Problem Fig

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Clinton Collins
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1 Problem 9.53 A flexible circular loop 6.50 cm in diameter lies in a magnetic field with magnitude T, directed into the plane of the page, as shown. The loop is pulled at the points indicated by the arrows, forming a loop of zero area in 0.50 s. (a) Find the average induced emf in the circuit. (b) What is the direction of the current in R, from a to b, or from b to a? xplain your reasoning. Fig PHYS W 1

2 A dc generator and back emf in a motor Fig In this example, steps are taken to make the emf which is generated always positive. Due to the commutator, the brushes switch sides when the induced emf changes direction, and hence the current produced will always be in the same directions. The bumpiness can be smoothed out by using a large number of coils and commutator segments. When using the slip rings (ac alternator), the emf had both positive and negative values, and ε ave =0. With the commutator, the emf is never negative, and the average value is positive. PHYS W

3 This motor s back emf is just the emf induced by the changing magnetic flux through the rotating coil. For N coils in the loop, ε = N ωba sinωt ave T / ave sinω 0 sinω t t ave= π where T = π ω ω π ω π ω 0 = = sinωt ( 1 )cosωt ω 0 = π π ω ω ( 1 )[cosπ cos0] ω π ω π ε ave = NωBA π Slidewire generator (motional emf) This is a simple version of a dc generator, and although it is not very useful practically, it is useful pedagogically. Fig PHYS W 3

4 The cause of the induced emf is the increase in area. Choose A to point into the page, parallel to B. ΦB = B A= BAcosφ = BA ε = d B Φ = B da = B Lv = BLv Note, to oppose the change that is causing the induced emf, the current must flow counterclockwise. (Check this with the R.H. rule). If v is constant, and the rod stays on the rails, the emf generated will be constant. This is a d.c. generator. But we now have a current in a conductor which is moving in a magnetic field. Hence There will be a force on it pointing opposite to v. Thus, work must be done to move the rod to the right. R For a resistance in the circuit, energy will be dissipated from the I R losses. Fig. 9.1 PHYS W 4

5 How does the work done compare to the energy losses? To keep the rod moving, a force F, equal to the resisting force, must be applied in the direction of. v The force is given by F = ILB ε = LB= R Hence, the rate at which work is done is The energy dissipated is I BLv R Fv BLv R=( ) R= R B = B LB Lv R The work done is equal to the energy dissipated. = Lv R B Lv R The situation of the sliding rod in the slidewire generator can be analyzed in a different way. As the rod in Fig moves to the right, charged particles in it experience a force F = qv xb which creates an excess of positive charge at the top of the rod. PHYS W 5

6 (Assume that the rails on which the rod slides are insulators for the moment). An electric field will then be created which points from top to bottom on the rod. In equilibrium, the downward electrostatic force will equal the upward magnetic force. q=qvb The difference in potential between top and bottom, V = Vtop Vbottom = L= vbl The moving rod has become like a battery, with a potential difference between the ends, and hence a source of emf. When the rest of the circuit (here, just a rectangle) is a conductor, this emf drives current around it, in the counterclockwise direction shown. V = ε =vbl (motional emf) The emf ε (magnitude and direction) has been determined now in two ways, Faraday s law / Lenz s law, and by analysis of the motion of the rod (motional emf). Both are the same. PHYS W 6

7 We can generalize this. Suppose the circuit is not a simple rectangle, and that the magnetic field is not perpendicular to the plane of the area, and not uniform. From above, = vxb d ε = dl = ( vxb) dl ε = (v xb) dl (motional emf) This is equivalent to the original statement of Faraday s law, ε = dφ B B Consider again a conducting loop which encloses a magnetic field which can vary with time. I PHYS W 7

8 If the strength of the field shown is increasing in time, then an emf will be induced which drives a current in the direction shown. What makes the charges move so that a current flows? The conductor is not moving in a magnetic field, so there are no magnetic forces. The conductor may not even be in a magnetic field (but just encloses it). What is the only thing that makes electric charges move? It is an electric field. So the changing magnetic field must induce an electric field. lectric fields are not only produced by charges, but by time varying magnetic fields. Integrating around the loop, we get d l = ε = dφ If the loop is a conductor, and a charge q moves once around the loop, then there is work done. q dl = qε B B PHYS W 8

9 This electric field is very different from the electrostatic field that we first studied. Recall, for an electrostatic field, d = 0 l Such a field is conservative, and for conservative fields we could define a potential. Our new electric field, produced by a changing magnetic field, is nonconservative, and is a nonelectrostatic field. F = q However, the fundamental force law,, applies to both fields., conservative and nonconservative. Consider this. Suppose the loop in the previous slide is nonconducting. In fact, suppose the loop is just imaginary. Is an electric field still produced? Applications of magnetically induced electric fields. Power generating stations Alternators in cars Computer disk drives Magnetic tapes PHYS W 9

10 Problem 9.7 The current in the long straight wire AB is upward and is increasing steadily at a rate di (a) At an instant when the current is i, what are the magnitude and direction of the field B at a distance r to the right of the wire? (b) What is the flux dφ B through the narrow shaded strip? (c) What is the total flux through the loop? (d) What is the induced emf in the loop? What is the direction? (e) If a=1.0 cm, b=36.0 cm, L=4.0 cm, and di = 9.60 A/s, calculate the value of the induced emf. Fig. 9.7 PHYS W 10

11 Problem 9.67 The magnetic field B, at all points within a circular region of radius R, is uniform in space and directed into the plane of the page. (the region could be a cross section inside the windings of a long, straight solenoid). If the magnetic field is increasing at a rate db, what are the magnitude and direction of the force on a stationary positive point charge q located at points a, b, and c? Fig PHYS W 11

