Electromagnetic Induction

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1 . Electromagnetic Induction Concepts and Principles Creating Electrical Energy When electric charges move, their electric fields vary. In the previous two chapters we considered moving electric charges as the source of magnetic fields, but we could just as easily have considered the variation in the electric field as the source of the magnetic field. This leads to an interesting question: If varying electric fields can create magnetic fields, can varying magnetic fields create electric fields? The answer is yes, and this process, termed electromagnetic induction, is at the heart of almost all electrical power generation worldwide. In addition to incredible technological importance, electromagnetic induction hints at a deep inter-relationship and symmetry between electric and magnetic fields that will explored more fully in later chapters. Imagine a region of space with a magnetic field. Surrounding a portion of this region is a hypothetical closed path. (Often, a real loop of wire will be the closed path of interest, but induction occurs whether or not a real wire loop is present.) First, let me define magnetic flux. In analogy to electric flux, introduced in conjunction with Gauss Law, magnetic flux is defined to be: da This equation involves the vector dot product between the magnetic field and an infinitesimally small area within the area bounded by the closed path. This dot product between magnetic field and area is often visualized as the amount of field that passes through the little piece of area. The integral simply tells us to add up all of these infinitesimal magnetic fluxes to get the total flux through the area enclosed by the path. 1

2 Second, let me define emf. (Actually, let me apologize. Emf used to stand for electromotive force, even though it is not a force. In light of this misleading name, emf now, officially, stands for emf. It s not short for anything. I m not making this up.) Emf is the name for the electrical energy per unit charge created by changing magnetic flux. In general, any process that generates electrical energy creates emf. In addition to changing magnetic flux, chemical batteries and some solar cells create emf. Mathematically, emf is defined by integrating the electric field around the closed path described above: E dl (Don t worry, you will never actually calculate this integral.) All you need to know is that emf is the energy created per unit charge. The units of emf, joules per coulomb, are given the name volts (V). In crude language, an emf is a voltage. Let s put this all together. The central relationship describing electromagnetic induction, termed Faraday s Law, claims that: where d dt is the emf induced (the voltage created) in the closed path, is the magnetic flux that passes through the closed path, and the negative sign indicates that the emf s direction in the closed path is to oppose the change in magnetic flux. (If the closed path is a real loop of wire, the emf will drive an induced current whose direction is such that the magnetic field produced by this induced current is opposite to the change in magnetic field that produces the induced current. Crystal clear?) 2

3 . Electromagnetic Induction Analysis Tools Abrupt Change in Flux An N-turn loop of radius R is a distance r from a very long straight wire carrying current i. If the current is reduced to zero in a time, what is the average induced emf in the loop? What is the direction of the induced current? Assume r >>R. i When the current is reduced, the magnetic flux through the small loop will change. y Faraday s Law, this change in flux will create an emf in the loop. The first step toward finding the emf involves finding the magnetic flux through the loop of interest. For this example, the direction of the magnetic field and the direction of the loop s area are parallel, so the dot product reduces to: da ( )( da) Moreover, since the loop is far from the wire, and the loop s diameter is small, the magnetic field from the wire is approximately constant over the area of the loop, and the flux is given by: A 0i 2 ( )( N R ) 2 r 2 N 0iR 2r The N indicates that there are N loops of wire through which the flux passes. 3

4 This flux is reduced to zero over some finite time interval. Interpreting Faraday s Law over a finite time interval results in: d dt t ( f i ) t N 0iR (0 2r 2 N 0iR 2rT This is the average emf induced during the time period under investigation. 2 ) Since initially the magnetic field through the loop was in the +z-direction, and was then removed, the induced emf will drive an induced current counterclockwise around the loop in an attempt to counter the reduction in flux. In other words, the induced current will try to maintain a constant magnetic flux through the loop. Continuous Change in Flux A 1000-turn secondary coil of radius 2.0 cm is concentric with a 400-turn primary coil of radius 20 cm carrying AC current at 60 Hz with peak current 2.0 A. (AC current can be modeled as i(t) = i 0 cos (2 ft).) What is the induced emf in the secondary coil as a function of time? current source i Again, the first step toward finding the emf involves finding the magnetic flux through the secondary loop. For this example, the direction of the magnetic field and the direction of the loop s area are parallel, flux reduces to: ( )( da) 4

