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Chapter 20

Chapter 20. Induced Voltages and Inductance. Michael Faraday. 1791 – 1867 Great experimental scientist Invented electric motor, generator and transformers Discovered electromagnetic induction Discovered laws of electrolysis. Faraday’s Experiment – Set Up.

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Chapter 20

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  1. Chapter 20 Induced Voltages and Inductance

  2. Michael Faraday • 1791 – 1867 • Great experimental scientist • Invented electric motor, generator and transformers • Discovered electromagnetic induction • Discovered laws of electrolysis

  3. Faraday’s Experiment – Set Up • A current can be produced by a changing magnetic field • First shown in an experiment by Michael Faraday • A primary coil is connected to a battery • A secondary coil is connected to an ammeter

  4. Faraday’s Experiment • The purpose of the secondary circuit is to detect current that might be produced by the magnetic field • When the switch is closed, the ammeter reads a current and then returns to zero • When the switch is opened, the ammeter reads a current in the opposite direction and then returns to zero • When there is a steady current in the primary circuit, the ammeter reads zero

  5. Faraday’s Conclusions • An electrical current is produced by a changing magnetic field • The secondary circuit acts as if a source of emf were connected to it for a short time • It is customary to say that an induced emf is produced in the secondary circuit by the changing magnetic field

  6. Magnetic Flux • The emf is actually induced by a change in the quantity called the magnetic flux rather than simply by a change in the magnetic field • Magnetic flux is defined in a manner similar to that of electrical flux • Magnetic flux is proportional to both the strength of the magnetic field passing through the plane of a loop of wire and the area of the loop

  7. Magnetic Flux, 2 • You are given a loop of wire • The wire is in a uniform magnetic field • The loop has an area A • The flux is defined as • ΦB = BA = B A cos θ • θ is the angle between B and the normal to the plane

  8. Magnetic Flux, 3 • When the field is perpendicular to the plane of the loop, as in a, θ = 0 and ΦB = ΦB, max = BA • When the field is parallel to the plane of the loop, as in b, θ = 90° and ΦB = 0 • The flux can be negative, for example if θ = 180° • SI units of flux are T. m² = Wb (Weber)

  9. Magnetic Flux, final • The flux can be visualized with respect to magnetic field lines • The value of the magnetic flux is proportional to the total number of lines passing through the loop • When the area is perpendicular to the lines, the maximum number of lines pass through the area and the flux is a maximum • When the area is parallel to the lines, no lines pass through the area and the flux is 0

  10. Example 1 A square loop 2.00 m on a side is placed in a magneticfield of magnitude 0.300 T. If the field makes an angle of50.0° with the normal to the plane of the loop, find themagnetic flux through the loop.

  11. Example 2 A solenoid 4.00 cm in diameter and 20.0 cm long has250 turns and carries a current of 15.0 A. Calculate themagnetic flux through the circular cross-sectional areaof the solenoid.

  12. Practice 1 Find the flux of the Earth’s magnetic field of magnitude5.00 × 10−5 T through a square loop of area 20.0 cm2(a) when the field is perpendicular to the plane of theloop, (b) when the field makes a 40.0° angle with the normalto the plane of the loop, and (c) when the fieldmakes a 90.0° angle with the normal to the plane.

  13. Electromagnetic Induction –An Experiment • When a magnet moves toward a loop of wire, the ammeter shows the presence of a current (a) • When the magnet is held stationary, there is no current (b) • When the magnet moves away from the loop, the ammeter shows a current in the opposite direction (c) • If the loop is moved instead of the magnet, a current is also detected

  14. Electromagnetic Induction – Results of the Experiment • A current is set up in the circuit as long as there is relative motion between the magnet and the loop • The same experimental results are found whether the loop moves or the magnet moves • The current is called an induced current because is it produced by an induced emf

  15. Faraday’s Law and Electromagnetic Induction • The instantaneous emf induced in a circuit equals the time rate of change of magnetic flux through the circuit • If a circuit contains N tightly wound loops and the flux changes by ΔΦB during a time interval Δt, the average emf induced is given by Faraday’s Law:

  16. Faraday’s Law and Lenz’ Law • The change in the flux, ΔΦB, can be produced by a change in B, A or θ • Since ΦB = B A cos θ • The negative sign in Faraday’s Law is included to indicate the polarity of the induced emf, which is found by Lenz’ Law • The current caused by the induced emf travels in the direction that creates a magnetic field with flux opposing the change in the original flux through the circuit

  17. Example 3 A 300-turn solenoid with a length of 20 cm and a radiusof 1.5 cm carries a current of 2.0 A. A second coil of fourturns is wrapped tightly about this solenoid so that it canbe considered to have the same radius as the solenoid.Find (a) the change in the magnetic flux through the coiland (b) the magnitude of the average induced emf in thecoil when the current in the solenoid increases to 5.0 A ina period of 0.90 s.

