# Physics 1302W.400 Lecture 33 Introductory Physics for Scientists and Engineering II

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1 Physics 1302W.400 Lecture 33 Introductory Physics for Scientists and Engineering II In today s lecture, we will discuss generators and motors. Slide 30-1

2 Announcement Quiz 4 will be next week. The Final Exam will be Thursday, May 10, 6:30-930pm. Requests for Spring make-up finals must be completed no later than 4pm on Wednesday, April 25, 2018, using an online form (see link on our course site). Common Make up exams (Phys 1101W, 1201W, 1301W, 1302W) will be Friday, May 11 from 8:00am-11:00am. Other make up final times will be determined by the course faculty member. All requests for make up final exams will be addressed after the April 25th deadline. Slide 30-2

3 Induced emf Recall the definition of magnetic flux through a loop: Φ B B d A, If we apply this definition to the situation shown in the figure, and if we let the time interval Δt approach zero, we get = dφ B dt, This is a quantitative statement of Faraday s law. The negative sign ensures that the direction of the induced emf is in accordance with Lenz s law. Slide 30-3

4 Clicker Question A long, straight wire carries a steady current I. A rectangular conducting loop lies in the same plane as the wire, with two sides parallel to the wire and two sides perpendicular. Suppose the loop is pushed toward the wire as shown. Given the direction of I, the induced current in the loop is 1. clockwise. 2. counterclockwise. 3. We need more information. Slide 30-4

5 Generator In an electric generator, a solenoid that contains N windings (each of area A) is rotated at constant rotational speed ω in a uniform magnetic field of magnitude B. What is the emf induced in the solenoid? Slide 30-5

6 Generator As the solenoid rotates, the magnetic flux through it changes, and this changing flux causes an emf in the solenoid. The magnetic flux through the solenoid is obtained by multiplying the magnetic flux through a single winding, Φ B B A, by the number of windings N. Let s assume that the plane of the winding is perpendicular to the direction of the magnetic field at time t = 0. Slide 30-6

7 Generator As the solenoid rotates, the scalar product B A changes with time. At instant t, the angle between A and B is ωt, and so the magnetic flux through a single winding is: Φ B B A = BA cos ωt. Through the N windings of the solenoid, the magnetic flux is Φ B NBA cos ωt. So the magnitude of the induced emf oscillates sinusoidally: ind = d dt (NBA cos ωt) = ωnba sin ωt. Note: The emf has a different sign for half of each rotation! Slide 30-7

8 Generator and Capacitor Coupled Motor/Generator Description: A Genecon hand-held generator is connected to a 1 F capacitor. By turning the handle, the capacitor is charged. When the handle is released, it continues to rotate as the capacitor discharges. Description: The outputs of two Genecon generators (hand-held generator) are coupled. When the handle of one generator is turned by hand, the handle of the second generator will also rotate. Slide 30-8

9 Generators Slide 30-9

10 Motors Slide 30-10

11 Motors A generator run backwards is a motor! Apply an alternating current (ac current) to the coil in the magnetic field and it will experience a torque τ = µ x B The magnetic moment for a solenoid is simply: µ = N I A (A is the vector area of a single loop) A direct current (dc current) would produce an oscillating coil, but an ac current (reversing direction right when the coil is vertically aligned in the B field) will flip µ so that the torque remains in the same direction, yielding a net rotation. Slide 30-11

12 Motors When the coil rotates, there is a back emf (Lenz s law) that acts like a drag on the coil When the motor is first turned on, there is no back emf as coil is not yet in motion, and the initial rotation is faster than in normal operation Slide 30-12

13 Generators and Motors Φ m = N B A = N B A cos θ θ = ω t Φ m = N B A cos ω t = - dφ m /dt = - N B A d (cos ω t)/dt = + N B A ω sin ω t = max sin ω t max = N B A ω Slide 30-13

14 Electric field accompanying a changing magnetic field Consider a conducting circular loop in a cylindrical uniform magnetic field. Suppose the magnetic field is directed out of the page and is steadily increasing. What is the direction of the induced current in the loop? Clockwise (Lenz s law)! The electric field responsible for this current must form loops. These electric field lines must be oriented clockwise (Lenz s law). Slide 30-14

