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CHAPTER 20: INDUCED VOLTAGES AND INDUCTANCE HERRIMAN HIGH AP PHYSICS 2.

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Presentation on theme: "CHAPTER 20: INDUCED VOLTAGES AND INDUCTANCE HERRIMAN HIGH AP PHYSICS 2."— Presentation transcript:

1 CHAPTER 20: INDUCED VOLTAGES AND INDUCTANCE HERRIMAN HIGH AP PHYSICS 2

2 ELECTROMAGNETISM When charge is moving (current) it produces a magnetic field. So if current produces a magnetic field can a magnetic field produce current? HERRIMAN HIGH AP PHYSICS B

3 ELECTROMAGNETIC INDUCTION A changing magnetic field can produce an induced emf (potential difference) A potential difference can cause current to flow HERRIMAN HIGH AP PHYSICS 2

4 SECTION 20.1 MAGNETIC FLUX It is actually the change in something called the magnetic flux that induces an emf Magnetic flux is defined as the value of a perpendicular magnetic field multiplied by the area. ***NOTE θ is the angle between B and the perpendicular (or “normal”) to the loop*** ________________________________________________ B is perpendicular to the plane of the loop with an area of A, and θ is the angle between B and the normal. HERRIMAN HIGH AP PHYSICS 2

5 SAMPLE PROBLEM A conducting circular loop of radius 0.250 m is place d in the xy plane of a uniform magnetic field of 0.360 T that points in the positive z direction, the same direction as the normal to the plane. a)Calculate the magnetic flux through the loop. b)Suppose the loop is rotated clockwise around the x axis so that the normal direction now points at a 45° angle with respect to the z axis. Recalculate the magnetic flux of the loop. c)What is the change in flux due to the rotation of the loop? HERRIMAN HIGH AP PHYSICS 2

6 SOLUTION: a)A = πr 2 =π(0.250 m) 2 = 0.196 m 2 Φ B = AB cos θ = (0.196 m 2 )(0.360 T)(cos 0°) = 0.0706 Wb b)Φ B = AB cos θ = (0.196 m 2 )(0.360 T)(cos 45°) = 0.0499 Wb c)ΔΦ B = 0.0499 Wb - 0.0706 Wb = -0.0207 Wb HERRIMAN HIGH AP PHYSICS B

7 SECTION 20.2 FARADAY’S LAW The instantaneous emf induced in a circuit equals the rate of change of the magnetic flux through the circuit ε =Emf (We also call this voltage!) N the number of turns ∆Φ=Change in flux ∆t=change in time HERRIMAN HIGH AP PHYSICS 2

8 WHEN IS EMF INDUCED? If the magnitude of B changes If the direction of B changes If the area enclosed by the loop changes HERRIMAN HIGH AP PHYSICS 2

9 MICHAEL FARADAY Faraday is the scientist that was credited with the discovery of electromagnetic induction. He also invented the electric motor, electric generator, and transformer. HERRIMAN HIGH AP PHYSICS 2

10 LENZ’S LAW The polarity of the induced emf is such that it produces a current whose magnetic field opposes the change in magnetic flux through the loop. This means the current tends to keep the original flux though the circuit HERRIMAN HIGH AP PHYSICS 2

11 AN EXAMPLE As the magnetic field goes into the loop of wire more magnetic field is going through the wire so the flux gets bigger The current will flow in order to decrease the change in the flux so it will flow clockwise HERRIMAN HIGH AP PHYSICS 2

12 AN EXAMPLE As the magnetic field goes out of the loop of wire less magnetic field is going through the wire so the flux gets smaller The current will flow in order to decrease the change in the flux so it will flow Counterclockwise HERRIMAN HIGH AP PHYSICS 2

13 SAMPLE PROBLEM A coil with 25 turns of wire is wrapped in a frame with a cross sectional area of 1.8 cm on a side. Each turn has the same area, equal to that of the frame, and the total resistance of the coil is 0.350 Ω. An applied uniform magnetic field is perpendicular to the plane of the coil. a)If the field changes uniformly from 0.0 T to 0.5 T in 0.8 seconds, what is the induced EMF on the coil while the field is changing? Find the: b)Magnitude c)And direction of the induced current in the coil while the field is changing. HERRIMAN HIGH AP PHYSICS 2

14 SOLUTION a)A = L 2 = (0.018 m) 2 = 3.24 x 10 —4 m 2 Φ B i = 0 Wb Φ B f = AB cos θ = (3.24 x 10 —4 m 2 )(0.50 T)(cos 0°) = 1.62 x 10 -4 Wb ε = -(25 turns)(1.62 x 10 -4 Wb/0.8 sec) = -5.06 x 10 -3 V b) I = V/R = 5.06 x 10 -3 V/ 0.35 Ω = 0.0145 Amps c) Magnetic field is up, so put your palm in the opposite direction, and your thumb will point in the direction of the current (clockwise). HERRIMAN HIGH AP PHYSICS 2

15 SECTION 20.3 MOTIONAL EMF CONDUCTING BAR As the bar moves through the magnetic field electrons move down because of the magnetic force When there is a charge separation like that there is a potential difference and an Electric Field is created The charge will build until the downward magnetic force qvB is balanced by the upward electric force qE x x x x x x x v + – HERRIMAN HIGH AP PHYSICS 2

16 CONDUCTING BAR IN CLOSED PATH Because of the moving electrons again a voltage and electric field will be produced Because it is a closed path current will flow Diagram acquired at http://www.physics.upenn.edu/courses/gladney/phys151/lectures/lecture_mar_21_2003.shtml through www.google.com.http://www.physics.upenn.edu/courses/gladney/phys151/lectures/lecture_mar_21_2003.shtmlwww.google.com HERRIMAN HIGH AP PHYSICS 2

17 SAMPLE PROBLEM The sliding bar has a length of 0.5 m and moves at 2.0 m/s in a magnetic field of 0.25 T. a)Find the induced voltage in the moving rod. b)If the resistance of the circuit is 0.5 Ω, find the current in the circuit and the power delivered to the resistor. c)Calculate the Magnetic force on the bar. x x x x x x x v + – HERRIMAN HIGH AP PHYSICS 2

18 SOLUTION a)ε= Blv = (0.25 T)((0.5 m)(2.0 m/s) = 0.25 V b)I = ε/R = 0.25 V / 0.5 Ω = 0.5 A P = IV = (0.5 A)(0.25 V) = 0.125 W c) F in = IBl = (0.5 A)(0.25 T)(0.5 m) = 0.0625 N HERRIMAN HIGH AP PHYSICS 2


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