PHYSICS - CLUTCH CH 28: INDUCTION AND INDUCTANCE.

Similar documents
PHYSICS - CLUTCH 1E CH 28: INDUCTION AND INDUCTANCE.

Physics 1202: Lecture 11 Today s Agenda

PHYSICS - GIANCOLI CALC 4E CH 29: ELECTROMAGNETIC INDUCTION.

PHY2049 Exam 2 solutions Fall 2016 Solution:

EMF induced in a coil by moving a bar magnet. Induced EMF: Faraday s Law. Induction and Oscillations. Electromagnetic Induction.

ELECTROMAGNETIC INDUCTION AND FARADAY S LAW

Physics 4B. Question and 3 tie (clockwise), then 2 and 5 tie (zero), then 4 and 6 tie (counterclockwise) B i. ( T / s) = 1.74 V.

Physics Electricity and Magnetism Lecture 12 - Inductance, RL Circuits. Y&F Chapter 30, Sect 1-4

Physics 114 Exam 3 Spring Name:

PHYS 1101 Practice problem set 12, Chapter 32: 21, 22, 24, 57, 61, 83 Chapter 33: 7, 12, 32, 38, 44, 49, 76

INDUCTANCE. RC Cicuits vs LR Circuits

Physics 114 Exam 2 Fall 2014 Solutions. Name:

Physics 4B. A positive value is obtained, so the current is counterclockwise around the circuit.

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

Physics 114 Exam 2 Spring Name:

Inductor = (coil of wire)

Week3, Chapter 4. Position and Displacement. Motion in Two Dimensions. Instantaneous Velocity. Average Velocity

PHYS 202 Notes, Week 6

Lecture #4 Capacitors and Inductors Energy Stored in C and L Equivalent Circuits Thevenin Norton

Energy Storage Elements: Capacitors and Inductors

DC Circuits. Crossing the emf in this direction +ΔV

Chapter 6 Electrical Systems and Electromechanical Systems

UNIVERSITY OF UTAH ELECTRICAL & COMPUTER ENGINEERING DEPARTMENT. 10k. 3mH. 10k. Only one current in the branch:

Complex Numbers, Signals, and Circuits

Handout 10: Inductance. Self-Inductance and inductors

ˆ (0.10 m) E ( N m /C ) 36 ˆj ( j C m)

( ) = ( ) + ( 0) ) ( )

Chapter 30 Inductance and Electromagnetic Oscillations

Electricity and Magnetism Review Faraday s Law

Vote today! Physics 122, Fall November (c) University of Rochester 1. Today in Physics 122: applications of induction

Physics 2102 Spring 2007 Lecture 10 Current and Resistance

General Physics II. Electromagnetic Induction and Electromagnetic Waves

Electromagnetic Induction (Chapters 31-32)

i I (I + i) 3/27/2006 Circuits ( F.Robilliard) 1

Chapter 23 Magnetic Flux and Faraday s Law of Induction

Coupling Element and Coupled circuits. Coupled inductor Ideal transformer Controlled sources

(b) i(t) for t 0. (c) υ 1 (t) and υ 2 (t) for t 0. Solution: υ 2 (0 ) = I 0 R 1 = = 10 V. υ 1 (0 ) = 0. (Given).

Slide 1 / 26. Inductance by Bryan Pflueger

Part 4: Electromagnetism. 4.1: Induction. A. Faraday's Law. The magnetic flux through a loop of wire is

Electrical Circuits 2.1 INTRODUCTION CHAPTER

Advanced Circuits Topics - Part 1 by Dr. Colton (Fall 2017)

MAE140 - Linear Circuits - Winter 16 Final, March 16, 2016

CONDUCTORS AND INSULATORS

Chapter 30 Inductance

Chapter 32. Inductance

Inductance, Inductors, RL Circuits & RC Circuits, LC, and RLC Circuits

Induction_P1. 1. [1 mark]

Chapter 31. Induction and Magnetic Moment

Kirchhoff second rule

JEE ADVANCE : 2015 P1 PHASE TEST 4 ( )

Introduction to circuit analysis. Classification of Materials

TUTORIAL PROBLEMS. E.1 KCL, KVL, Power and Energy. Q.1 Determine the current i in the following circuit. All units in VAΩ,,

Self-inductance A time-varying current in a circuit produces an induced emf opposing the emf that initially set up the time-varying current.

