Electricity and Magnetism Applying Faraday s Law

Transcription

1 Electricity and Magnetism Applying Faraday s Law Lana heridan De Anza College Nov 24, 2015

2 Last time Ampere s law Faraday s law Lenz s law

3 Overview induction and energy transfer induced electric fields inductance self-induction RL Circuits

4 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

5 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

6 esult n the e sysnsisfore, Faraday s and Lenz s Laws What about this case? We found the current should flow counterclockwise. ire in to I v Figure (Quick Quiz 31.3)

7 esult n the e sysnsisfore, Faraday s and Lenz s Laws What about this case? We found the current should flow counterclockwise. ire in to I v The flux from Figure the wire is into the (Quick page and Quiz increasing. 31.3) The field from the current is out of the page. There is an upward resistive force on the ring.

8 Faraday s and Lenz s Laws A counterclockwise current I is induced in the loop. The magnetic Consider a conducting force bar F B placed on the bar oncarrying conducting this rails in a magnetic field, with a resistor current (outside opposes the the field) motion. completing the circuit. B in I R F B v F app R e nducting ity v along nder the rce F app. cuit diawn in (a). a I x Using the motional emf approach, what is the induced emf across the bar? Using Faraday s law, what is the induced emf across the bar? b I

9 ase, the rate of change of magnetic flux through the circuit and the corresponding Faraday s and Lenz s Laws Motional emf: A counterclockwise current I is induced in the loop. The magnetic force on the bar carrying this current opposes the motion. E F= B Bvl upwards Faraday s Law: E = Φ B t = Bl x t = Bvl B in I R F B v F app R e B v I x I a A counterclockwise current begins to flow as the rod moves. (Opposes the field.) b

10 current because the charges are free to move in the closed conducting path. In this case, the rate of change of magnetic flux through the circuit and the corresponding Faraday s and Lenz s Laws A counterclockwise current I is induced in the loop. The magnetic force F B on the bar carrying this Power is delivered to the resistor as current flows. That power current opposes must come the motion. from the force needed to keep the rod in motion. B in I R F B v F app R e B v I x I a Prove they are equal. b

11 Faraday s and Lenz s Laws Power delivered to resistor: P = E2 R = B2 v 2 l 2 R Power supplied by applied force needed to keep rod moving with constant velocity v: F net = 0 F app = F B

12 Faraday s and Lenz s Laws Power delivered to resistor: P = E2 R = B2 v 2 l 2 R Power supplied by applied force needed to keep rod moving with constant velocity v: F net = 0 F app = F B P = F app v = (IlB)v = E R Bvl = B2 v 2 l 2 R

13 Faraday s and Lenz s Laws Power delivered to resistor: P = E2 R = B2 v 2 l 2 R Power supplied by applied force needed to keep rod moving with constant velocity v: F net = 0 F app = F B P = F app v = (IlB)v = E R Bvl = B2 v 2 l 2 R

14 Faraday s and Lenz s Laws Implication: it is possible to turn mechanical power into electrical power.

15 Electric Generators 1 Figure from hyperphysics.phys-arstr.gsu.edu

16 Electric Guitar Pickups trings are made of ferrous metal: steel (iron) or nickel, which become magnetized by the permanent magnets. Plucked strings create a changing magnetic field that produces a current in the pickup coil. 1 Figure from HowtuffWorks.

17 Induction and Energy Transfer to a Wire Loop When you move a magnet near a loop of wire or a loop of wire near a magnetic field, a force resists the motion. This is due to the induced magnetic dipole in the loop. ince you must apply a force to overcome the resistance, you do work moving the magnet / loop. This energy goes to heating the loop of wire.

