Faraday s Law. To be able to find. motional emf transformer and motional emf. Motional emf

Similar documents
Ans: In the rectangular loop with the assigned direction for i2: di L dt , (1) where (2) a) At t = 0, i1(t) = I1U(t) is applied and (1) becomes

Circuits 24/08/2010. Question. Question. Practice Questions QV CV. Review Formula s RC R R R V IR ... Charging P IV I R ... E Pt.

PHYSICS 102. Intro PHYSICS-ELECTROMAGNETISM

ENGI 4430 Advanced Calculus for Engineering Faculty of Engineering and Applied Science Problem Set 9 Solutions [Theorems of Gauss and Stokes]

Science Advertisement Intergovernmental Panel on Climate Change: The Physical Science Basis 2/3/2007 Physics 253

Physics 201, Lecture 5

Control Volume Derivation

Chapter 4 Circular and Curvilinear Motions

Ch.4 Motion in 2D. Ch.4 Motion in 2D

Reinforcement learning

() t. () t r () t or v. ( t) () () ( ) = ( ) or ( ) () () () t or dv () () Section 10.4 Motion in Space: Velocity and Acceleration

LECTURE 5. is defined by the position vectors r, 1. and. The displacement vector (from P 1 to P 2 ) is defined through r and 1.

Classification of Equations Characteristics

ME 141. Engineering Mechanics

Chapter 25 Electric Potential

ELECTROSTATICS. 4πε0. E dr. The electric field is along the direction where the potential decreases at the maximum rate. 5. Electric Potential Energy:

Outline. Part 1, Topic 3 Separation of Charge and Electric Fields. Dr. Sven Achenbach - based on a script by Dr. Eric Salt - Outline

Today - Lecture 13. Today s lecture continue with rotations, torque, Note that chapters 11, 12, 13 all involve rotations

Lecture 3. Electrostatics

Average & instantaneous velocity and acceleration Motion with constant acceleration

DEPARTMENT OF CIVIL AND ENVIRONMENTAL ENGINEERING FLUID MECHANICS III Solutions to Problem Sheet 3

f(x) dx with An integral having either an infinite limit of integration or an unbounded integrand is called improper. Here are two examples dx x x 2

An object moving with speed v around a point at distance r, has an angular velocity. m/s m

Collection of Formulas

Rotations.

SOME USEFUL MATHEMATICS

2-d Motion: Constant Acceleration

PHYSICS 1210 Exam 1 University of Wyoming 14 February points

Physics 101 Lecture 4 Motion in 2D and 3D

Motion. ( (3 dim) ( (1 dim) dt. Equations of Motion (Constant Acceleration) Newton s Law and Weight. Magnitude of the Frictional Force

Ch.9. Electromagnetic Induction

Physics Worksheet Lesson 4: Linear Motion Section: Name:

( )( )( ) ( ) + ( ) ( ) ( )

Motion. Part 2: Constant Acceleration. Acceleration. October Lab Physics. Ms. Levine 1. Acceleration. Acceleration. Units for Acceleration.

Derivation of the differential equation of motion

Motion on a Curve and Curvature

r r r r r EE334 Electromagnetic Theory I Todd Kaiser

Go over vector and vector algebra Displacement and position in 2-D Average and instantaneous velocity in 2-D Average and instantaneous acceleration

PHYS PRACTICE EXAM 2

Axis. Axis. Axis. Solid cylinder (or disk) about. Hoop about. Annular cylinder (or ring) about central axis. central axis.

Ch 26 - Capacitance! What s Next! Review! Lab this week!

A 1.3 m 2.5 m 2.8 m. x = m m = 8400 m. y = 4900 m 3200 m = 1700 m

1 Using Integration to Find Arc Lengths and Surface Areas

The shortest path between two truths in the real domain passes through the complex domain. J. Hadamard

1. Kinematics of Particles

1. Consider a PSA initially at rest in the beginning of the left-hand end of a long ISS corridor. Assume xo = 0 on the left end of the ISS corridor.

Electric Potential. and Equipotentials

Lecture 5. Chapter 3. Electromagnetic Theory, Photons, and Light

Relative and Circular Motion

3 Motion with constant acceleration: Linear and projectile motion

Homework 5 for BST 631: Statistical Theory I Solutions, 09/21/2006

Magnetostatics Bar Magnet. Magnetostatics Oersted s Experiment

Invert and multiply. Fractions express a ratio of two quantities. For example, the fraction

D zone schemes

QUESTION PAPER PHYSICS-PH

Section 35 SHM and Circular Motion

TIME VARYING MAGNETIC FIELDS AND MAXWELL S EQUATIONS

4. Compare the electric force holding the electron in orbit ( r = 0.53

Electric Potential and Gauss s Law, Configuration Energy Challenge Problem Solutions

ARC 202L. Not e s : I n s t r u c t o r s : D e J a r n e t t, L i n, O r t e n b e r g, P a n g, P r i t c h a r d - S c h m i t z b e r g e r

Electric Potential. Outline. Potential Energy per Unit Charge. Potential Difference. Potential Energy Difference. Quiz Thursday on Chapters 23, 24.