15 Electric field accompanying a changing magnetic field The work done on a charged particle as it travels around the conducting loop is W q (closed path) = q E d The work done by the electric field per unit charge is thus W q q = E d The work per unit charge is the induced emf, so we get E d = dφ B dt This is an alternative statement of Faraday s law. Slide 30-15

16 Electric field magnitude If the uniform cylindrical magnetic field has a radius R = 0.20 m and increases at a steady rate of T/s. What is the magnitude of the electric field at a radial distance r = 0.10 m from the center? Slide 30-16

17 Electric field magnitude Because of the circular symmetry, the magnitude E of the electric field cannot vary along any given electric field line, i.e., it can not vary along any circular path centered on the axis of the cylindrical magnetic field. E E d = E d Because is always parallel to d, we have and E d = 2πrE. Slide 30-17

18 Electric field magnitude Now we need to calculate the magnetic flux through the circular clockwise integration path. The magnetic flux is negative (curling the fingers of your right hand in the direction of the integration path makes your thumb point in the direction opposite the direction of ). So we get B Φ B = BA = πr 2 B. and dφ B dt = +πr 2 db dt. Slide 30-18

19 Electric field magnitude Therefore, Faraday s law gives: and 2πrE = πr 2 db dt E = 1 r db (0.10 m)(0.050 T/s) = 2 dt 2 = T m/s = N/C. The electric field has cylindrical symmetry. Its magnitude is proportional to r and to the time derivative of B. Slide 30-19

20 Vertical Transformer Light Under Water Description: Power is turned on to a large coil (primary). A small coil (secondary), connected to a light bulb, is placed around the iron core. The bulb lights. Description: A small coil and light socket are encased in wax. The coil and bulb are placed in a beaker of water. This beaker is then placed on the iron core of a powered coil. The bulb lights. A transformer is a static electrical device that transfers electrical energy between two or more circuits through electromagnetic induction. A varying current in one coil of the transformer produces a varying magnetic field, which in turn induces a varying electromotive force (emf) or "voltage" in a second coil. Power can be transferred between the two coils through the magnetic field, without a metallic connection between the two circuits. Faraday's law of induction discovered in 1831 described this effect. Transformers are used to increase or decrease the alternating voltages in electric power applications. (Wikipedia) Slide 30-20

21 Physics 1302W.400 Lecture 34 Introductory Physics for Scientists and Engineering II In today s lecture, we will discuss magnetic induction. Slide 30-21

22 Induced emf Recall that magnetic flux through a loop is given by: Φ B B d A, We found that induced emf in a loop is related to the magnetic flux through the loop via = dφ B dt This is a quantitative statement of Faraday s law. The negative sign ensures that the direction of the induced emf is in accordance with Lenz s law. A more general statement of Faraday s law is: E d = dφ B dt Slide 30-22

23 Inductance An important consequence of Faraday s law is that when a current through a conducting loop changes, the change induces an emf in the loop itself. For example, disconnecting a battery connected to a solenoid causes the current and the magnetic flux to decrease. Faraday s law tells us that an emf is induced in the solenoid that opposes the change in flux. Consequently, a current is induced. This current has the same direction as the original current. Slide 30-23

24 Inductance The induced emf is proportional to the rate of change in the magnetic flux, and the magnetic field is proportional to I: = dφ B db ind dt dt di dt The self-inductance L of the loop or solenoid is defined as the constant of proportionality: ind An equivalent definition is: = L di dt Φ B = LI SI units of L: 1 Henry = 1 H 1 V s/a = 1 Wb/A = 1 kg m 2 /C 2 A device that has an appreciable inductance is called an inductor. Slide 30-24

25 Inductance of a Solenoid For a very long solenoid, the magnetic field due to a current I is uniform inside the solenoid (B = µ 0 ni) and zero outside. Recall that n is the number of windings per unit length n = N/l. The magnetic flux is: Φ = NBA = µ N 2 IA 0 Since Φ B = LI L = µ 0 N 2 A, we have: Similar to capacitance, self-inductance only depends on geometric factors! Slide 30-25