Chapter 30. Inductance

8.022 (E&M) Lecture 8

INDUCTANCE Self Inductance

General Physics (PHY 2140)

Conservation of Angular Momentum = "Spin"

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

AP Physics C. Inductance. Free Response Problems

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

Designing Information Devices and Systems II Spring 2018 J. Roychowdhury and M. Maharbiz Discussion 3A

Chapter 22. Induction

MAGNETISM MAGNETIC DIPOLES

General Physics (PHY 2140)

Boise State University Department of Electrical and Computer Engineering ECE 212L Circuit Analysis and Design Lab

Chapter 21 Magnetic Induction Lecture 12

Moments of Inertia. and reminds us of the analogous equation for linear momentum p= mv, which is of the form. The kinetic energy of the body is.

Electricity and Magnetism Lecture 13 - Physics 121 Electromagnetic Oscillations in LC & LCR Circuits,

Sections begin this week. Cancelled Sections: Th Labs begin this week. Attend your only second lab slot this week.

E40M Device Models, Resistors, Voltage and Current Sources, Diodes, Solar Cells. M. Horowitz, J. Plummer, R. Howe 1

Electromagnetic Induction and Waves (Chapters 33-34)

EE215 FUNDAMENTALS OF ELECTRICAL ENGINEERING

Chapter 23: Magnetic Flux and Faraday s Law of Induction

Class: Life-Science Subject: Physics

Electricity and Magnetism - Physics 121 Lecture 10 - Sources of Magnetic Fields (Currents) Y&F Chapter 28, Sec. 1-7

Chapter 31. Faraday s Law

Physics 111: Mechanics Lecture 11

Gravitational Acceleration: A case of constant acceleration (approx. 2 hr.) (6/7/11)

Chapter 31. Faraday s Law

Electromagnetic Induction Faraday s Law Lenz s Law Self-Inductance RL Circuits Energy in a Magnetic Field Mutual Inductance

Chapter 11 Angular Momentum

Louisiana State University Physics 2102, Exam 3 April 2nd, 2009.

Week 11: Chapter 11. The Vector Product. The Vector Product Defined. The Vector Product and Torque. More About the Vector Product

N S. 4/4/2006 Magnetic Fields ( F.Robilliard) 1

Chapter 20: Electromagnetic Induction. PHY2054: Chapter 20 1

PHYSICS Fall Lecture 15. Electromagnetic Induction and Faraday s Law

Physics 2A Chapter 3 HW Solutions

Chapter 11: Angular Momentum

Spring 2002 Lecture #13

PES 1120 Spring 2014, Spendier Lecture 6/Page 1

CHAPTER 5: ELECTROMAGNETIC INDUCTION

Dr. Fritz Wilhelm, Physics 230 E:\Excel files\230 lecture\ch26 capacitance.docx 1 of 13 Last saved: 12/27/2008; 8:40 PM. Homework: See website.

RLC Circuit (3) We can then write the differential equation for charge on the capacitor. The solution of this differential equation is

6.01: Introduction to EECS 1 Week 6 October 15, 2009

Inductance. Slide 2 / 26. Slide 1 / 26. Slide 4 / 26. Slide 3 / 26. Slide 6 / 26. Slide 5 / 26. Mutual Inductance. Mutual Inductance.

r where the electric constant

10/9/2003 PHY Lecture 11 1

Boise State University Department of Electrical and Computer Engineering ECE 212L Circuit Analysis and Design Lab

Chapter 30 INDUCTANCE. Copyright 2012 Pearson Education Inc.

