniform, we start with a thin ring and then sum (via integration) the ircled area. Magnetic Field inside and around a solenoid P e central axis of a e turns are shown, as are the solenoid. Each turn pror the solenoid s axis, the s directed along the axis. ng magnetic field. Outside field there is very weak. Each turn of wire locally has a circular magnetic field around it. The fields from all the wires add together to create very dense field lines inside the solenoid.

25 Magnetic Field of a solenoid IELD DUE TO CURRENT P 2 P 1 Fig Magnetic field lines for a real solenoid of finite length. The field is strong and uniform at interior points such as P 1 but relatively weak at external points such as P 2. The wires on opposite sides (top and bottom in the picture) have currents in opposite directions. The fields add up between them, vidual turns (windings) that make up the solenoid. For points very close to a turn, but cancel out outside of them. the wire behaves magnetically almost like a long straight wire, and the lines of B : there are almost concentric circles. Figure suggests that the field tends to cancel between adjacent turns. It also suggests that, at points inside the solenoid :

26 Magnetic Field B : ds of an ideal solenoid : 0 i enc, (29-21) In an ideal solenoid (with infinite length) the field outside is enoid negligible of Fig , andwhere insideb : is is uniform. (imilar within the tosolenoid a capacitor!) and using the rectangular Amperian loop abcda. We write B : ds : as Can use an Amperian loop to find the B-field inside: d h c i B ds = µ 0 I enc B a b plication of Ampere s law to a section of a long ideal solenoid carrying mperian loop is the rectangle abcda.

27 Magnetic Field B : ds of an ideal solenoid : 0 i enc, (29-21) In an ideal solenoid (with infinite length) the field outside is enoid negligible of Fig , andwhere insideb : is is uniform. (imilar within the tosolenoid a capacitor!) and using the rectangular Amperian loop abcda. We write B : ds : as Can use an Amperian loop to find the B-field inside: d h c i B ds = µ 0 I enc B a b plication Here, of Ampere s supposelaw there to a section are n of turns a long per ideal unit solenoid lengthcarrying the solenoid, mperian then loop Iis enc the = rectangle Inh abcda. Bh = µ 0 Inh Inside an ideal solenoid: B = µ 0 In

28 Question For what current through a solenoid with 50 turns per centimeter will the magnetic field be 20 mt?

29 Question For what current through a solenoid with 50 turns per centimeter will the magnetic field be 20 mt? I = 3.18 A

30 Induction and Inductance Changing magnetic fields can put forces on charges. To see how, we start by looking again at conductors moving in magnetic fields.

31 Motional EMF In steady state, the electric and If a conductor moves magnetic through forces a magnetic on an electron field at an angle to the field, an emf is in induced the conductor across the are conductor. balanced. F e F B E B in v The strai form magne is moving in influence of model, the that is direc influence of accumulate charge sepa the electron model. The on charges r The electron tion for equ There are two ways Due to see the this: magnetic force on 1 force on conduction electrons, charges the ends F = of qthe v B 2 in the restconductor frame of thebecome conductor oppositely there is also an electric field charged, which establishes an

32 Motional EMF In steady state, the electric and magnetic forces on an electron in the conductor are balanced. F e F B E B in v ue to the magnetic force on lectrons, the ends of the The straight conductor o form magnetic field directed is moving in a direction per influence of some external a model, the electrons in the that is directed along the F influence net = q(e + v B) = 0 of this force, the accumulate there, leaving a charge E = separation, vb (v B) an electr E the electrons = vb are also desc l model. The charges accumu E = vbl on charges remaining in th The electrons are then desc tion for equilibrium require Once the charge distribution reaches equilibrium, the net force on each charge:

33 Motional emf and loops Imagine a loop of wire that moves in a uniform magnetic field, B, directed into the page. a b v Imagine the loop is composed of a pair of curved rods cut along the lines shown. Which way (left or right) the emf directed in the top half? In the bottom? How do the magnitudes compare?

34 Motional emf and loops Imagine a loop of wire that moves in a uniform magnetic field, B, directed into the page. a b v In this case, part of the loop near a develops a negative charge and the part near b a positive charge, but overall no steady current flows around the loop.

35 Motional emf and loops Now imagine a loop of wire that moves in a non-uniform magnetic field falling towards a wire. a b v I How do the magnitudes of the emfs in the top and bottom compare?

36 Motional emf and loops Now imagine a loop of wire that moves in a non-uniform magnetic field falling towards a wire. a b v I How do the magnitudes of the emfs in the top and bottom compare? They are not the same! A current can flow.

