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Ch. 28

0. Ch. 28. Electromagnetic Induction. 0. Chapter Overview. Motional EMF Faraday’s Law Lenz’s Law Magnetic Flux Electric Generator Transformers. 0. Motional EMF. If a conductor moves perpendicular to a magnetic field, a potential difference is induced across the conductor

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Ch. 28

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  1. 0 Ch. 28 Electromagnetic Induction

  2. 0 Chapter Overview • Motional EMF • Faraday’s Law • Lenz’s Law • Magnetic Flux • Electric Generator • Transformers

  3. 0 Motional EMF • If a conductor moves perpendicular to a magnetic field, a potential difference is induced across the conductor • Moving the conductor in the B-field produces magnetic force • Charge separation from magnetic force produces electric force • At equilibrium electric force balances magnetic force

  4. 0 Motional EMF • qvB = qE • Multiply both sides by L • vBL = qEL • ΔV = vBL

  5. 0 A bar of length 10 cm closes a circuit as shown. The bar moves at 2.0 m/s perpendicular to a B-field of strength .25 T. a) Find the current in the circuit if the light bulb has a resistance of 5.0 Ω. b) Ignoring the resistance of the connecting wires find the potential difference across the light bulb. Label the higher and lower potential side of the light bulb.

  6. 0 Faraday’s Law • In 1820 Oersted demonstrated that a current could create a magnetic field – i.e. that an E-field could produce a B-field • Almost immediately Faraday asked the opposite question – Could a B-field produce an E-field?

  7. 0 Faraday’s Law • Faraday worked on the question for 11 years until he accidentally found his answer while disconnecting an unsuccessful experiment • Faraday had been searching for a current produced by a steady field, but it was a changing field that produced the current and he produced the changing magnetic field as he disconnected his experiment

  8. 0 Magnetic Flux • Magnetic flux through a single surface • ΦB = BA cosφ • If a coil has N turns • ΦB = NBA cosφ

  9. 0 Faraday’s Law • An EMF (potential difference) is induced when the magnetic flux through a surface changes in time

  10. 0 Which of the following can produce a changing magnetic flux? • A changing B-field • A changing area • A changing angle • 1,2, and 3 • None of the above

  11. 0 Lenz’s Law • What pole is induced near coil when a N pole of a bar magnet is inserted into a coil? (OBS) • What pole is induced if the N pole is removed form a coil? (OBS) • Reverse the magnet. What poles are observed now? (OBS)

  12. 0 Lenz’s Law • The induced EMF is produced so as to oppose the change in magnetic flux producing it.

  13. 001 002 003 004 005 006 007 008 009 010 011 012 013 014 015 016 017 018 019 020 021 022 023 024 0 A conducting coil is placed over an AC electromagnet. When the magnet is turned on what will the coil do? • Nothing • It will attract to the magnet • It will repel from the magnet • Cannot be determined

  14. 0 Lenz’s Law Tube • Two identically sized disks are dropped. One is dropped inside a tube in which it fits. Which will hit the ground first?

  15. 0 A coil is oriented perpendicular to a B-field directed into the page. The coil is at rest. What is the direction of the induced current? • Clockwise • Counterclockwise • There is no induced current • Cannot be determined

  16. 0 For the situation depicted, what will be the direction of the induced current? • Clockwise • Counterclockwise • There is no induced current • Cannot be determined

  17. 0 A wire is formed into a loop of radius .050 m. The coil is oriented perpendicular to a uniform B-field of strength .075 T. a) Sketch the situation. b) The ends of the wire are pulled so that the wire collapses in a time of .060 s. Is an EMF induced? Explain. c) Find the magnitude of the induced EMF. d) If a 2.0 Ω resistor is connected across the ends of the wire, how much power is dissipated. Where did the power come from?