12 ddy Currents ddy currents are currents set up in a piece of bulk metal (as opposed to a clearly defined circuit, for example), by a changing magnetic flux. In the figure shown, a piece of metal sheet swings into and out of a magnetic field (it could be between the poles of an electromagnet). As the sheet enters or leaves the field, induced emfs are generated to oppose the change. These set up currents as shown by the big arrows. If the moving piece of metal is big, then the section of metal which enters the field first has an induced current, and hence a braking force applied to it. The return current is generally outside the field and that section of metal experiences no force. x x x x x x x x x x v 1 F F x x x x x B in A metal piece swings in a magnetic field. The metal piece is slowed considerably by the magnetic force on the induced eddy currents as the metal piece enters of leaves the field. v PHYS W 1

13 If the piece of metal has slots cut into it, the large eddy current loops are broken up, and the braking action greatly reduced. ddy currents can have undesirable effects. ddy currents in cores of transformers (which we will study later) waste energy through I R heating losses. Advantages: damp unwanted oscillations, for example in sensitive mechanical balances. magnetic braking in some rapid transit cars. lectromagnets positioned in the car over the rails can induce eddy currents in the rails. One particularly interesting example of the effects if eddy currents occurs in Io, the large moon of Jupiter. If the metal sheet has of slots cut into it, the eddy currents are confined to narrow slots and are greatly reduced. The braking on the pendulum is greatly reduced. PHYS W 13

14 Io is the 4 th largest moon in the solar system and the most volcanic of any body. Io moves rapidly through Jupiter s magnetic field and this produces eddy currents in Io s interior. These currents dissipate energy at the rate of 10 1 W. This energy helps in a minor way to keep Io s interior hot, and is an indirect cause of volcanic eruptions. [Jupiter s varying gravitational pull produces friction in Io s interior which causes tidal heating. This is an even bigger cause of heating and volcanic activity]. Displacement current We are slowly acquiring the 4 equations known as Maxwell s equations, and which are needed for a complete understanding of electromagnetism and the fundamental nature of light. So far we have: PHYS W 14

da= Q encl ε 0 B da=0 B d l =µ 0i encl dφ ε = B = dl (Gauss s law) (Gauss s law for magnetism) (Ampere s law) (Faraday s law) It turns out that the 3 rd equation (Ampere s law) is incomplete and we

15 da= Q encl ε 0 B da=0 B d l =µ 0i encl dφ ε = B = dl (Gauss s law) (Gauss s law for magnetism) (Ampere s law) (Faraday s law) It turns out that the 3 rd equation (Ampere s law) is incomplete and we must generalize Ampere s law by completing it. Imagine a circuit with a charging capacitor. Current i C flows into the left hand side, and out the right. The electric field increases in the space between the capacitor plates. Fig. 9. PHYS W 15

16 Apply Ampere s law to the circular path shown. B dl around the path shown equals µ (where is the conduction C current flowing through the area bounded by that path). In this case, the conduction current i flows through the flat area bounded by the path. C But the bulging surface to the right is also an area bounded by the same path. The current through that surface is zero. So, Ampere s law seems to be fine for the flat surface but not for the bulging surface. Ampere s law is incomplete. There is an equivalent current between the capacitor plates and it is related to the changing field. For a parallel plate capacitor, with plates of area A separated by d, and filled with material of permittivity C= εa d The charge on the capacitor at any time is ε 0i C i PHYS W 16

17 q( t) εa = CV( t) = ( d) = εa= εφ d The current then between the capacitor plates which represents the continuation of the conduction current is dq dφ =ε This current was called the displacement current by Maxwell. i D dφ =ε (displacement current) Ampere s law then becomes B d l = µ 0 ( i C + i D ) encl The displacement current density is j. D PHYS W 17

18 j D dφ d = ε = ε A Is the displacement current really there? Can we demonstrate its existence? The answer is yes. If there is a current between the charging capacitor plates, then there will have to be an associated magnetic field. Fig. 9.3 shows a capacitor charging. Apply Ampere s law to a circle of radius r<r. The current enclosed by this circle is j D A = id ( πr πr ) r R r R B dl = π rb= µ 0 id = µ 0 i C i = D ic Fig. 9.3 ( for the charging capacitor) PHYS W 18

19 µ 0 r π R µ 0 πr B = B i C i C (r<r) = (r>r) These results are confirmed by experiment. Maxwell s equations of electromagnetism Maxwell s equations in integral form for free space are: da= Q encl ε 0 (Gauss s law for ) B da=0 (Gauss s law for B ) d B dl ( Φ = 0 ic + ε0 ) encl d ΦB d l = ε = (Faraday s law) µ (Ampere s law) ---() ---(1) ---(4) ---(3) PHYS W 19

20 Note that the in (1) is the conservative one, and that the in (4) is the nonconservative on. n But plays no part in Gauss s law, and plays no part in Faraday s law. n So we can take the total electric field to be the electric field used in Maxwell s equations. c c = c+ n Note the symmetry in Maxwell s equations. In free space, Q encl =0 and equations (1) and () on the previous slide are Identical in form. Similarly, conduction current i C =0 and equations (3) and (4) become: dφ d B dl = ε0 µ 0 = ε0µ 0 da dφ B d dl = = B da More on this later. PHYS W 0

Chapter 29 Electromagnetic Induction

Chapter 29 Electromagnetic Induction In this chapter we investigate how changing the magnetic flux in a circuit induces an emf and a current. We learned in Chapter 25 that an electromotive force (E) is