18 Energy Transfer TION Easy AND INDUCTANCE to see for the case of a loop being pulled out of a B-field. F 2 x B i Decreasing the area decreases the flux, inducing a current. F 1 L F 3 v b Fig You pull a closed conducting loop out of a magnetic field at constant : velocity v. While the loop is moving, a clockwise current i is induced in the loop, and the loop segments still within the magnetic field experience forces F :, F :, and F : The B-field is uniform within the rectangle and zero outside it. Figure 30-8 shows another situation involving induced current. A rectan-

19 is more apparent in the arrangement of Fig b; a conducting plate,free to rotate about a pivot, is allowed to swing down through a magnetic field like apendulum.each time the plate enters and leaves the field,a portion of its mechanical energy is transferred to its thermal energy. After several swings, no Loops in B-Fields Question The figure shows four wire loops, with edge lengths of either L or mechanical energy remains and the warmed-up plate just hangs from its pivot. 2L. All four loops will move through a region of uniform magnetic fieldcheckpoint B (directed out 3 of the page) at the same constant velocity. The figure shows four wire loops, with edge lengths of either L or 2L. All four loops Rank will move the through four loops a region according of uniform magnetic to the maximum field B : (directed magnitude out of the page) of the at emf the induced same constant as velocity. they move Rank the into four the loops field, according greatest to the maximum first. magnitude of the emf induced as they move through the field, greatest first. B a b c d A a, b, c, d 30-6 Induced Electric Fields B (b and c), (a and d) Let us place a copper ring of radius r in a uniform external magnetic field, as in C Fig. (c 30-11a.The and d), field neglecting (a and b) fringing fills a cylindrical volume of radius R. D uppose (a and that b), we (c increase and d) the strength of this field at a steady rate, perhaps by increasing in an appropriate way the current in the windings of the electromagnet 1 that produces the field. The magnetic flux through the ring will then Halliday, Resnick, Walker, page

20 is more apparent in the arrangement of Fig b; a conducting plate,free to rotate about a pivot, is allowed to swing down through a magnetic field like apendulum.each time the plate enters and leaves the field,a portion of its mechanical energy is transferred to its thermal energy. After several swings, no Loops in B-Fields Question The figure shows four wire loops, with edge lengths of either L or mechanical energy remains and the warmed-up plate just hangs from its pivot. 2L. All four loops will move through a region of uniform magnetic fieldcheckpoint B (directed out 3 of the page) at the same constant velocity. The figure shows four wire loops, with edge lengths of either L or 2L. All four loops Rank will move the through four loops a region according of uniform magnetic to the maximum field B : (directed magnitude out of the page) of the at emf the induced same constant as velocity. they move Rank the into four the loops field, according greatest to the maximum first. magnitude of the emf induced as they move through the field, greatest first. B a b c d A a, b, c, d 30-6 Induced Electric Fields B (b and c), (a and d) Let us place a copper ring of radius r in a uniform external magnetic field, as in C Fig. (c 30-11a.The and d), field neglecting (a and b) fringing fills a cylindrical volume of radius R. D uppose (a and that b), we (c increase and d) the strength of this field at a steady rate, perhaps by increasing in an appropriate way the current in the windings of the electromagnet 1 that produces the field. The magnetic flux through the ring will then Halliday, Resnick, Walker, page

21 31.4 Induced emf and Electric Fields 947 Loop moving into and out of a B-field,, v a w 0 3w B in x B v B v c e x B F x B w B 2 2 v R. b 0 w 3w 4w x d 0 w 3w 4w x

22 Energy Transfer TION If AND weinductance pull a loop out of a magnetic field, a current flows in the loop. F 2 x B i Decreasing the area decreases the flux, inducing a current. F 1 L F 3 v b Fig You pull a closed conducting loop out of a magnetic field at constant : velocity v. While the loop is moving, a clockwise current i is induced in the loop, and the loop segments still within the magnetic field experience forces F :, F :, and F : The B-field is uniform within the rectangle and zero outside it. Figure 30-8 shows another situation involving induced current. A rectan-