  18. 0 Applications of Faraday’s Law • Pick up coils • Generators • Transformers

  19. 0 Pick up coils

  20. 0 Generators • In a generator, a coil is turned at a constant angular frequency in a B-field • What produces the changing flux to generate the EMF? (GR)

  21. 0 Generators • Flux is produced by changing angle • θ = ωt • ΦB = NAB cos ωt • ΔV = -dΦ/dt = NABω sin ωt • What does the output of a generator look like? (GR)

  22. 0 AC Generators • Output of AC Generator is sine wave • Can we make a DC Generator? (BRST)

  23. 0 DC Generator • To create a DC generator (such as Genecon) use a split ring • Output for a simple DC generator is shown • Output can be made smoother by more sophisticated combinations of rings and magnets

  24. 0 a) What must be the magnetic field strength so that a generator consisting of 1000 turns of a coil of radius 25 cm produces a peak output of 160 V when turned at a frequency of 60 Hz? b) Sketch a graph of the output of the generator.

  25. 0 Why AC? • In the early days of electrification, there was a debate of whether power companies should supply ac or dc • Edison favored dc whereas Westinghouse favored ac • Clearly Westinghouse won the debate. Why?

  26. Transformers • Electromagnetic induction can be used to change ac electric potential from one value to another

  27. Transformers • Flux through one coil is the same everywhere • dΦB/dt is the same for each coil • Vp = -Np dΦB/dt and • Vs = -Ns dΦB/dt • After a little algebra

  28. Transformers • If the number of turns in the secondary is greater than in the primary, is the secondary potential greater or less than the primary? (GR) • If the number of turns in the secondary is less than in the primary, is the secondary potential greater or less than the primary? (GR)

  29. Transformers • If the number of turns in the secondary is greater than in the primary, the secondary potential is greater. This is called a step up transformer • If the number of turns in the secondary is less than in the primary, the secondary potential is less than the primary. This is called a step down transformer

  30. Ex. To operate a neon sign a potential difference of 12000 V is needed. What must be the ratio of turns if the primary potential is 120 V?

  31. A DC potential of 100 V is applied to the primary of a step-up transformer with turns ratio of 500. What is the potential difference across the secondary? • .2 V • 50,000 V • 0 V • Not enough information given

  32. 120 V ac is applied across the primary of a step down transformer with turns ratio 1/50. How does the power applied at the primary compare to that at the secondary? (Assume a lossless transformer) • It is 1/50 th as big • It is 50 x bigger • It is the same • Not enough information to answer

  33. Power and Current in Transformers • Conservation of Energy implies power at primary is the same as power at secondary • What happens to current? • IsVs = IpVp so

  34. Current in Transformers • In a step up transformer, current is decreased • Is step down transformer current is increased

  35. 0 Application to Power Generation • Need to minimize losses in transmission • Losses given by i2R, so lower current means less loss • Transmitting at higher potential decreases current

  36. Self Inductance • If you apply current to a coil, you will induce an EMF • V = - dΦ/dt = -d(μ0niNA)/dt = - μ0nINA di/dt • The induced EMF is proportional to the change in current V = -Ldi/dt • Where L = μ0n2Al is called the inductance of the coil A A l

  37. RL Circuits EX. a) Use Kirchhoff’s Voltage Law to find an equation for the current in the circuit. b) Find a solution if I(0) = 0.

  38. Current grows with exponential asymptote to steady value of E0/R. Time Constant is given by L/R

  39. Ex. An RL circuit w/o a battery has an initial current I0. Sketch the circuit. b) Use KVL to find an equation for the current. c) Find I(t).

  40. Magnetic Energy • Circuits do not turn on or off instantaneously. • Inductance means that circuits turn on and off with a time constant L/R • This is due to the fact that energy is stored in the inductor. It takes time to initially store the energy in the inductor when the switch is closed, it takes time to remove the energy from the inductor when the switch is opened.

  41. Energy in an Inductor • P = IV = I LdI/dt • For a coil I = B/(μ0N/l) and L = μ0N2Al • So Energy = 1/(2μ0)B2LA

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