Chapter 31A - Electromagnetic Induction. A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University

Transcription

1 Chapter 31A - Electromagnetic Induction A PowerPoint Presentation by Paul E. Tippens, Professor of Physics Southern Polytechnic State University 2007

2 Objectives: After completing this module, you should be able to: Calculate the magnitude and direction of the induced current or emf in a conductor moving with respect to a given -field. Calculate the magnetic flux through an area in a given -field. Apply Lenz s s law and the right-hand hand rule to determine directions of induced emf. Describe the operation and use of ac and dc generators or motors.

3 Induced Current When a conductor moves across flux lines, magnetic forces on on the free electrons induce an an electric current. Right-hand hand force rule shows current outward for down and inward for up up motion. (Verify) Down Up I I Down Up v F F v

4 Induced EMF: Observations Faraday s s observations: Relative motion induces emf. Flux lines in Wb Direction of emf depends on direction of motion. Emf is proportional to rate at which lines are cut ( (v). Emf is proportional to the number of turns N. N turns; velocityv Faraday s Law: E= -N t The negative sign means that E opposes its cause.

5 Magnetic Flux Density Magnetic flux lines are continuous and closed. Direction is that of the vector at any point. A Magnetic Flux density: A When area A is is perpendicular to to flux: A ; = A The unit of flux density is the weber per square meter.

6 Calculating Flux When Area is Not Perpendicular to Field The flux penetrating the area A when the normal vector n makes an angle of with the -field is: Acos A n The angle is the complement of the angle that the plane of the area makes with field. (Cos( = Sin )

7 Example 1: A current loop has an area of 40 cm 2 and is placed in a 3-T -field at the given angles. Find the flux through the loop in each case. x x x x x x x x A x x x x x x x x n n n A = 40 cm 2 (a) = 0 0 (b) = 90 0 (c) = 60 0 (a) = A cos 0 0 = (3 T)(0.004 m 2 )(1); 12.0 mwb (b) = A cos 90 0 = (3 T)(0.004 m 2 )(0); 0 0 mwb (c) = A cos 60 0 = (3 T)(0.004 m 2 )(0.5); 6.00 mwb

8 Application of Faraday s s Law Faraday s Law: E= -N t A change in flux can occur by a change in area or by a change in the -field: = A = A Rotating loop = A n n Loop at rest = A n

9 Example 2: A coil has 200 turns of area 30 cm 2. It flips from vertical to horizontal position in a time of 0.03 s. What is the induced emf if the constant -field is 4 mt? A A = 30 cm 2 0 = 30 cm 2 = A A = (3 mt)(30 cm 2 ) = (0.004 T)( m 2 ) N N = 200 turns n S = 1.2 x 10-5 Wb = 4 mt; 0 0 to 90 0 E N t x 10 Wb (200) 0.03 s E = V The negative sign indicates the polarity of the voltage.

10 Lenz s s Law Lenz s s law: An An induced current will be be in in such a direction as as to to produce a magnetic field that will oppose the motion of of the magnetic field that is is producing it. it. Induced Left motion Induced I Right motion I N Flux increasing to left induces loop flux to the right. S Flux decreasing by right move induces loop flux to the left. N S

11 Example 3: Use Lenz s s law to determine direction of induced current through R if switch is closed for circuit below ( ( increasing). Close switch. Then what is direction of induced current? R The rising current in in right circuit causes flux to to increase to to the left,, inducing current in in left circuit that must produce a rightward field to to oppose motion.. Hence current I through resistor Ris is to to the right as as shown.

12 Directions of Forces and EMFs An An emf E is is induced by by moving wire at at velocity v in in constant field. Note direction of of I. I. x x x x x x x x x x x x x x x I x x x x x x x x x x x x x I x x x x v x vx x x x x x Lx x x x x x x x x x x x x x x x x x x x x x From Lenz s s law, we see that a reverse field (out) is is created. This field causes a leftward force on on the wire that offers resistance to to the motion. Use right-hand hand force rule to to show this. I v Induced emf x x x x x x x x x x x x x x x x x x Lenz s law I v

13 Motional EMF in a Wire Force F on charge q in wire: F = qv; Work = FL = qvl Work E= q EMF: E= qvl q Lv x x x x x x I x x x x x x x x x x x x x x x x I x x x x x x x x x L x v x x x x x x x x x x x x x x x F v If wire of length L moves with velocity v an angle with : E=Lvsin v sin v Induced Emf E

