Lecture 29: MON 03 NOV

Similar documents
Lecture 29: MON 02 NOV

Lecture 30: WED 04 NOV

Electricity & Optics

LECTURE 23 INDUCED EMF. Instructor: Kazumi Tolich

FARADAY S AND LENZ LAW B O O K P G

Electromagnetic Induction Practice Problems Homework PSI AP Physics B

Faraday's Law ds B B G G ΦB B ds Φ ε = d B dt

PHYSICS Fall Lecture 15. Electromagnetic Induction and Faraday s Law

Physics 202 Chapter 31 Oct 23, Faraday s Law. Faraday s Law

Introduction. First Experiment

Chapter 27, 28 & 29: Magnetism & Electromagnetic Induction. Magnetic flux Faraday s and Lenz s law Electromagnetic Induction Ampere s law

Induction and Inductance

Physics 9 Wednesday, April 2, 2014

PHY 1214 General Physics II

Chapter 12. Magnetism and Electromagnetism

Magnetic flux. where θ is the angle between the magnetic field and the area vector. The unit of magnetic flux is the weber. 1 Wb = 1 T m 2.

Demo: Solenoid and Magnet. Topics. Chapter 22 Electromagnetic Induction. EMF Induced in a Moving Conductor

Physics 1402: Lecture 18 Today s Agenda

Chapter 23 Magnetic Flux and Faraday s Law of Induction

Faraday s Law. Underpinning of Much Technology

F A C U L T Y O F E D U C A T I O N. Physics Electromagnetism: Induced Currents Science and Mathematics Education Research Group

PHYS 1444 Section 003 Lecture #18

Chapter 9 FARADAY'S LAW Recommended Problems:

21 MAGNETIC FORCES AND MAGNETIC FIELDS

Electromagnetic Induction and Faraday s Law

Faraday s Law of Electromagnetic Induction

Faraday s Law. Lecture 17. Chapter 33. Physics II. Course website:

Induction and Inductance

Faraday s Law. Faraday s Law of Induction Motional emf. Lenz s Law. Motors and Generators. Eddy Currents

Can a Magnetic Field Produce a Current?

PHYSICS 1B. Today s lecture: Motional emf. and. Lenz s Law. Electricity & Magnetism

Chapter 21 Magnetic Induction Lecture 12

Chapter 30. Induction and Inductance

Physics 201. Professor P. Q. Hung. 311B, Physics Building. Physics 201 p. 1/1

Physics 54 Lecture March 1, Micro-quiz problems (magnetic fields and forces) Magnetic dipoles and their interaction with magnetic fields

EXAM 3: SOLUTIONS. B = B. A 2 = BA 2 cos 0 o = BA 2. =Φ(2) B A 2 = A 1 cos 60 o = A 1 2 =0.5m2

AAST/AEDT. Electromagnetic Induction. If the permanent magnet is at rest, then - there is no current in a coil.

MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Physics Spring Experiment 5: Faraday s Law

Slide 1 / 50. Electromagnetic Induction and Faraday s Law

Faraday s Law; Inductance

Can a Magnetic Field Produce a Current?

Faraday s Law. Lecture 17. Chapter 33. Physics II. Course website:

PHYS 1442 Section 004 Lecture #14

Physics for Scientists & Engineers 2

University Physics (Prof. David Flory) Chapt_31 Tuesday, July 31, 2007

Chapter 5: Electromagnetic Induction

Lecture 10 Induction and Inductance Ch. 30

PHYS102 Previous Exam Problems. Induction

Induced Electric Field

Physics 11b Lecture #13

Slide 1 / 50. Slide 2 / 50. Slide 3 / 50. Electromagnetic Induction and Faraday s Law. Electromagnetic Induction and Faraday s Law.