  18. Example 4 A circular coil enclosing an area of 100 cm2 is made of200 turns of copper wire. The wire making up the coil hasresistance of 5.0 Ω, and the ends of the wire are connectedto form a closed circuit. Initially, a 1.1-T uniform magneticfield points perpendicularly upward through the plane ofthe coil. The direction of the field then reverses so thatthe final magnetic field has a magnitude of 1.1 T andpoints downward through the coil. If the time requiredfor the field to reverse directions is 0.10 s, what averagecurrent flows through the coil during that time?

  19. Application of Faraday’s Law – Motional emf • A straight conductor of length ℓ moves perpendicularly with constant velocity through a uniform field • The electrons in the conductor experience a magnetic force • F = q v B • The electrons tend to move to the lower end of the conductor

  20. Motional emf • As the negative charges accumulate at the base, a net positive charge exists at the upper end of the conductor • As a result of this charge separation, an electric field is produced in the conductor • Charges build up at the ends of the conductor until the downward magnetic force is balanced by the upward electric force • There is a potential difference between the upper and lower ends of the conductor

  21. Motional emf, cont • The potential difference between the ends of the conductor can be found by • ΔV = B ℓ v • The upper end is at a higher potential than the lower end • A potential difference is maintained across the conductor as long as there is motion through the field • If the motion is reversed, the polarity of the potential difference is also reversed

  22. Motional emf in a Circuit • Assume the moving bar has zero resistance • As the bar is pulled to the right with a given velocity under the influence of an applied force, the free charges experience a magnetic force along the length of the bar • This force sets up an induced current because the charges are free to move in the closed path

  23. Motional emf in a Circuit, cont • The changing magnetic flux through the loop and the corresponding induced emf in the bar result from the change in area of the loop • The induced, motional emf, acts like a battery in the circuit

  24. Example 5 A conducting rod of length ℓ moves on two horizontalfrictionless rails, as in Figure P20.18. A constant force ofmagnitude 1.00 N moves the bar at a uniform speed of2.00 m/s through a magnetic field that is directed intothe page. (a) What is the current in an 8.00-Ω resistor R?(b) What is the rate of energy dissipation in the resistor?(c) What is the mechanical power delivered by theconstant force?

  25. Example 6 A helicopter has blades of length 3.0 m, rotating at2.0 rev/s about a central hub. If the vertical component ofEarth’s magnetic field is 5.0 × 10−5 T, what is the emfinduced between the blade tip and the central hub?

  26. Practice 2 A 12.0-m-long steel beam is accidentally dropped by aconstruction crane from a height of 9.00 m. The horizontalcomponent of the Earth’s magnetic field over theregion is 18.0 μT. What is the induced emf in the beamjust before impact with the Earth? Assume the longdimension of the beam remains in a horizontal plane,oriented perpendicular to the horizontal component ofthe Earth’s magnetic field.

  27. Lenz’ Law – Moving Magnet Example • A bar magnet is moved to the right toward a stationary loop of wire (a) • As the magnet moves, the magnetic flux increases with time • The induced current produces a flux to the left, so the current is in the direction shown (b)

  28. Lenz’ Law, Final Note • When applying Lenz’ Law, there are two magnetic fields to consider • The external changing magnetic field that induces the current in the loop • The magnetic field produced by the current in the loop

  29. Example 7 A copper bar is moved to the right while its axis is maintainedin a direction perpendicular to a magnetic field, asshown in Figure P20.27. If the top of the bar becomespositive relative to the bottom, what is the direction of themagnetic field?