26 Problem A solenoid has 2760 windings of radius 50 mm and is 0.60 m long. If the current through the solenoid is increasing at a rate of 0.10 A/s, what is the magnitude of the induced emf? V V V V V Slide 30-26

27 Clicker Question A coil with self-inductance L carries a current I given by I = I 0 sin 2πft. Which graph best describes the self-induced emf as a function of time? Slide 30-27

28 Calculating Inductances The self-inductance of a current-carrying device or current loop is a measure of the emf induced in the device or loop when current is changed. To determine this self-inductance, follow these four steps: 1. Derive an expression for the magnitude of the magnetic field in the current-carrying device or current loop as a function of the current. Your expression should depend only on the current I and possibly but not necessarily the position within the device or current loop. 2. Calculate the magnetic flux Φ B through the device or current loop. If the expression you derived in step 1 depends on position, you will have to integrate that expression over the volume of the device or circuit. Use symmetry to simplify the integral and divide the device into segments on which B is constant. Slide 30-28

29 Calculating Inductances Φ B 3. Substitute the resulting expression you obtained for into = dφ B = L di ind dt dt As you take the derivative with respect to time, keep in mind that only the current varies with respect to time, so you should end up with an expression that contains the derivative di/dt on both sides of the equal sign. 4. Solve your expression for L after eliminating di/dt. Slide 30-29

30 Mutual Inductance of two circuits Application in, e.g., transformers (recall yesterday s demos) Slide 30-30

31 Vertical Transformer Light Under Water Description: Power is turned on to a large coil (primary). A small coil (secondary), connected to a light bulb, is placed around the iron core. The bulb lights. Description: A small coil and light socket are encased in wax. The coil and bulb are placed in a beaker of water. This beaker is then placed on the iron core of a powered coil. The bulb lights. A transformer is a static electrical device that transfers electrical energy between two or more circuits through electromagnetic induction. A varying current in one coil of the transformer produces a varying magnetic field, which in turn induces a varying electromotive force (emf) or "voltage" in a second coil. Power can be transferred between the two coils through the magnetic field, without a metallic connection between the two circuits. Faraday's law of induction discovered in 1831 described this effect. Transformers are used to increase or decrease the alternating voltages in electric power applications. (Wikipedia) Slide 30-31

32 Induction Coils with Battery Induction Coils with Radio Description: A coil of wire is connected to a car battery through a tap switch. A second coil is connected to a galvanometer. When the switch is tapped, the galvanometer deflects. Description: A coil is connected to the earphone output of a radio. A second coil is connected to a audio amplifier/speaker. When the radio is turned on, the speaker will broadcast. Try to rotate one coil by 90 degrees. The signal will disappear. Slide 30-32

33 Mutual Inductance Φ B = L I B 1 ~ I 1 B 2 ~ I 2 Φ B1,2 = M 12 I 1 Φ B2,1 = M 21 I 2 M 12 = M 21 = M = Mutual Inductance Example: two co-axial solenoids Slide 30-33

34 Mutual Inductance of Two Coaxial Solenoids For r < r 1 : B 1 = µ o n 1 I 1 For r > r 1 : B 1 ~ 0 The magnetic flux through a single winding of the outer solenoid is just the flux inside the inner solenoid Φ B1,2 = N 2 B 1 π r 1 2 = n 2 l B 1 π r 1 2 = n 2 l µ o n 1 I 1 π r 1 2 So, we have: M 12 = Φ B1,2 / I 1 = µ o n 1 n 2 l π r 1 2 Slide 30-34

35 Clicker Question What properties of the conducting loop or device does inductance not depend on? 1. Size 2. Shape 3. Number of windings 4. Current Slide 30-35

36 Physics 1302W.400 Lecture 35 Introductory Physics for Scientists and Engineering II In today s lecture, we will discuss magnetic energy. Slide 30-36

37 Announcements - Quiz #4 will be next week. It will cover: Chapters 28 and 29 and the related discussion problems - Homework #12 is due at 5 pm on Wednesday, April 18 It covers Chapters 28, 29 and the beginning of Chapter 30 Slide 30-37

38 Magnetic energy Work must be done on an inductor to establish the current through it because the change in current causes an induced emf that opposes this change. This work increases the magnetic potential energy stored in the magnetic field of the inductor. The work dw done on the inductor when a charge dq moves through it is dw = ind dq Using the definition of current, we can write Slide 30-38

