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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.

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Presentation on theme: "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."— Presentation transcript:

1 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 2 30-2: Two symmetric situations YES!! This is formulated in Faraday ’ s law. It is the basis for the electric generator!! 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?

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

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

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

8 8 Ans. 4.05 V, 2.03 A. 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?

9 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 10 Checkpoint #2 Another example: a bar magnet is moved to the left/ right toward a stationary loop of wire.

11 11 30.5: Induction and energy transfers q E = q v B 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:

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

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

14 14 The power applied: P app = F v The induced emf: E = - d  B /dt = B l v i = B l v/R 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).

15 15 Therefore, the force exerted on the wire is: F = i l B = B 2 l 2 v/R The power delivered/ applied due to the wire is: P app = B 2 l 2 v 2 /R But, the power dissipated is: P diss = i 2 R = B 2 l 2 v 2 /R Therefore, P diss = P app That is, the work you do in pulling the loop through the magnetic field appears as thermal energy in the loop!

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


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