W10D1: Inductance and Magnetic Field Energy

1. W10D1:Inductance and Magnetic Field Energy Today’s Reading Assignment W10D1 Inductance & Magnetic Energy Course Notes: Sections 11.1-3

2. Math Review Week 10 Tuesday from 9-11 pm in 32-082 PS 7 due Week 10 Tuesday at 9 pm in boxes outside 32-082 or 26-152 Next Reading Assignment W10D2 DC Circuits & Kirchhoff’s Loop Rules Course Notes: Sections 7.1-7.5 Exam 3 Thursday April 18 7:30 pm –9:30 pm Announcements

3. Outline Faraday Law Problem Solving Faraday Law Demonstrations Mutual Inductance Self Inductance Energy in Inductors Transformers

4. Faraday’s Law of Induction If C is a stationary closed curve and S is a surface spanning C then The changing magnetic flux through S induces a non-electrostatic electric field whose line integral around C is non-zero

5. Problem: Calculating Induced Electric Field Consider a uniform magnetic field which points into the page and is confined to a circular region with radius R. Suppose the magnitude increases with time, i.e. dB/dt > 0. Find the magnitude and direction of the induced electric field in the regions (i) r < R, and (ii) r > R. (iii) Plot the magnitude of the electric field as a function r.

6. Faraday’s LawDemonstrations

7. Demonstration: Electric Guitar H32 Pickups http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H%2032&show=0

8. Electric Guitar

9. Demonstration:32-082 Aluminum Plate between Pole Faces of a Magnet H 1426-152 Copper Pendulum Between Poles of a Magnet H13 http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 14&show=0 http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 13&show=0

10. What happened to kinetic energy of pendulum? Eddy Current Braking

11. Eddy Current Braking The magnet induces currents in the metal that dissipate the energy through Joule heating: • Current is induced counter-clockwise (out from center) • Force is opposing motion (creates slowing torque)

12. Eddy Current Braking The magnet induces currents in the metal that dissipate the energy through Joule heating: • Current is induced clockwise (out from center) • Force is opposing motion (creates slowing torque) • EMF proportional to angular frequency

13. Demonstration:26-152 Levitating Magnet H2832-082 Levitating Coil on an Aluminum Plate H15 http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 28&show=0 http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 15&show=0

14. Mutual Inductance

15. Demonstration:Two Small Coils and Radio H31 http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 31&show=0

16. Mutual Inductance Current I2 in coil 2, induces magnetic flux F12 in coil 1. “Mutual inductance” M12: Change current in coil 2? Induce EMF in coil 1:

17. Group Problem: Mutual Inductance An infinite straight wire carrying current I is placed to the left of a rectangular loop of wire with width w and length l. What is the mutual inductance of the system?

18. Self Inductance

19. Self Inductance What if is the effect of putting current into coil 1? There is “self flux”: Faraday’s Law

20. Calculating Self Inductance Unit: Henry • Assume a current I is flowing in your device • Calculate the B field due to that I • Calculate the flux due to that B field • Calculate the self inductance (divide out I)

21. Worked Example: Solenoid Calculate the self-inductance L of a solenoid (n turns per meter, length , radius R)

22. Solenoid Inductance

23. Concept Question: Solenoid A very long solenoid consisting of N turns has radius R and length d, (d>>R).  Suppose the number of turns is halved keeping all the other parameters fixed. The self inductance • remains the same. • doubles. • is halved. • is four times as large. • is four times as small. • None of the above.

24. Concept Q. Ans.: Solenoid Solution 5. The self-induction of the solenoid is equal to the total flux through the object which is the product of the number of turns time the flux through each turn. The flux through each turn is proportional to the magnitude of magnetic field which is proportional to the number of turns per unit length or hence proportional to the number of turns. Hence the self-induction of the solenoid is proportional to the square of the number of turns. If the number of turns is halved keeping all the other parameters fixed then he self inductance is four times as small.

25. Group Problem: Toroid Calculate the self-inductance L of a toroid with a square cross section with inner radius a, outer radius b = a+h, (height h) and N square windings . • REMEMBER • Assume a current I is flowing in your device • Calculate the B field due to that I • Calculate the flux due to that B field • Calculate the self inductance (divide out I)

26. Energy in Inductors

27. Inductor Behavior L I Inductor with constant current does nothing

28. Back EMF I I

29. Demos: 26-152 Back “emf” in Large Inductor H17 32-082 Marconi Coil H12 http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 17&show=0 http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 12&show=0

30. Marconi Coil: On the Titanic Another ship Same era Titanic Marconi Telegraph

31. Marconi Coil: Titanic Replica

32. The Point: Big EMF Big L Big dI Small dt Huge EMF

33. Energy To “Charge” Inductor • 1. Start with “uncharged” inductor • Gradually increase current. Must do work: • 3. Integrate up to find total work done:

34. Energy Stored in Inductor But where is energy stored?

35. Energy Density Volume Example: Solenoid Ideal solenoid, length l, radius R, n turns/length, current I:

36. Energy Density Energy is stored in the magnetic field Magnetic Energy Density Energy is stored in the electric field Electric Energy Density

37. Worked Example: Energy Stored in Toroid Consider a toroid with a square cross section with inner radius a, outer radius b = a+h, (height h) and N square windings with current I. Calculate the energy stored in the magnetic field of the torus.

38. Solution: Energy Stored in Toroid The magnetic field in the torus is given by The stored energy is then The self-inductance is

39. I I X Group Problem: Coaxial Cable • How much energy is stored per unit length? • What is inductance per unit length? HINTS: This does require an integral The EASIEST way to do (2) is to use (1) Inner wire: r = a Outer wire: r = b

40. Transformer Step-up transformer Flux F through each turn same: Ns > Np: step-up transformer Ns < Np: step-down transformer

41. Demonstrations:26-152 One Turn Secondary: Nail H10 26-152 Many Turn Secondary: Jacob’s Ladder H1132-082 Variable Turns Around a Primary Coil H9 http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 10&show=0 http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 11&show=0 http://tsgphysics.mit.edu/front/?page=demo.php&letnum=H 9&show=0

42. House=Left, Line=Right Line=Left, House=Right I don’t know Concept Question: Residential Transformer If the transformer in the can looks like the picture, how is it connected?

43. Answer: Residential Transformer Answer: 1. House on left, line on right The house needs a lower voltage, so we step down to the house (fewer turns on house side)

44. Transmission of Electric Power Power loss can be greatly reduced if transmitted at high voltage

45. Electrical Power Power is change in energy per unit time So power to move current through circuit elements:

46. Power - Resistor Moving across a resistor in the direction of current decreases your potential. Resistors alwaysdissipate power

47. Example: Transmission lines An average of 120 kW of electric power is sent from a power plant. The transmission lines have a total resistance of 0.40 W. Calculate the power loss if the power is sent at (a) 240 V, and (b) 24,000 V. (a) 83% loss!! (b) 0.0083% loss

48. Transmission lines We just calculated that I2R is smaller for bigger voltages. What about V2/R? Isn’t that bigger? Why doesn’t that matter?