39 Magnetic energy So work done on the inductor to create a current I is W = dw = L I di = 1 LI 2 2 One typically chooses the zero of magnetic potential energy to be zero when there is no current through the inductor. Therefore, the magnetic potential energy stored in an inductor when there is a current through it is equal to the work done in increasing the current from zero to I: U B = 1 2 LI 2 Slide 30-39

40 Magnetic energy Notice the analogy between the energy stored in an inductor, U B = 1 LI 2 and the energy stored in a capacitor, 2 U E = 1 CV 2 2 In the case of the capacitor, we can think of the potential energy as stored in the electric field, with the energy density (energy per unit volume): u E = ½ ε 0 E 2 It is possible to show that, quite generally, the potential energy density of the magnetic fields is: u B 1 2 The total magnetic potential energy is obtained by integrating over the relevant volume. B 2 µ 0. Slide 30-40

41 Brief Review The self-inductance L of the loop or solenoid: ind = L di dt or Φ B = LI Work must be done on an inductor to establish the current through it because the change in current causes an induced emf that opposes this change. This work increases the magnetic potential energy stored in the magnetic field of the inductor Potential energy stored in an inductor and capacitor: U B = 1 LI 2 2 and U E = 1 CV 2 2 Energy density of electric and magnetic fields: u E = ½ ε 0 E 2 and u B = B 2 /(2µ 0 ) Slide 30-41

42 Magnetic energy stored in a square toroid Consider a toroid with 200 square windings of width w = 30 mm and inner radius is R = 60. The current is 1.5 ma. The current causes a magnetic field inside the toroid. This magnetic field stores magnetic potential energy. What is the magnetic potential energy? Slide 30-42

43 Magnetic Field inside a Toroid Recall: a solenoid bent into a circle is a toroid. Note the cylindrical symmetry. Apply Ampère s law to a circular path of radius r that coincides with the magnetic field lines. Because the field is tangential to the integration path, we get B d = B d = B d = B(2πr) If there are N windings, then I enc = NI Ampère s law give us NI B = µ 0 2πr (toroid) Slide 30-43

44 Magnetic energy stored in a square toroid We previously found that the magnetic field inside a toroid is given by B = µ 0 NI 2πr, where r is the radial distance from the center of the ring formed by the toroid. B 2 The magnetic potential energy density is: u B 1. 2 µ 0 U B = u B dv = 1 2µ B 2 dv = µ N 2 I π dv. 2 r 2 The volume element is: dv = 2πrwdr to obtain the self- U B = µ N 2 I 2 R+w 0 2π wr dr 8π = µ N 2 I 2 w 0 2 R r 2 4π = µ N 2 I 2 w! 0 ln 1+ w \$ # &. 4π " R% Note: We can now use the formula inductance L of the toroid! R+w dr. R r U B = 1 2 LI 2 Slide 30-44

45 Magnetic energy stored in a square toroid Substituting the values for µ 0, N, I, R, and w, we obtain for the magnetic potential energy stored in the toroid U B = (10 7 T m/a)(200) 2 ( A) 2 30 mm (30 mm)ln mm = T m 2 A = J This is a very small amount of energy; a significant increase in I and N would lead to a much larger magnetic potential energy stored in the system. (1 T m 2 A = 1 [N/(A m)] m 2 A = 1 N m = 1 J) Slide 30-45

46 Dipole Radiation Radio Waves from a Spark Description: A 230 MHz oscillator is set on a board in order to get it away from metal. The receiving antenna consists of adjustable rods and a light bulb. The bulb lights when the receiving antenna is parallel to the transmitter. Description: A small radio is set to an unused AM frequency. When sparks are produced by shorting a battery, clicks can be heard from the radio. Slide 30-46

47 Light Fluorescent Bulb Description: The field from a high voltage Tesla coil lights a fluorescent bulb 4 feet away. (A Tesla coil is an electrical resonant transformer circuit designed by inventor Nikola Tesla in It is used to produce highvoltage, low-current, high frequency alternating-current electricity. Wikipedia) Slide 30-47

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