37 Motional emf and loops Now imagine a loop of wire that moves in a non-uniform magnetic field falling towards a wire. (Quiz 31.3) a b v I What is the direction of the induced current in the loop of wire? (A) clockwise (B) counterclockwise (C) zero (D) impossible to determine

38 Motional emf and loops Now imagine a loop of wire that moves in a non-uniform magnetic field falling towards a wire. (Quiz 31.3) a b v I What is the direction of the induced current in the loop of wire? (A) clockwise (B) counterclockwise (C) zero (D) impossible to determine

39 Motional emf and loops What was different in the two cases (uniform vs. non-uniform field)?

40 Motional emf and loops What was different in the two cases (uniform vs. non-uniform field)? The field at different parts of the loop was different. The magnetic flux through the loop was changing.

41 Reminder: (30.18) Magnetic Flux field B that makes an case is (30.19), then u and the he plane as in Figure aximum value). weber (Wb); 1 Wb 5 Magnetic flux plane is a agnetic to the plane. Definition of magnetic flux B u d A Figure The magnetic flux through an area element da is B? d A 5 B da cos u, where d A is a vector perpendicular to the surface. The magnetic flux of a magnetic field through a surface A is Φ B = B ( A) da Units: Tm 2 If the surface is a flat plane and B is uniform, that just reduces to: Φ B = B A

42 a magnet is moved When the magnet is a loop of wire When the magnet is held moved away from the cted Changing to a sensitive flux andstationary, emf there is no loop, the ammeter sh ter, the ammeter induced current in the that the induced curr that a current When ais magnet is atloop, rest even nearwhen a loop the of wire there is no is opposite potential that shown d in the difference loop. across the magnet ends ofis the inside wire. the loop. part a. I N N I N b c

43 toward a loop of wire When the magnet is connected to a sensitive stationary, there is no ammeter, the ammeter induced current in th When the north pole of shows the that magnet a current is moved is towards the loop, loop, even the when the magnetic flux increases. induced in the loop. magnet is inside the Changing flux and emf 1 A simple experiment t a current is induced en a magnet is moved way from the loop. I N a A current flows clockwise in the loop. b N

44 he magnet is held moved away from the ry, Changing there is no flux andloop, emf the ammeter shows current in the that the induced current en when When the the north pole of is opposite the magnet that shown is moved in away from the loop, is inside the the magnetic loop. flux decreases. part a. I N N c A current flows counterclockwise in the loop.

45 Faraday s Law Faraday s Law If a conducting loop experiences a changing magnetic flux through the area of the loop, an emf E F is induced in the loop that is directly proportional to the rate of change of the flux, Φ B with time. Faraday s Law for a conducting loop: E = Φ B t

46 Faraday s Law Faraday s Law for a coil of N turns: E F = N Φ B t if Φ B is the flux through a single loop.

47 Changing Magnetic Flux The magnetic flux might change for any of several reasons: the magnitude of B can change with time, the area A enclosed by the loop can change with time, or the angle θ between the field and the normal to the loop can change with time.

48 Lenz s Law The magnet's motion creates a magnetic dipole that opposes the motion. Additional examples, video, and practice available at Lenz s Law 30-4 Lenz s Law An induced current has a direction such that the magnetic fieldoon dueafter to the Faraday current proposed opposes his the law of induc devised a rule for determining the direction of an change in the magnetic flux that induces the current. An induced current has a direction such that the ma opposes the change in the magnetic flux that induces t Furthermore, the direction of an induced emf is th a feel for Lenz s law,let us apply it in two different where the north pole of a magnet is being moved to N Basically, Lenz s 1. Opposition law let s us to Pole interpret Movement. the minus The approach N Fig increases the magnetic flux through µ sign in the equation we write to represent current in the loop. From Fig , we know t Faraday s Law. netic dipole with a south pole and a north po moment : is directed from south to north. i increase E = being Φ caused B by the approaching mag thus : ) must t face toward the approaching no 30-4). Then the curled straight right-hand rul Fig Lenz s law at work.as the the current induced in the loop must be counte magnet is 1 moved toward the loop, a current Figure from Halliday, Resnick, Walker, 9th If we ed. next pull the magnet away from th

49 ummary Ampère s law motional emf Faraday s law Lenz s law Homework Halliday, Resnick, Walker: PREVIOU: Ch 29, onward from page 783. Questions: 3; Problems: 1, 11, 21, 23 NEW: Ch 29, Questions: 7; Problems: 35, 43, 45, 51, 53 NEW: Ch 30, onward from page 816. Questions: 1, 3; Problems: 1, 3, 4.