University Physics Volume II Unit 2: Electricity and Magnetism Chapter 13: Electromagnetic Induction Conceptual Questions

Problem Solving: Faraday s Law & Inductance. Faraday s Law

Magnets. Domain = small magnetized region of a magnetic material. all the atoms are grouped together and aligned

Chapter 30. Induction and Inductance

Chapter 5. Electromagnetic Induction

DO PHYSICS ONLINE MOTORS AND GENERATORS FARADAY S LAW ELECTROMAGNETIC INDUCTION

Electromagnetic Induction

Induced Electric Field

General Physics (PHY 2140)

PHY122 Physics for the Life Sciences II

Lecture 33. PHYC 161 Fall 2016

Motional EMF. Toward Faraday's Law. Phys 122 Lecture 21

Last Homework. Reading: Chap. 33 and Chap. 33. Suggested exercises: 33.1, 33.3, 33.5, 33.7, 33.9, 33.11, 33.13, 33.15,

Concept Questions with Answers. Concept Questions with Answers W11D2. Concept Questions Review

Application Of Faraday s Law

Chapter 20: Electromagnetic Induction. PHY2054: Chapter 20 1

PS I AP Physics 2 Electromagnetic Induction Multiple Choice Questions

Ch. 23 Electromagnetic Induction, AC Circuits, And Electrical Technologies

Electromagnetics in Medical Physics

mag ( ) 1 ). Since I m interested in the magnitude of the flux, I ll avoid the minus sign by taking the normal to point upward.

CHAPTER 5: ELECTROMAGNETIC INDUCTION

Chapter 23: Magnetic Flux and Faraday s Law of Induction

Electromagnetic Induction! March 11, 2014 Chapter 29 1

Lecture 18: Faraday s Law & Motional EMF

(a) zero. B 2 l 2. (c) (b)

Revision Guide for Chapter 15

Induced Electric Field

Induction and Inductance

Agenda for Today. Elements of Physics II. Lenz Law. Emf opposes change in flux Faraday s Law Induced EMF in a conducting loop

C. Incorrect! Use the formula for magnetic flux. This is the product of magnetic field, times area, times the angle between them.

Physics 4. Magnetic Induction. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Physics 240 Fall 2005: Exam #3 Solutions. Please print your name: Please list your discussion section number: Please list your discussion instructor:

Problem Solving 6: Ampere s Law and Faraday s Law. Part One: Ampere s Law

PHYSICS. Chapter 30 Lecture FOR SCIENTISTS AND ENGINEERS A STRATEGIC APPROACH 4/E RANDALL D. KNIGHT

Magnetism. and its applications

Electromagnetic Induction

Physics 2B Winter 2012 Final Exam Practice

REVIEW SESSION. Midterm 2

PHY101: Major Concepts in Physics I

General Review. LECTURE 16 Faraday s Law of Induction

Solve: From Example 33.5, the on-axis magnetic field of a current loop is

Physics 132: Lecture 15 Elements of Physics II Agenda for Today

Induction and inductance

Physics 115. Induction Induced currents. General Physics II. Session 30

Our goal for today. 1. To go over the pictorial approach to Lenz s law.

CURRENT-CARRYING CONDUCTORS / MOVING CHARGES / CHARGED PARTICLES IN CIRCULAR ORBITS

PHY 1214 General Physics II

AP Physics C - E & M

AP Physics 2 Electromagnetic Induction Multiple Choice

Transcription:

Physics 2113 Jonathan Dowling Lecture 29: MON 03 NOV Ch30.1 4 Induction and Inductance I Fender Stratocaster Solenoid Pickup

Magnetic Circuit Breaker As the normal operating or "rated" current flows through the sensing coil, a magnetic field is created around that coil. When the current flow increases, the strength of the magnetic field increases, drawing the spring-biased, movable magnetic core toward the pole piece. As the core moves inward, the efficiency of the magnetic circuit is increased, creating an even greater electromagnetic force. When the core is fully "in", maximum electromagnetic force is attained. The armature is attracted to the pole piece, unlatching a trip mechanism thereby opening the contacts.