5 Moreover, since the magnetic field is approximately constant over the relatively small area of the secondary loop, the flux is given by: Plugging in values yields: A N p 0i 2 ( )( N s r ) 2R 2 N p N s 0i0 r cos(2 ft) 2R ( Tm 2 )cos(2 ft) Applying Faraday s Law: d dt d ( ) cos(2 ft) dt ( )(2 f )sin(2 ft) (1.19 V) sin(2 ft) Thus, the secondary coil of wire has a 60 Hz AC voltage 90 out of phase with the voltage in the primary coil. (Sine and cosine functions are 90 out of phase with each other.) The maximum emf in the secondary coil is 1.19 V. 5

6 . Electromagnetic Induction Activities 6

7 For each of the actions below, indicate the direction of the induced current in the loop. The magnet and loop center are in the same plane. a. N S v b. S N v c. N v = 0 S d. N S v e. v N S v 7

8 For each of the actions below, indicate the direction of the induced current in the circular loop. The magnet and loop center are in the same plane. a. N v = 0 S b. N v S c. S v d. N N S e. N S 8

9 For each of the actions below, indicate the direction of the induced current in the left loop. The loops are parallel and their centers are coplanar. a. close switch b. open switch c. v d. open switch e. v 9

10 For each of the actions below, indicate the direction of the induced current in the left loop. The loops are perpendicular and their centers are coplanar. a. close switch b. open switch c. close switch d. close switch e. v 10

11 For each of the actions below, indicate the direction of the induced current in the secondary loop (the loop without the battery). The loops are either parallel or perpendicular and their centers are coplanar. a. close switch b. c. v close switch d. close switch e. open switch 11

12 The device below consists of a primary coil (containing an adjustable current source) and a secondary coil. For the given primary current (i p ) vs. time graph, construct the corresponding secondary current (i s ) vs. time graph. Current to the right is considered positive. a. i p i current source i s Positive current in the primary creates flux toward the right through the secondary. As this primary current (and flux) increases, a secondary current will be induced to create flux to oppose this change. Since the induced emf is the derivative of the change in flux, the induced current is proportional to the derivative of the primary current. t t b. i p i s t t c. i p i s t t 12

13 The device below consists of a primary coil (containing an adjustable current source) and a secondary coil. For the given secondary current (i s ) vs. time graph, construct the corresponding primary current (i p ) vs. time graph. Current to the right is considered positive. a. i p i current source i s t t b. i p i s t t c. i p i s t t 13

14 The device below consists of a primary loop (containing an adjustable current source) and a secondary loop. For the given graph of either primary current (i p ) or secondary current (i s ) vs. time, construct the corresponding graph. Counterclockwise current is considered positive. a. current source i i p i s t t b. i p i s t t c. i p i s t t 14

15 Six identical, circular loops of wire are rotating at the same angular frequency in the uniform magnetic field shown below. The rotation axes of A, and C are perpendicular to the plane of the page and the rotation axes of D, E and F are parallel to the plane of the page. A C D E F a. At the instant shown, rank these scenarios on the basis of the flux through the loop. Largest Smallest The ranking cannot be determined based on the information provided. b. At the instant shown, rank these scenarios on the basis of emf induced in the loop. Largest Smallest The ranking cannot be determined based on the information provided. Explain the reason for your rankings: 15

16 Six new prototypes for electrical generators are tested in the laboratory. Each generator consists of N circular loops of wire with radius R rotated at frequency f in the same uniform magnetic field. N R (cm) f (Hz) A C D E F Rank these generators on the basis of the maximum emf produced. Largest Smallest The ranking cannot be determined based on the information provided. Explain the reason for your ranking: 16