Transcription:

!! www.clutchprep.com

CONCEPT: ELECTROMAGNETIC INDUCTION A col of wre wth a VOLTAGE across each end wll have a current n t - Wre doesn t HAVE to have voltage source, voltage can be INDUCED V Common ways to INDUCE a voltage on a col of wre: - Move a bar magnet - Vary current n electromagnet - Turn electromagnet on and off Induced voltage known as INDUCED EMF INDUCED CURRENT - Process known as ELECTROMAGNETIC INDUCTION EXAMPLE of current nducton: vሬԧ vሬԧ = 0 vሬԧ CURRENT INDUCED NO CURRENT INDUCED CURRENT INDUCED CERTAIN changes wll nduce a current, and the magntude of the current depends on the rate of these changes - Bar magnet movng nto col Faster t goes, larger the nduced current - Current changng n electromagnet near a col Faster the current changes, larger the nduced current Page 2

CONCEPT: MAGNETIC FLUX Remember! Electrc flux s just the amount of passng through a surface - MAGNETIC FLUX s just the amount of passng through a surface MAGNETIC FLUX s gven by A θ Normal Φ B = (UNITS 1 Wb = 1 ) B Magnetc flux changes wth,, and - IMPORTANT to remember ths changes n magnetc flux wll be mportant later! EXAMPLE: What s the magnetc flux through the square surface depcted n the followng fgure, f B = 0.05 T? Assume the sde length of the square s 5 cm. Surface B 30 o Page 3

PRACTICE: MAGNETIC FLUX THROUGH A RING A rng of radus 0.5m les n the xy-plane. If a magnetc feld of magntude 2 T ponts at an angle of 22 o above the x-axs, what s the magnetc flux through the rng? EXAMPLE: ROTATING RING A rng of radus 2 cm s n the presence of a 0.6 T magnetc feld. If the rng begns wth ts plane parallel to the magnetc feld, and ends wth the plane of the rng perpendcular to the magnetc feld, what s the change n the magnetc flux? Page 4

CONCEPT: FARADAY S LAW Changng magnetc feld through conductng loops - Ths s actually due to a changng MAGNETIC FLUX A changng MAGNETIC FLUX leads to an nduced EMF: Ɛ nd = - Ths s known as Faraday s Law Remember! Φ = BA cos θ - So, magnetc flux changes wth,, and EXAMPLE 1: A square conductng wre of sde length 4 cm s n a 2 T magnetc feld. It rotates such that the angle of the magnetc feld to the normal of the square ncreases from 30 o to 60 o n 2 s. What s the nduced current on the wre f ts resstance s 5 Ω? Page 5

PRACTICE: FARADAY S LAW AND TWO SOLENOIDS Two solenods are placed end to end, wth one solenod connected to a varable power source, and the other solenod connected to a 10 Ω resstor. The frst solenod has 10 turns per cm and has as an ntal current of 2 A, and the second solenod has 5 turns and a radus of 2 cm. a) What s the change n magnetc feld emtted by the frst solenod f the current ncreases from 2 A to 5 A n 1 s? b) What s the change n the magnetc flux through the other solenod durng ths 1 s? c) What s the nduced EMF on the second solenod? d) What s the nduced current on the second solenod? EXAMPLE: CURRENT IN A CIRCUIT WITH A CHANGING, EXTERNAL MAGNETIC FIELD What current does the ammeter read f the followng crcut, wth an area of 50 cm 2, s placed n a magnetc feld that s changng at 0.05 T/s? Note that the resstor has a resstance of 2 Ω. BሬሬԦ A Page 6

CONCEPT: LENZ S LAW Faraday s Law tells us the magntude of the nduced EMF magntude of nduced current - To fnd DIRECTION of nduced current, we use Lenz s Law LENZ s LAW states: A conductor wll nduce a magnetc feld on tself to changes n ts magnetc flux B v v B nd v B B nd v B B Once the drecton of the nduced magnetc feld s known, rght hand rule gves drecton of nduced current B B v B nd v nd EXAMPLE: In the followng scenaros, fnd the drecton of the current nduced on the conductors. v v v Page 7