14 Example 4: A 0.20-m m length of wire moves at a constant speed of 5 m/s in at with a 0.4-T -Field. What is the magnitude and direction of the induced emf in the wire? E= E=Lvsin 0 (0.4 T)(0.20 m)(5 m/s)sin140 v North E = V South Using right-hand hand rule,, point fingers to to right, thumb along velocity, and hand pushes in in direction of of induced emf to the north in in the diagram. v South I North

15 The AC Generator An alternating AC current is produced by rotating a loop in a constant -field. Current on left is outward by right-hand hand rule. The right segment has an inward current. Rotating Loop in -field v I I v When loop is vertical,, the current is zero. I in R is right, zero, left, and then zero as loop rotates.

16 Operation of AC Generator I=0 I=0

17 Calculating Induced EMF Rectangular loop a x b loop a x b Each segment a has constant velocity v. a n b Area A = ab. n b/2 x v oth segments a moving with v at angle with gives emf: E= avsin ; vr b 2 E 2a b sin T 2 E A sin T v = r r = b/2 v sin n x v

18 Sinusoidal Current of Generator +E xx.. -E The emf varies sinusoidally with max and min emf For N turns, the EMF is: E NA sin

19 Example 5: An ac generator has 12 turns of wire of area 0.08 m 2. The loop rotates in a magnetic field of 0.3 T at a frequency of 60 Hz.. Find the maximum induced emf. = 2 2f f = 2(60 Hz) = 377 rad/s Emf is maximum when = E max =NA ; Since sin 1 E = max 2 (12)(0.3 T)(.08 m )(377 rad/s). n x f = 60 Hz The maximum emf generated is therefore: E max = 109 V If If the resistance is is known, then Ohm s s law (V (V V = IR) can be be applied to to find the maximum induced current.

20 The DC Generator The simple ac generator can be converted to a dc generator by using a single split-ring commutator to reverse connections twice per revolution. Commutator E t DC Generator For the dc dc generator: The emf fluctuates in in magnitude, but always has the same direction (polarity).

21 The Electric Motor In In a simple electric motor,, a current loop experiences a torque which produces rotational motion. Such motion induces a back emf to to oppose the motion. Applied voltage back emf = net voltage V E b b = IR IR Since back emf E b increases with rotational frequency,, the starting current is is high and the operating current is is low: E b = NA sin sin E b I V Electric Motor

22 Armature and Field Windings In the commercial motor, many coils of wire around the armature will produce a smooth torque. (Note directions of I in wires.) Series-Wound Motor: The field and armature wiring are connected in series. Motor Shunt-Wound Motor: The field windings and the armature windings are connected in parallel.

23 Example 6: A series-wound dc motor has an internal resistance of 3.. The 120-V supply line draws 4 A when at full speed. What is the emf in the motor and the starting current? E b Recall that: V E b b = IR IR I V 120 V E b = (4 A)(3 The back emf in motor: E b = 108 V The starting current I s is found by noting that E b = 0 in beginning (armature has not started rotating). 120 V 0 = I s (3 I s = 40 A

24 Summary Faraday s Law: E= -N t A change in flux can occur by a change in area or by a change in the -field: = A = A Calculating flux through an area in a -field: A ; = A Acos

25 Summary (Cont.) Lenz s s law: An An induced current will be be in in such a direction as as to to produce a magnetic field that will oppose the motion of of the magnetic field that is is producing it. it. Induced Left motion Induced I Right motion I N Flux increasing to left induces loop flux to the right. S Flux decreasing by right move induces loop flux to the left. N S

26 Summary (Cont.) An emf is induced by a wire moving with a velocity v at an angle with a -field. v sin E= Lvsin Induced Emf E In In general for for a coil of of N turns of of area A rotating with a frequency in in a -field, the generated emf is is given by by the following relationship: v For N turns, the EMF is: E NA sin

27 Summary (Cont.) The ac ac generator is is shown to to the right. The dc dc generator and a dc dc motor are shown below: DC Generator V Electric Motor

28 Summary (Cont.) The rotor generates a back emf in in the operation of of a motor that reduces the applied voltage. The following relationship exists: Applied voltage back emf = net voltage V E b b = IR IR Motor

29 CONCLUSION: Chapter 31A Electromagnetic Induction