17 Six identical vertical metal bars start at the position shown and move at constant velocity through identical magnetic fields. The bars make electrical contact with and move along frictionless metal rods attached to identical light bulbs. A 10 cm/s 10 cm/s C D 5 cm/s 20 cm/s E F 10 cm/s 0 cm/s a. At the instant shown, rank these scenarios on the basis of the flux through the loop. Largest Smallest The ranking cannot be determined based on the information provided. b. At the instant shown, rank these scenarios on the basis of the magnitude of the current in the light bulb. Largest Smallest The ranking cannot be determined based on the information provided. Explain the reason for your rankings: 17

18 Six vertical metal bars start at the position shown and move at constant velocity through identical magnetic fields. The bars are currently 5, 10 or 15 cm long and have extendable sliders that make electrical contact with and move along frictionless metal rods attached to identical light bulbs. A 10 cm/s 10 cm/s C D 10 cm/s 20 cm/s E F 20 cm/s 20 cm/s At the instant shown, rank these scenarios on the basis of the magnitude of the current in the light bulb. Largest Smallest The ranking cannot be determined based on the information provided. Explain the reason for your rankings: 18

19 Six identical horizontal metal bars start at rest and fall vertically through identical magnitude magnetic fields (except A). The bars make electrical contact with and move along frictionless metal rods attached to identical light bulbs. A C D E F Rank these scenarios on the time it takes the horizontal bar to reach the bottom of the rods. Largest Smallest The ranking cannot be determined based on the information provided. Explain the reason for your ranking: 19

20 Six identical horizontal metal bars start at rest and fall vertically through identical magnitude magnetic fields (except A). The bars make electrical contact with and move along frictionless metal rods attached to identical batteries. A C D E F Rank these scenarios on the time it takes the horizontal bar to reach the bottom of the rods. Largest Smallest The ranking cannot be determined based on the information provided. Explain the reason for your ranking: 20

21 The 100 turn, 10 cm square loop at right is located in a 10 mt magnetic field. The magnitude of the magnetic field is increased over a period of 0.50 s. The bulb's resistance is 30. Qualitative Analysis On the diagram above, indicate the direction of the induced current in the loop. Mathematical Analysis To what value should the field be increased in order to supply the 10 ma necessary to light the bulb? 21

22 The 5.0 cm square loop at right consists of 100 turns of wire and is located in a 0.10 T magnetic field. The polarity of the magnetic field is reversed. The bulb's resistance is 30. Qualitative Analysis On the diagram above, indicate the direction of the induced current in the loop. Mathematical Analysis Over what time interval must the reversal take place in order to supply the 10 ma necessary to light the bulb? 22

23 A 200-turn loop of radius 1.0 cm is 50 cm from a very long straight wire carrying 250 ma. The current is reduced linearly to zero in a time of 3.0 s. i Qualitative Analysis On the diagram above, indicate the direction of the induced current in the loop. Mathematical Analysis What is the induced emf in the loop? 23

24 The near side of a 1000 turn, 10 cm square loop is 1.0 cm from a very long straight wire carrying 250 ma. The current in the wire is reduced linearly to zero in a time of 0.1 s. i Qualitative Analysis On the diagram above, indicate the direction of the induced current in the loop. Mathematical Analysis What is the induced emf in the loop? 24

25 A 2000-turn secondary coil of radius 2.0 cm is concentric with a 200-turn primary coil of radius 30 cm carrying current 300 ma. The primary current is reduced to zero. i current source Qualitative Analysis On the diagram above, indicate the direction of the induced current in the loop. Mathematical Analysis Over what time interval must the primary current be reduced to zero to induce 6.0 V in the secondary coil? 25