23 Eddy Currents If the wire is replaced by a solid conducting plate, circulations of ON AND current INDUCTANCE form in the plate. - y l B Eddy current loop Pivot n- B (a) ince the cross section of the plate is larger than that of a similar Eddy wire, Currents the resistance will be low, but the current can be high. uppose we replace the conducting loop of Fig with a solid conducting plate. If we then move the plate out of the magnetic field as we did the loop (Fig a), The plate will heat. the relative motion of the field and the conductor again induces a current in the conductor. Thus, we again encounter an opposing force and must do work because of the induced current. With the plate, however, the conduction electrons making (b)

24 Induced Electric Fields If moving a conductor in a magnetic field causes a current to flow, if must be because the process has created an electric field across the conductor. The fact that in a conducting plate circulations of current appear tells us that the electric field lines must also makes these circles. Another way to cause a current and electric field is to change the flux by increasing or decreasing the magnetic field INDUCED ELEC Copper ring Circular path E (a) R i B r (b) E R B E r E

25 Induced Electric Fields Electric field lines R R B 3 (c) (d) The circulation Fig E-field(a) occurs If the whether magnetic or field not aincreases conductor at a issteady rate, a present: rent it isappears, the direct as shown, result of in the changing copper ring magnetic of radius flux. r.(b) An induce even when the ring is removed; the electric field is shown at four po picture of the induced electric field, displayed as field lines. (d) Four Faraday s Law of Induction (in words) enclose identical areas. Equal emfs are induced around paths 1 and 2 A changing the region magnetic of changing field gives magnetic rise to field.a an electric smaller field. emf is induced aroun partially lies in that region. No net emf is induced around path 4, whic

26 Induced emf and the Electric Field For a closed path, s, E = E ds Notice that by definition V = E ds = 0. Emf does not have this property. When a charge is moved around a closed path in an electrostatic electric field the work done is zero: W app = q( V ) = 0

27 Induced emf and the Electric Field When a charge is moved around a closed path in an electrostatic electric field the work done is zero: W app = q( V ) = 0 For the induced E-field from a changing magnetic flux, the associated force F = qe is not conservative. We say the E-field is nonconservative. This is no longer the electrostatic case.

28 Faraday s Law Faraday s Law for a conducting loop: E = Φ B t (Differential form:) E = dφ B dt

29 Faraday s Law Faraday s Law for a conducting loop: E = Φ B t (Differential form:) E = dφ B dt Faraday s law reformulated: E ds = dφ B dt

30 g that an elecetic Induced flux. emf and the Electric Field the loop. lectric field is suggests that E E generates an atic field pronducting loop r to the plane time, an emf the loop. The d electric field in which the k done by the o qe. Because lectric field in tangent to the circumference of E r E B in Figure A conducting loop We can also write Faraday s Law as: of radius r in a uniform magnetic field perpendicular to the plane of the loop. E ds = dφ B dt

31 Reformulated Faraday s Law We can use this value of E together with E = dφ B dt, to reformulate Faraday s Law. Faraday s law E ds = Φ B t (Differential form:) E ds = dφ B dt

32 Faraday s Law Examples E B Faraday s law: E ds = Φ B (b) E t E (Differential form:) E ds = dφ B dt lectric field lines 4 R 2 1 B (d) 3 ic field increases at a steady rate, a constant induced cur-

33 Electric Potential Electric potential has meaning only for electric fields that are the result of static charges; it has no meaning for electric fields that are produced by induction. For E-fields produced by static charges E ds = 0 For induced E-fields, the integral may not be zero. E = E ds

34 Inductors A capacitor is a device that stores an electric field as a component of a circuit. inductor a device that stores a magnetic field in a circuit It is typically a coil of wire.