F a r a d a y ' s E x p e r i m e n t s I n a s e r i e s o f e x p e r i m e n t s, M i c h a e l F a r a d a y i n E n g l a n d a n d J o s e p h H e n r y i n t h e U. S. w e r e a b l e t o g e n e r a t e e l e c t r i c c u r r e n t s w i t h o u t t h e u s e o f b a t t e r i e s. T h e c i r c u i t s h o w n i n t h e f i g u r e c o n s i s t s o f a w i r e l o o p c o n n e c t e d t o a s e n s i t i v e a m m e t e r ( k n o w n a s a " g a l v a n o m e t e r " ). I f w e a p p r o a c h t h e l o o p w i t h a p e r m a n e n t m a g n e t w e s e e a c u r r e n t b e i n g r e g i s t e r e d b y t h e g a l v a n o m e t e r. 1. A c u r r e n t a p p e a r s o n l y i f t h e r e i s r e l a t i v e m o t i o n b e t w e e n t h e m a g n e t a n d t h e l o o p. 2. F a s t e r m o t i o n r e s u l t s i n a l a r g e r c u r r e n t. 3. I f w e r e v e r s e t h e d i r e c t i o n o f m o t i o n o r t h e p o l a r i t y o f t h e m a g n e t, t h e c u r r e n t r e v e r s e s s i g n a n d f l o w s i n t h e o p p o s i t e d i r e c t i o n. T h e c u r r e n t g e n e r a t e d i s k n o w n a s " i n d u c e d c u r r e n t "; t h e e m f t h a t a p p e a r s i s k n o w n a s " i n d u c e d e m f "; t h e w h o l e e f f e c t i s c a l l e d " i n d u c t i o n."

Changing B-Field Induces a Current in a Wire Loop

Note Current Changes Sign With Direction No Current When Magnet Stops

loop 2 loop 1 In the figure we show a second type of experiment in which current is induced in loop 2 when the switch S in loop 1 is either closed or opened. When the current in loop 1 is constant no induced current is observed in loop 2. The conclusion is that the magnetic field in an induction experiment can be generated either by a permanent magnet or by an electric current in a coil. Faraday summarized the results of his experiments in what is known as " Faraday's law of induction. " An emf is induced in a loop when the number of magnetic field lines that pass through the loop is changing.

Loop Two is Connected To A Light Bulb. Loop One Has a 60 Hz Alternating Current The Current in Loop One Produces a Rapidly Changing Magnetic Field in Loop Two That Induces a Current in Loop Two Lighting the Bulb!

Faraday s Law: What? The Flux! A time varying magnetic FLUX creates an induced EMF Definition of magnetic flux is similar to definition of electric flux. Units of Magnetic Flux: Weber! Φ B = S! B d! A E EMF = dφ B dt d! A da B Take note of the MINUS sign!! The induced EMF acts in such a way that it OPPOSES the change in magnetic flux ( Lenz s Law ).

E EMF = dφ B dt = A db dt EMF is proportional to the slope db/dt b > d = e > a = c = 0

Lenz s Law The Loop Current Produces a B Field that Opposes the CHANGE in the bar magnet field. Upper Drawing: B Field from Magnet is INCREASING so Loop Current is Clockwise and Produces an Opposing B Field that Tries to CANCEL the INCREASING Magnet Field Lower Drawing: B Field from Magnet is DECREASING so Loop Current is Counterclockwise and Tries to BOOST the Decreasing Magnet Field.

a = b > c = 0

30.4.1. Consider the situation shown. A triangular, aluminum loop is slowly moving to the right. Eventually, it will enter and pass through the uniform magnetic field region represented by the tails of arrows directed away from you. Initially, there is no current in the loop. When the loop is entering the magnetic field, what will be the direction of any induced current present in the loop? a) clockwise b) counterclockwise c) No current is induced.

30.4.1. Consider the situation shown. A triangular, aluminum loop is slowly moving to the right. Eventually, it will enter and pass through the uniform magnetic field region represented by the tails of arrows directed away from you. Initially, there is no current in the loop. When the loop is entering the magnetic field, what will be the direction of any induced current present in the loop? a) clockwise b) counterclockwise c) No current is induced.

30.4.2. Consider the situation shown. A triangular, aluminum loop is slowly moving to the right. Eventually, it will enter and pass through the uniform magnetic field region represented by the tails of arrows directed away from you. Initially, there is no current in the loop. When the loop is exiting the magnetic field, what will be the direction of any induced current present in the loop? a) clockwise b) counterclockwise c) No current is induced.

30.4.2. Consider the situation shown. A triangular, aluminum loop is slowly moving to the right. Eventually, it will enter and pass through the uniform magnetic field region represented by the tails of arrows directed away from you. Initially, there is no current in the loop. When the loop is exiting the magnetic field, what will be the direction of any induced current present in the loop? a) clockwise b) counterclockwise c) No current is induced.

30.4.3. A rigid, circular metal loop begins at rest in a uniform magnetic field directed away from you as shown. The loop is then pulled through the field toward the right, but does not exit the field. What is the direction of any induced current within the loop? a) clockwise b) counterclockwise c) No current is induced.

