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Fields and Waves I Lecture 18 K. A. Connor Y. Maréchal

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1 Fields and Waves I Lecture 18 K. A. Connor Y. Maréchal
Magnetic Force and Energy K. A. Connor Electrical, Computer, and Systems Engineering Department Rensselaer Polytechnic Institute, Troy, NY Y. Maréchal Power Engineering Department Institut National Polytechnique de Grenoble, France Welcome to Fields and Waves I Before I start, can those of you with pagers and cell phones please turn them off? Thanks.

2 J. Darryl Michael – GE Global Research Center, Niskayuna, NY
These Slides Were Prepared by Prof. Kenneth A. Connor Using Original Materials Written Mostly by the Following: Kenneth A. Connor – ECSE Department, Rensselaer Polytechnic Institute, Troy, NY J. Darryl Michael – GE Global Research Center, Niskayuna, NY Thomas P. Crowley – National Institute of Standards and Technology, Boulder, CO Sheppard J. Salon – ECSE Department, Rensselaer Polytechnic Institute, Troy, NY Lale Ergene – ITU Informatics Institute, Istanbul, Turkey Jeffrey Braunstein – Chung-Ang University, Seoul, Korea Materials from other sources are referenced where they are used. Those listed as Ulaby are figures from Ulaby’s textbook. 20 September 2018 Fields and Waves I

3 Overview Review Energy and Force Applications Magnetic Materials
Magnetic Circuits Energy and Force Energy Force Applications DC motors Induction heating 20 September 2018 Fields and Waves I

4 Magnetic Force and Energy
Magnetic circuits and Reluctances

5 Boundary Conditions Arguing from analogy with Electric Fields
20 September 2018 Fields and Waves I

6 Introduction MAGNETIC CIRCUITS used to analyze relays, switches, speakers... In a simple experiment: R Flux stays in TOROID 20 September 2018 Fields and Waves I

7 Magnetic Flux - TOROID Flux is a constant =
Flux stays in toroid - so area is a constant B and H are constant along the path L is much higher that with air 20 September 2018 Fields and Waves I

8 Magnetic Flux - TOROID with GAP
Introduce an air gap to toroid: g Apply boundary conditions across gap: Can get very large H in gap Gap has very strong effect on H and on energy consumption 20 September 2018 Fields and Waves I

9 Magnetic Circuits - Analogy to E-circuits
Reluctance of iron Reluctance of gap Magneto -motive force enables us to draw analogy to electric circuits 20 September 2018 Fields and Waves I

10 Magnetic Circuits & Electric Circuits
V or e.m.f NI or m.m.f Magneto motive force I Flux High m - low reluctance path Magnetic circuits model: 20 September 2018 Fields and Waves I

11 Example 4 a. Evaluate  H  dl around the dashed line in the figure on the left below. Then, determine |H| and |B| in the iron core. Make reasonable approximations. b. What is the inductance, L? c. For the figure on the left, what are the reluctance and magnetomotive force? Draw a magnetic circuit equivalent and show how to solve for the inductance using the circuit. d. Analyze the situation on the right using magnetic circuits. Determine the flux through the iron core. What is the inductance? What is H in the core and in the gap? e. Calculate numerical values for L, |H|gap and |H|core when N = 1000, I = 1 A, w = 5 cm, g = 1 cm, l = 20 cm, and 20 September 2018 Fields and Waves I

12 Example 4 20 September 2018 Fields and Waves I

13 Example 4 – Continued Lwa=19H 20 September 2018 Fields and Waves I

14 Flux distribution in a relay
High reluctance, low flux density Low reluctance, High flux density 20 September 2018 Fields and Waves I

15 Magnetic Force and Energy

16 Energy Power in inductor:
energy in Inductor Can we obtain energy in terms of B and H fields ? Flux linkage: Also, 20 September 2018 Fields and Waves I

17 Magnetic Energy Energy = Energy Density: (per unit volume) VOLUME
Energy stored in Magnetic field Energy Density: (per unit volume) 20 September 2018 Fields and Waves I

18 Example 1 – Coaxial Cable
One of the three standard configurations Detailed Solution for Coax: Ampere’s Law Contours Ampere’s Law Left Hand Side: a Right Hand Side: 20 September 2018 Fields and Waves I

19 Example 1 – Coaxial Cable
Assume the outer conductor is very thin 20 September 2018 Fields and Waves I

20 Example 1 – Coaxial Cable
The energy in the magnetic field can be divided into two terms: r 20 September 2018 Fields and Waves I

21 Example 1 – Coaxial Cable
20 September 2018 Fields and Waves I

22 Example 1 – Coaxial Cable
Ulaby To compute the inductance per unit length, we need to determine the magnetic flux through the area S between the conductors 20 September 2018 Fields and Waves I

23 Example 1 – Coaxial Cable
The flux through the surface S: Note that the flux is linked only once since there is only one turn. Thus, the inductance is given by: or 20 September 2018 Fields and Waves I

24 Problem Using the flux: External inductance Using the energy: ???
Total inductance Additional term 20 September 2018 Fields and Waves I

25 Example 1 – Coaxial Cable
Note that this analysis does not incorporate the flux inside the center conductor so it does not give us the total inductance. However, figuring out the flux linking this current is difficult. Thus we leave this to our method based on energy. Ulaby External Inductance: What we have determined is called the external inductance, since it is inductance due to the magnetic field external to the current-carrying wires. Internal Inductance: What we have neglected is the contribution to the inductance from the field inside the wires. 20 September 2018 Fields and Waves I