PRACTICE: DIRECTION OF INDUCED CURRENT IN A RING What s the drecton of the nduced current n the nner rng shown n the followng fgure? For ths problem, consder the battery s voltage as contnuously INCREASING. Note: the arrow strkng through the battery n the crcut dagram ndcates that the voltage of the battery s varable (.e. t can be changed). EXAMPLE: BAR MAGNET VS CURRENT-CARRYING WIRE A bar magnet moves relatve to a col of wre as ndcated n the fgure below and nduces a current n the col. A current carryng wre carres a current relatve to a col as shown n the second fgure. Would you need to ncrease or decrease the magntude of the current n the wre to nduce a current n the col that moves n the SAME drecton as the current nduced by the bar magnet? Scenaro 1 v Scenaro 2 Page 8

CONCEPT: MOTIONAL EMF Remember! A changng magnetc feld can produce an EMF - BUT so can moton. Ths s referred to as a MOTIONAL EMF. If a conductor moves through a magnetc feld, charges feel a Postve charges feel the force [ UPWARD / DOWNWARD] Separaton of charges L FሬԦ B BሬሬԦ vሬԧ Separaton of charges E feld Electrc force that magnetc force - To balance, E = vb Induced EMF Ɛ = EL = EXAMPLE: If a conductor of length 10 cm moves wth a velocty of 20 m/s n a magnetc feld of 0.05 T, what s the current through the conductor f ts resstance s 15 Ω If a conductor moves along U-shaped wre, MAGNETIC FLUX changes - Change n Change n magnetc flux Producton of L BሬሬԦ vሬԧ Change n area of Change n magnetc flux of Induced EMF Ɛ = ΔΦ B Δt = EXAMPLE: In the crcut below, f the wre has a resstance of 10 Ω, what s the current nduced f the length of the bar s 10 cm, the speed of the bar s 25 cm/s, and the magnetc feld s 0.02 T? What about the power generated by the crcut? a BሬሬԦ vሬԧ b Page 9

PRACTICE: BAR MOVING IN UNKNOWN MAGNETIC FIELD A thn rod moves n a perpendcular, unknown magnetc feld. If the length of the rod s 10 cm and the nduced EMF s 1 V when t moves at 5 m/s, what s the magntude of the magnetc feld? Page 10

CONCEPT: TRANSFORMERS Power n North Amerca s delvered to outlets n homes at 120 V. - Ths s too large to operate many delcate electroncs, such as computers. Remember! A col wth a changng magnetc feld can nduce an EMF on a second col - Ths nduced EMF can be as small as needed. A TRANSFORMER does exactly ths t uses Faraday s law to convert a large voltage to a small EMF: V 1 V 2 The rato of the VOLTAGES n a transformer depends upon the rato of the TURNS: V 2 V 1 = N 2 N 1 EXAMPLE: You need to buld a transformer that drops the 120 V of a regular North Amercan outlet to a much safer 15 V. You already have a solenod wth 50 turns made, but you need to make a second solenod to complete your transformer. What s the least number of turns the second solenod could have? Page 11

PRACTICE: OPERATING A LAPTOP An outlet n North Amerca outputs electrcty at 120 V, but a typcal laptop needs to operate at around 20 V. In order to do so, a transformer s placed n a laptop s power supply. If the col n the crcut connected to the laptop has 20 turns, how many turns must the col n the crcut wth the outlet have? Page 12