30.4.3. A rigid, circular metal loop begins at rest in a uniform magnetic field directed away from you as shown. The loop is then pulled through the field toward the right, but does not exit the field. What is the direction of any induced current within the loop? a) clockwise b) counterclockwise c) No current is induced.

Example d! A B A closed loop of wire encloses an area of A = 1 m 2 in which in a uniform magnetic field exists at 30 0 to the PLANE of the loop. The magnetic field is DECREASING at a rate of db/dt = 1T/s. The resistance of the wire is 10 Ω. What is the induced current? Is it clockwise or counterclockwise? 60 Φ B =! B d! A = BAcos(60 0 ) = BA /2 E = dφ B dt i = E R = A 2R i = (1m2 ) 2(10Ω) = A 2 db dt db dt 30 0 (1T/s) = 0.05A

Example 3 loops are shown. B = 0 everywhere except in the circular region I where B is uniform, pointing out of the page and is increasing at a steady rate. Rank the 3 loops in order of increasing induced EMF. (a) III < II < I? (b) III < II = I? I B III encloses no flux so EMF=0 I and II enclose same flux so EMF same. Are Currents in Loops I & II Clockwise or Counterclockwise? II III (c) III = II = I?

30.4.4. A coil of wire that forms a complete loop is moving with a constant speed v toward a very long, current carrying wire, only a portion of which is shown. What affect, if any, does the current carrying wire have on the coil of wire? a) Since the magnetic field increases as the coil approaches the wire, a current is induced in the coil. b) The rectangle will be distorted as it is pulled in the direction of the current in the wire. B 1/ r r! c) Close to the wire, a magnetic force acts on the loop that accelerates the loop away from the wire. d) Since the magnetic field around the wire is not changing, there is no effect on the coil. e) Since the coil and the wire are not touching, there is no effect.

30.4.4. A coil of wire that forms a complete loop is moving with a constant speed v toward a very long, current carrying wire, only a portion of which is shown. What affect, if any, does the current carrying wire have on the coil of wire? a) Since the magnetic field increases as the coil approaches the wire, a current is induced in the coil. b) The rectangle will be distorted as it is pulled in the direction of the current in the wire. c) Close to the wire, a magnetic force acts on the loop that accelerates the loop away from the wire. d) Since the magnetic field around the wire is not changing, there is no effect on the coil. e) Since the coil and the wire are not touching, there is no effect.

Example An infinitely long wire carries a constant current i as shown A square loop of side L is moving towards the wire with a constant velocity v. What is the EMF induced in the loop when it is a distance R from the loop? Φ B = = = L 0 µ 0 ildx 2π(R + x) µ 0 il ln( R + x) 2 π µ 0iL R + ln 2π R L L 0 B = dr/dt=v L L Choose a strip of width dx located as shown. Flux thru this strip dφ = BLdx = E = dφ B dt = µ 0Li 2π µ 0i 2πr d dt R x µ 0 ildx 2π(R + x) ln 1 + L R i r=r+x

Example i E = dφ B dt dr/dt=v L R x B µ = 0Li d L ln 1 + 2π dt R = = µ 0Li dr R L 2π dt R + L R 2 µ 0i L v 2π ( R + L) R 2 What is the DIRECTION of the induced current? Magnetic field due to wire points INTO page and gets stronger as you get closer to wire So, flux into page is INCREASING Hence, current induced must be counter clockwise to oppose this increase in flux = CCW

30.3.2. A circular ring is rotated clockwise at a constant rate for an extended period of time using the apparatus shown. Which of the graphs below correctly shows the magnetic flux through the ring as a function of time? Note: At time t = 0 s, the plane of the ring is perpendicular to the direction of the magnetic field.

30.3.2. A circular ring is rotated clockwise at a constant rate for an extended period of time using the apparatus shown. Which of the graphs below correctly shows the magnetic flux through the ring as a function of time? Note: At time t = 0 s, the plane of the ring is perpendicular to the direction of the magnetic field.

Example : The Generator A square loop of wire of side L is rotated at a uniform frequency f in the presence of a uniform magnetic field B as shown. Describe the EMF induced in the loop. Φ B = = BL 2 S! B d! A cos( θ ) E = dφ B dt = BL 2 dθ L θ = ωt f = 2πω B dt sin(θ) 2 = BL ( 2πf ) sin(2πft) B