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Chapter 30: Induction and Inductance

Chapter 30: Induction and Inductance. Introduction What are we going to talk about in chapter 31:. A change of magnetic flux through a conducting loop produces a current! What is lenz ’ s law? What is the relation between induction and energy transfer? What are eddy currents?.

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Chapter 30: Induction and Inductance

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  1. Chapter 30: Induction and Inductance Introduction What are we going to talk about in chapter 31: • A change of magnetic flux through a conducting loop produces a current! • What is lenz’s law? • What is the relation between induction and energy transfer? • What are eddy currents?

  2. 30-2: Two symmetric situations We have seen (ch. 29) that: Current loop in a magnetic field leads to torque (the basis for the electric motor). Is the opposite also true? Does a torque on a loop in a magnetic field lead to a current? YES!! This is formulated in Faraday’s law. It is the basis for the electric generator!!

  3. 30-2: Two experiments: Experiment #1: Loop of wire connected to a galvanometer. A magnet is moved towards or away from the loop. Result: an induced current is set up in the circuit as long as there is relative motion between the magnet and the coil (w/o a battery!!). The work per unit charged to produce the current is called the induced emf.

  4. Experiment #2: Primary circuit has an emf, secondary circuit has no emf. Result: an induced emf (and current) is produced in the secondary circuit only when the current (and hence the magnetic flux) is changing.

  5. 30-3: Faraday’s law of induction The emf induced in a circuit is directly proportional to the time rate of change of magnetic flux through the circuit. E = - dFB/dt where FB is the magnetic flux through the circuit.

  6. If there are N loops, all of the same area: E = - N dFB/dt where FB is the magnetic flux through one loop circuit. • What factors effect the emf? • The magnitude of B may vary with time • The area of the circuit can change with time • The angle (q) between B and the  plane can change • A combination of the above Checkpoint #1

  7. Some applications: • Cooking utensils • Ground fault interrupter (GFI) • Microphone (or electronic guitar!!)

  8. Example: A coil is wrapped with 200 turns of wire on the perimeter of a square frame of side 18 cm. The total resistance of the coil is 2W. B is  the plane of the coil and changes linearly from 0 to 0.5 T in 0.80 seconds. Find the emf in the coil while the field is changing. What is the induced current? Ans. 4.05 V, 2.03 A.

  9. 30.4: Lenz’s law: Lenz’s law says: The polarity of the induced emf is such that it tends to produce a current that will create a magnetic flux to oppose the change in magnetic flux through the loop. For example: What is the direction of the induced current in the figure? Why? The current is clockwise. What happens if you stop? What happens if you reverse direction?

  10. Another example: a bar magnet is moved to the left/ right toward a stationary loop of wire. Checkpoint #2

  11. 30.5: Induction and energy transfers Consider a straight conductor (length l) moving with constant velocity (v to the right) in a perpendicular magnetic field (B into the page). Electrons will move towards the bottom and accumulate there leaving a net positive charge at the top until: q E = q v B

  12. Therefore, a potential difference V will be created across the conductor: V = E l = B l v Which end is at a higher potential? The upper end is at higher potential. If the direction of motion is reversed, the polarity of V is also reversed. Notice that: FB = B l x

  13. Therefore, the induced emf E is: E = -d FB/dt = - B l v The induced current is: i = B l v/R The power (P) delivered by the applied force is (from phys-101): P = Fappv = i l B v = (Blv)2/R =E2/R This power is dissipated in the resistor (i2 R)!!

  14. If there is a rectangular circuit part of which is in perpendicular magnetic field and is being pulled out of the field, you must apply a constant force (F) in order for the circuit to move with constant speed (v). The power applied: Papp = F v The induced emf: E = - dFB/dt = B l v i = B l v/R

  15. Therefore, the force exerted on the wire is: F = i l B = B2 l2 v/R The power delivered/ applied due to the wire is: Papp = B2 l2 v2/R But, the power dissipated is: Pdiss = i2 R = B2 l2 v2/R Therefore, Pdiss = Papp That is, the work you do in pulling the loop through the magnetic field appears as thermal energy in the loop!

  16. Eddy currents: Checkpoint #3 و آخر دعوانا أن الحمد لله رب العالمين

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