CONCEPT: MUTUAL INDUCTANCE If two cols are brought close, and the current changes through one col, an EMF s nduced on the other col - Ths s known as MUTUAL INDUCTANCE - The col wth the changng current s known as the, the other the Magnetc feld B depends on Flux through 2 depends on Total flux through 2 s to current through 1 - We call the proportonalty constant MUTUAL INDUCTANCE, M Col 1 N 1 1 Col 2 N 2 The MUTUAL INDUCTANCE between two cols s M = (UNITS: Henry, 1 H = 1 / ) EXAMPLE: What s the mutual nductance of two solenods of length L and area A, one wth N1 turns and the other wth N2? L N 1 N 2 Induced EMF on second col s dependent upon the mutual nductance: - Ɛ 2 = M Δ 1 Δt EXAMPLE: Whch of the followng prmary cols would nduce the largest EMF on a secondary col of wre f they all had ther current changng at the same rate. 10 turns 10 cm 20 turns 2.5 cm Page 13

PRACTICE: MUTUAL INDUCTION BETWEEN TWO SOLENOIDS A solenod of 25 turns, wth an area of 0.005 m 2 s wound around a 10.0 cm solenod wth 50 turns, as shown n the fgure below. If, at some nstant n tme, the current through the 10.0 cm solenod s 0.5 A and changng at 50 ma/s, what s the nduced EMF on the 25 turn solenod? L N 1 N 2 Page 14

CONCEPT: SELF INDUCTANCE When consderng two cols, one acted as source of magnetc feld and one felt a changng magnetc flux - Sngle col wll also experence an nduced EMF due to changng ts own magnetc flux SELF-INDUCED EMF We defne SELF INDUCTANCE, L, dentcally to mutual nductance L = (UNITS are also H) Self-nduced EMF s then - Ɛ = L Δ Δt EXAMPLE: What s the self-nductance of a col of wre, wth N turns and a radus R? EXAMPLE: A solenod has 500 turns, wth a current of 0.5 A producng an average flux through col of 0.005 Wb. If the selfnduced EMF on ths solenod s 10 mv, how quckly must the current be changng? Page 15

PRACTICE: SELF-INDUCED EMF IN A COIL OF WIRE A crcular col of wre wth 20 turns has a current changng at a rate of 0.12 A/s. If the radus of the col s 20 cm, what s the nduced EMF on the col? Page 16

CONCEPT: INDUCTORS A col of wre placed n a crcut s known as an INDUCTOR or In order to use nductors n crcuts, we need to know how to apply KIRCHHOFF S LOOP RULE to them: a b a b Remember! By Faraday s law, the nduced EMF on an nductor s Ɛ = EXAMPLE: Wrte out Krchhoff s loop rule for the followng crcut. Treat the capactor as ntally charged. +q -q Snce nductors oppose changes n current (by Lenz law), they re use n crcuts to do just that. - For nstance, they re used n power transmsson lnes, n the event of lghtnng strkes Page 17

CONCEPT: LR CIRCUITS LR Crcuts are, as the name mples, crcuts contanng and S 2 L R S 1 V There are two steps needed to analyze ths crcut: - CURRENT GROWTH: When S1 s closed, but S2 s open, the battery produces a current n the crcut - CURRENT DECAY: When S1 s open and S2 s closed, the current decays because the battery s removed CURRENT GROWTH n an LR crcut does not occur nstantly the nductor ressts changes to currents (t) = V R (1 e t/τ ) L R V CURRENT DECAY n an LR crcut does not occur nstantly the nductor ressts changes to currents (t) = V R e t/τ L R The TIME CONSTANT, τ = L, determnes the how quckly growth and decay occurs R Page 18

EXAMPLE: UNKOWN RESISTANCE IN LR CIRCUIT An LR crcut has a tme constant of 0.025 s and s ntally connected to a 10 V battery. If after 0.005 s of beng dsconnected from the battery, the current s 0.5 A, what s the resstance of the crcut? PRACTICE: TIME TO HALF MAXIMUM CURRENT An LR crcut wth L = 0.1 H and R = 10 Ω are connected to a battery wth the crcut ntally broken. When the crcut s closed, how much tme passes untl the current reaches half of ts maxmum value? Page 19