  30. Generators • Alternating Current (AC) generator • Converts mechanical energy to electrical energy • Consists of a wire loop rotated by some external means • There are a variety of sources that can supply the energy to rotate the loop • These may include falling water, heat by burning coal to produce steam

  31. AC Generators, cont • Basic operation of the generator • As the loop rotates, the magnetic flux through it changes with time • This induces an emf and a current in the external circuit • The ends of the loop are connected to slip rings that rotate with the loop • Connections to the external circuit are made by stationary brushes in contact with the slip rings

  32. AC Generators, final • The emf generated by the rotating loop can be found by ε =2 B ℓ v=2 B ℓ sin θ • If the loop rotates with a constant angular speed, ω, and N turns ε = N B A ω sin ω t • ε = εmax when loop is parallel to the field • ε = 0 when when the loop is perpendicular to the field

  33. DC Generators • Components are essentially the same as that of an ac generator • The major difference is the contacts to the rotating loop are made by a split ring, or commutator

  34. DC Generators, cont • The output voltage always has the same polarity • The current is a pulsing current • To produce a steady current, many loops and commutators around the axis of rotation are used • The multiple outputs are superimposed and the output is almost free of fluctuations

  35. Motors • Motors are devices that convert electrical energy into mechanical energy • A motor is a generator run in reverse • A motor can perform useful mechanical work when a shaft connected to its rotating coil is attached to some external device

  36. Motors and Back emf • The phrase back emf is used for an emf that tends to reduce the applied current • When a motor is turned on, there is no back emf initially • The current is very large because it is limited only by the resistance of the coil

  37. Motors and Back emf, cont • As the coil begins to rotate, the induced back emf opposes the applied voltage • The current in the coil is reduced • The power requirements for starting a motor and for running it under heavy loads are greater than those for running the motor under average loads

  38. Example 8 A 100-turn square wire coil of area 0.040 m2 rotates abouta vertical axis at 1 500 rev/min, as indicated in FigureP20.30. The horizontal component of the Earth’smagnetic field at the location of the loop is 2.0 × 10−5 T.Calculate the maximum emf induced in the coil by theEarth’s field.

  39. Example 9 A motor has coils with a resistance of 30 Ω and operatesfrom a voltage of 240 V. When the motor is operating atits maximum speed, the back emf is 145 V. Find thecurrent in the coils (a) when the motor is first turned onand (b) when the motor has reached maximum speed.(c) If the current in the motor is 6.0 A at some instant,what is the back emf at that time?

  40. Self-inductance • Self-inductance occurs when the changing flux through a circuit arises from the circuit itself • As the current increases, the magnetic flux through a loop due to this current also increases • The increasing flux induces an emf that opposes the change in magnetic flux • As the magnitude of the current increases, the rate of increase lessens and the induced emf decreases • This opposing emf results in a gradual increase of the current

  41. Self-inductance cont • The self-induced emf must be proportional to the time rate of change of the current • L is a proportionality constant called the inductance of the device • The negative sign indicates that a changing current induces an emf in opposition to that change

  42. Self-inductance, final • The inductance of a coil depends on geometric factors • The SI unit of self-inductance is the Henry • 1 H = 1 (V · s) / A • You can determine an expression for L

  43. Example 10 A coiled telephone cord forms a spiral with 70.0 turns, adiameter of 1.30 cm, and an unstretched length of60.0 cm. Determine the self-inductance of one conductorin the unstretched cord.

  44. Joseph Henry • 1797 – 1878 • First director of the Smithsonian • First president of the Academy of Natural Science • First to produce an electric current with a magnetic field • Improved the design of the electro-magnetic and constructed a motor • Discovered self-inductance

  45. Inductor in a Circuit • Inductance can be interpreted as a measure of opposition to the rate of change in the current • Remember resistance R is a measure of opposition to the current • As a circuit is completed, the current begins to increase, but the inductor produces an emf that opposes the increasing current • Therefore, the current doesn’t change from 0 to its maximum instantaneously

  46. RL Circuit • When the current reaches its maximum, the rate of change and the back emf are zero • The time constant, , for an RL circuit is the time required for the current in the circuit to reach 63.2% of its final value

  47. RL Circuit, cont • The time constant depends on R and L • The current at any time can be found by

  48. Example 11 Consider the circuit shown in Figure P20.46. Take ε = 6.00V, L = 8.00 mH, and R = 4.00 Ω. (a) What is the inductivetime constant of the circuit? (b) Calculate the current inthe circuit 250 μs after the switch is closed. (c) What is thevalue of the final steady-state current? (d) How long does ittake the current to reach 80.0% of its maximum value?

  49. Energy Stored in a Magnetic Field • The emf induced by an inductor prevents a battery from establishing an instantaneous current in a circuit • The battery has to do work to produce a current • This work can be thought of as energy stored by the inductor in its magnetic field • PEL = ½ L I2

  50. Example 12 A 300-turn solenoid has a radius of 5.00 cm and a lengthof 20.0 cm. Find (a) the inductance of the solenoid and(b) the energy stored in the solenoid when the current inits windings is 0.500 A.

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