35 Circuit substitution problem. of is superconducting magne Notice that this the as Equation 26.2 Notice that thisexpression expression is the samesame asthis Equation 26.2, the capa Capacitor section. Th approximately ten times g component symbols symbol initially unchar Use Equation to express the magnetic field in the tromagnets. uch superco In upercond studying e storing energy. interior of the base solenoid: 26.3circuit Combinations resonance imaging, or MR diagram " Battery 26.3 Combin Find the mutual inductance, noting that magnetic Twothe or more capacitors oftenfo a organs without the need elements. The symbol battery V the equivalent capacitance of c! coil caused flux F BH through the handle s by the magful radiation. orthroughout more capacito Capacitor this Two section. this se wires between symbol initially theuncharged. equivalent capac netic field the base coil is BA: The of direction of the effective flowcapacitor of positive and switches asw In studying electric circuits, this section. Through diagram. uch The a diagra "a number circuit witch Wireless charging is used ofinitially other cordless ds Battery in ure Open charge is clockwise. capacitor Csymbol 27.6 Electrical elements. Theuncharged. circuit symbolsp symbol! symbol used by some manufacturers of electricwires cars that avoids model for a dire cap In studying electric between the circuit elem I andcircuit switches as wellcircuits, as theuch colo diagram. In typical electric ing apparatus. Closed is at the higher " Battery witch ure The symbol for the ca Open elements. The circuit a source such as a battery symbol model for a capacitor, a pair of! switch Figure 26.6 symbol CircuitClosed symbols forisdetermine between the atwires the higher potential and ci is r Let s an express batteries, and switches. Parallel Com and switches as well b capacitors, c symbols Figure 26.6 Circuit for transfer. First, consider thea capacitors, batteries, and Parallel Combination Notice that capacitors areswitches. in witch uretwo The symbol # Open capacitors to a resistor. (Resistors are Notice that capacitors are in Rin green, Two capacitors as sho!v batteries blue, are and symbol model forconnected a capacitor blue, resistor R batteries are in green, andconnecting " wires also have nation of capacitors. Figure 26. nation of capac switches are in red. The closed switches are in red. The closed Closed is at the higher poten a d whereas capacitors. The left plates of the switch can carry current, some to the resistor. Unles capacitors. The switch can whereas the battery by a conducting wire thecarry open onecurrent, cannot.! Figure 26.6 Circuit symbols for ais capacitor When is conn wires small the open one batteries, cannot. the compared battery bywit a C L capacitors, and switches. Parallel Combinat " inductor L delivered to the wires is ne combination is an LC circu Q max Notice that capacitors are in Two capacitors following acurr pos blue, batteries are in green,then and Imagine closed, both theconne Figure A circuit consistnation of capacitors. switches are in red. The closedcircuit in Figure from late between maximum p ing of a resistor of can resistance R capacitors. The left pl switch carry current, whereas We identify the entire circu and a battery having a potential cuit is zero, no energy is tr the open one cannot Oscillations the battery by a condu

36 Inductance Just like capacitors have a capacitance that depends on the geometry of the capacitor, inductors have an inductance that depends on their structure. For a solenoid inductor: L = µ 0 n 2 Al where n is the number of turns per unit length, A is the cross sectional area, and l is the length of the inductor. Units: henries, H. 1 henry = 1 H = 1 T m 2 / A

37 Value of µ 0 : New units The magnetic permeability of free space µ 0 is a constant. µ 0 = 4π 10 7 T m / A It can also be written in terms of henries: µ 0 = 4π 10 7 H / m (Remember, 1 H = 1 T m 2 / A)

38 Inductance However, capacitance is defined as being the constant of proportionality relating the charge on the plates to the potential difference across the plates q = C ( V ). Inductance also is formally defined this way. inductance the constant of proportionality relating the magnetic flux linkage (NΦ B ) to the current: NΦ B = L I ; L = NΦ B I Φ B is the magnetic flux through the coil, and I is the current in the coil.

39 ummary applications of Faraday s law inductance self-induction RL Circuits Homework Halliday, Resnick, Walker: PREV: Ch 29, Questions: 7; Problems: 35, 43, 45, 51, 53 PREV: Ch 30, Questions: 1, 3; Problems: 1, 3, 4. NEW: Ch 30, onward from page 816. Problems: 29, 31, 35, 37