26 A coil of radius 2.0 cm is concentric with a 400-turn/cm solenoid carrying current 1800 ma. The current is linearly reduced to zero over a time interval of 0.05 s. iinduced Qualitative Analysis On the diagram above, indicate the direction of the induced current in the loop. i current source When current flows in the solenoid, a magnetic flux exists that points to the right through the loop. If the current is reduced to zero, this flux disappears and a current will be induced in the loop to create a replacement flux. This current must flow down the front of the loop to create this flux. Mathematical Analysis How many turns are needed in the secondary coil to develop an emf of 12 V? First, I ll determine the flux through the loop when the current flows in the solenoid. Since the field from the solenoid is uniform inside of the loop, I don t need to integrate to find the flux. A ( 0ni)( N R Since this flux drops abruptly to zero, the average emf induced is simply the change in flux divided by the time over which the flux disappears. ( f i ) t 2 (0 ( 0ni)( N R )) t 2 0niN R t ( t) N 2 ni R 0 N (1.26x10 N (12)(0.05) )(40000)(1.8) (0.02) You ll need about 5300 turns in the secondary coil to develop an emf of 12 V. 2 ) 2 26

27 A 200-turn loop of radius 1.0 cm is 50 cm from a very long straight wire carrying AC current at 60 Hz with peak current 250 ma. (AC current can be modeled as i(t) = i 0 cos (2 ft).) What is the induced emf in the loop as a function of time? i Mathematical Analysis 27

28 The near side of a rectangular loop of height a and width 2a is a distance a from a very long straight wire that carries AC current, i(t) = i 0 cos (2 ft). What is the induced emf in the loop as a function of time? i Mathematical Analysis 28

29 A 100-turn loop of radius 1.0 cm is initially 50 cm from a very long straight wire carrying 250 ma. If the loop is moved away from the wire with a initial speed of 10 cm/s, what is the initial induced emf in the loop? i Mathematical Analysis 29

30 A 100-turn loop of radius 1.0 cm is initially 50 cm from a very long straight wire carrying 250 ma. If the loop s position is varied according to the relation r(t) = (5.0 cm) sin (2 t) + 50 cm, what is the induced emf in the loop as a function of time? i Mathematical Analysis 30

31 An n-turn secondary coil of radius r is concentric with a N-turn primary coil of radius R. If the current source supplies AC current, i(t) = i 0 cos (2 ft), what is the induced emf in the secondary coil as a function of time? Assume r << R. i current source Mathematical Analysis 31

32 A 1000-turn coil of radius 2.0 cm is concentric with a 400-turn/cm solenoid carrying AC current at 60 Hz with peak current 2.0 A. (AC current can be modeled as i(t) = i 0 cos (2 ft).) What is the induced emf in the coil as a function of time? i current source Mathematical Analysis 32

33 An N-turn coil of radius of R rotates at frequency f in a magnetic field. Mathematical Analysis a. With the initial magnetic flux (as shown above) defined as negative, sketch a graph of one complete cycle of the magnetic flux through the loop vs. time. t b. With counterclockwise current defined as positive, sketch and label on the axes above a graph of one complete cycle of the induced current in the loop vs. time. c. Determine the emf as a function of time. 33

34 A car alternator consists of a 250-turn coil of radius 5.0 cm. The coil rotates at 1000 rpm in a uniform magnetic field. In order to generate a peak emf of 20 V, what value of magnetic field is necessary? Mathematical Analysis 34

35 A metal bar of length L moves at constant speed v through a magnetic field. The bar makes electrical contact with and moves along frictionless metal rods attached to a light bulb of resistance R. v Qualitative Analysis On the diagram above, indicate the direction of the induced current in the bulb. Mathematical Analysis Determine the current through the bulb, i, as a function of L, v,, and R. 35

36 A metal bar of length L moves at constant speed v through a magnetic field. v Qualitative Analysis On the diagram above, indicate which end of the bar is positive. Mathematical Analysis Determine the emf induced between the two ends of the bar,, as a function of L, v, and. 36

37 A horizontal metal bar of length L and mass m starts at rest and falls vertically through a magnetic field. The bar makes electrical contact with and moves along frictionless metal rods attached to a light bulb of resistance R. Qualitative Analysis On the diagram above, indicate the direction of the induced current in the bulb. Mathematical Analysis Determine the terminal velocity of the horizontal bar (v terminal ) as a function of L, m,, R, and g. 37

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