26 Magnetic Force and Energy

27 Protection relay High sensitivity relay 20 September 2018
Fields and Waves I

28 Protection relay 20 September 2018 Fields and Waves I

29 Protection relay 20 September 2018 Fields and Waves I

30 Force : first approach from energy
First approach - F does work and changes energy with the energy stored being : 20 September 2018 Fields and Waves I

31 Example – Simple Relay Consider a simple electromagnetic relay consisting of a solenoid and a moveable arm. In the region of the gap, the normal component of the magnetic field will be continuous. 20 September 2018 Fields and Waves I

32 Alternative Calculation – Magnetic Pressure
The magnetic field intensity H is very different in the gap and the core, since B is the same. The magnetic energy density is also very different. 20 September 2018 Fields and Waves I

33 Alternative Calculation – Magnetic Pressure
The difference in the magnetic field energy density on the two sides produces a pressure difference. Energy Density Pressure Force High Pressure Low Pressure Net Force 20 September 2018 Fields and Waves I

34 Alternative Calculation – Magnetic Pressure
Since the pressure is so much higher on the gap side than in the core we only need to evaluate the pressure on the gap side to figure out the force. S is the area of the gap and core. To figure out the force, we first need to find the magnetic field, which we can do using the magnetic circuits technique. 20 September 2018 Fields and Waves I

35 Alternative Calculation – Magnetic Pressure
To analyze this configuration, we will use the idealized version at the right. Assume that each leg has a length lo and the area of each leg is S. The gap length is The reluctances are 20 September 2018 Fields and Waves I

36 Alternative Calculation – Magnetic Pressure
20 September 2018 Fields and Waves I

37 Forces on current wires : why Beakman’s Motor turns ?
A simple DC motor with brushes made with a battery, two paperclips, a rubber band and about 1 meter of enameled wire. 20 September 2018 Fields and Waves I

38 Force on currents First approach - similar to that for individual particles For one particle: For many particles: For a wire in a magnetic field. 20 September 2018 Fields and Waves I

39 FORCE The force on a current loop in a magnetic field can result in rotational torque if the loop has a fixed axis as shown. This is the basic configuration for the Beakman’s motor. Figure reference placeholder 20 September 2018 Fields and Waves I

40 DC motor : the commutator
20 September 2018 Fields and Waves I

41 Example – Rail Gun If a sliding contact is placed across a two wire transmission line carrying a large current, a very large force can result on the contact. Assume that all the wires (including the slider) have a radius = a and that the transmission line wires are separated by a distance d. This material is discussed more extensively in Unit 9 of the class notes. 20 September 2018 Fields and Waves I

42 Example – Rail Gun The external inductance of a two wire line of length l is given by (one of many forms: where we have used the fact that typically d >> a. The force on the sliding conductor will be: If d/a = 5 and I = 105 A, F = 100 Newtons 20 September 2018 Fields and Waves I

43 Magnetic Force and Energy
Motors and other applications

44 DC Motors The stator is the stationary outside part of a motor. The rotor is the inner part which rotates. In the motor animations, red represents a magnet or winding with a north polarization, while green represents a magnet or winding with a south polarization. Opposite, red and green, polarities attract. 20 September 2018 Fields and Waves I

45 DC Motors Just as the rotor reaches alignment, the brushes move across the commutator contacts and energize the next winding. In the animation the commutator contacts are brown and the brushes are dark grey. A yellow spark shows when the brushes switch to the next winding. 20 September 2018 Fields and Waves I

46 DC Motor Applications Automobiles Elsewhere Cordless hand drill
Windshield Wipers Door locks Window lifts Antenna retractor Seat adjust Mirror adjust Anti-lock Braking System Elsewhere Cordless hand drill Electric lawnmower Fans Toys Electric toothbrush Servo Motor 20 September 2018 Fields and Waves I

47 DC Motor 20 September 2018 Fields and Waves I

48 Brushless DC Motors A brushless DC motor has a rotor with permanent magnets and a stator with windings. It is essentially a dc motor turned inside out. The control electronics replace the function of the commutator and energize the proper winding. 20 September 2018 Fields and Waves I

49 Full Stepper Motor This animation demonstrates the principle for a stepper motor using full step commutation. The rotor of a permanent magnet stepper motor consists of permanent magnets and the stator has two pairs of windings. Just as the rotor aligns with one of the stator poles, the second phase is energized. The two phases alternate on and off and also reverse polarity. There are four steps. One phase lags the other phase by one step. This is equivalent to one forth of an electrical cycle or 90°. 20 September 2018 Fields and Waves I

50 Half Stepper Motor Full stepper Half stepper This animation shows the stepping pattern for a half-step stepper motor. The commutation sequence for a half-step stepper motor has eight steps instead of four. The main difference is that the second phase is turned on before the first phase is turned off. Thus, sometimes both phases are energized at the same time. A half-step motor has twice the resolution of a full step motor. It is very popular for this reason. 20 September 2018 Fields and Waves I

51 Real DC brushless Motors
This stepper motor is very simplified. The rotor of a real stepper motor usually has many poles (hundred of them). The stator poles of a real stepper motor also has many teeth. Demo 20 September 2018 Fields and Waves I

52 Some Interesting Inductors
Induction Heating 20 September 2018 Fields and Waves I

53 Some Interesting Inductors
Induction Heating in Aerospace 20 September 2018 Fields and Waves I

54 Numerical simulation of Induction heating
Induction heating of a gear box gearing 20 September 2018 Fields and Waves I


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