PRACTICE: UNKNOWN CURRENT IN AN LR CIRCUIT Consder the LR crcut shown below. Intally, both swtches are open. Frst, swtch 1 s closed, and current s allowed to grow to ts maxmum value. Then, swtch two s closed and swtch 1 s open, and current s allowed to decay for 0.05 s. What s the maxmum current n the crcut? What s the fnal current n the crcut f V = 10 V, L = 0.02 H, and R = 5 Ω? S 2 L R S 1 V Page 20

CONCEPT: LC CIRCUITS LC Crcuts are made up of and, as ther name mples The current n ths crcut OSCILLATES: +q -q q = 0 -q +q q = 0 +q -q MATHEMATICALLY, the current and charge are represented by - (t) = (ω = 1/LC s the angular frequency of oscllaton) - q(t) = φ s known as the PHASE ANGLE, and t determnes what part of the oscllaton you begn at EXAMPLE: An LC crcut wth an nductor of 0.05 H and a capactor of 35 µf begns wth the current at half ts maxmum value. What s the phase angle of ths oscllaton? Page 21

EXAMPLE: OSCILLATIONS IN AN LC CIRCUIT An LC crcuts, wth L = 0.05 H and C = 50 mf, begns wth the capactor fully charged. After 0.1 s, the current s 0.2 A. Under these condtons, how many seconds does t take for a fully charged plate to transfer all of ts charge to the other plate? PRACTICE: MAXIMUM CURRENT IN LR CIRCUIT An LR crcut has a 0.5 mf capactor ntally charged to 1 mc. If t s connected to a 0.04 H nductor, what s the maxmum current n the crcut? Page 22

CONCEPT: ENERGY IN AN LC CIRCUIT An nductor s just a col of wre we don t assume t has any resstance So when the capactor loses ts charge, and therefore ts energy, where does t go? An nductor can store MAGNETC ENERGY - U = So long as the wres don t have any resstance, energy s conserved n an LC crcut: +q -q q = 0 -q +q q = 0 +q -q E U C U L E U C U L E U C U L E U C U L E U C U L EXAMPLE: An LC crcut has an 0.1 H nductor and a 15 nf capactor, and begns wth the capactor maxmally charged. After 0.1 s, how much energy s stored by the nductor? If the ntal charge on the capactor were 50 mc, what s the maxmum current n the crcut? Page 23

PRACTICE: ENERGY LOSS DUE TO RESISTANCE Let s say an LC crcut begns wth the capactors carryng a maxmum charge of 10 mc. After the capactor has lost half of ts charge, what s the current n the crcut f L = 0.01 H and C = 50 mf? If durng the tme for the capactor to lose half ts charge, resstance wthn the crcut dsspated 0.2 mj, what then would the current n the crcut be? Page 24

CONCEPT: LRC CIRCUITS As the name mples, an LRC crcut contans,, and In and LRC crcut, wth the capactor ntally charged, we have: - ΣV = = 0 = 0 = 0 +q -q There are 3 solutons to the equaton above: the UNDERDAMPED, CRITICALLY DAMPED, and OVERDRAMPED UNDERDAMPED CRITICALLY DAMPED OVERDAMPED - q(t) = Qe (R/2L)t cos (ω t + φ) - q(t) = Qe (R/2L)t - No smple equaton q t - Occurs for small R - Looks almost lke an LC crcut - But R s sappng energy - Occurs when R 2 = 4L/C - Looks lke an RC Crcut - Occurs for large R - Looks lke an RC Crcut The new angular frequency s - ω = 1 LC R2 4L 2 Page 25

EXAMPLE: AMPLITUDE DECAY IN AN LRC CIRCUIT An LRC crcut has an nductance of 10 mh, a capactance of 100 µf, and a resstance of 20 Ω. What type of LRC crcut s ths? How long wll t take for the maxmum charge stored on the capactor to drop by half? Page 26