Chapter 28 Sources of Magnetic Field

Physic 231 Lecture 4. Mi it ftd l t. Main points of today s lecture: Example: addition of velocities Trajectories of objects in 2 = =

Axis Thin spherical shell about any diameter

One of the common descriptions of curvilinear motion uses path variables, which are measurements made along the tangent t and normal n to the path of

2D Motion WS. A horizontally launched projectile s initial vertical velocity is zero. Solve the following problems with this information.

Exam 1 September 21, 2012 Instructor: Timothy Martin

Equations from The Four Principal Kinetic States of Material Bodies. Copyright 2005 Joseph A. Rybczyk

CHAPTER 29 ELECTRIC FIELD AND POTENTIAL EXERCISES

PHY 213. General Physics II Test 2.

Physics 11b Lecture #11

Energy Dissipation Gravitational Potential Energy Power

Q2.1 This is the x t graph of the motion of a particle. Of the four points P, Q, R, and S, the velocity v x is greatest (most positive) at

Solutions to Midterm Physics 201

Mechanics Acceleration The Kinematics Equations

For this purpose, we need the following result:

Homework 3 MAE 118C Problems 2, 5, 7, 10, 14, 15, 18, 23, 30, 31 from Chapter 5, Lamarsh & Baratta. The flux for a point source is:


Circular Motion. Radians. One revolution is equivalent to which is also equivalent to 2π radians. Therefore we can.

Chapter 2. Motion along a straight line. 9/9/2015 Physics 218

Field due to a collection of N discrete point charges: r is in the direction from

Faraday s Law. Faraday s Law. Faraday s Experiments. Faraday s Experiments. Magnetic Flux. Chapter 31. Law of Induction (emf( emf) Faraday s Law

( ) ( ) ( ) ( ) ( ) # B x ( ˆ i ) ( ) # B y ( ˆ j ) ( ) # B y ("ˆ ( ) ( ) ( (( ) # ("ˆ ( ) ( ) ( ) # B ˆ z ( k )

Physics 100: Lecture 1

That is, the acceleration of the electron is larger than the acceleration of the proton by the same factor the electron is lighter than the proton.

(a) Counter-Clockwise (b) Clockwise ()N (c) No rotation (d) Not enough information

Addition & Subtraction of Polynomials

Available online Journal of Scientific and Engineering Research, 2018, 5(10): Research Article

Chapter 10. Simple Harmonic Motion and Elasticity. Goals for Chapter 10

CHAPTER? 29 ELECTRIC FIELD AND POTENTIAL EXERCISES = 2, N = (5.6) 1 = = = = = Newton

Physics 122, Fall December 2012

MATH 124 AND 125 FINAL EXAM REVIEW PACKET (Revised spring 2008)

Sharif University of Technology - CEDRA By: Professor Ali Meghdari

INTEGRALS. Exercise 1. Let f : [a, b] R be bounded, and let P and Q be partitions of [a, b]. Prove that if P Q then U(P ) U(Q) and L(P ) L(Q).

School of Electrical and Computer Engineering, Cornell University. ECE 303: Electromagnetic Fields and Waves. Fall 2007

Class Summary. be functions and f( D) , we define the composition of f with g, denoted g f by

Solutions to Problems : Chapter 19 Problems appeared on the end of chapter 19 of the Textbook

Viscous Damping Summary Sheet No Damping Case: Damped behaviour depends on the relative size of ω o and b/2m 3 Cases: 1.

THERMAL PHYSICS. E nc T. W PdV. degrees of freedom. 32 m N V. P mv. Q c. AeT (emitted energy rate) E Ae T Tsurroundings. Q nc p

Magnetic Fields Due to Currents

Transcription:

Objecie F s w Tnsfome Moionl To be ble o fin nsfome. moionl nsfome n moionl. 331 1 331 Mwell s quion: ic Fiel D: Guss lw :KV : Guss lw H: Ampee s w Poin Fom Inegl Fom D D Q sufce loop H sufce H I enclose J enclose loop lecic i n mgneic fiels fil e no ele. l (is inepenenl. Then how oes n elecomgneic ( fiel eis? enclose F s peimen Hpohesis: If cuen coul inuce mgneic fiel (Ose s epeimen hen he mgneic fiel shoul be ble o inuce cuen. 3 ws h he mgneic fiel inuces cuen in coil. moing coil in h oes no chnges wih ime coil in he ime ing moing coil in ime ing 331 3 331 4

F sw Fom epeimen, cuen is inuce when o he posiion(e of he coil chnges wih ime. Quni ele o n e mgneic flu uen inuce b ing mgneic flu wih ime. Define h he cuen is inuce b elecic poenil iffeence clle elecomoie foce ( Ψ ol 331 5 Fs s s w( elecic poenil wihin he loop n is efine s follows. Ψ ol I is no necess h he close loop is he conucing loop. en s lw: The iecion of is in such iecion h euces he mgniue of he oiginl mgneic flu 331 6 lose oop in F s lw The following ssem is loce in he e whee he mgneic fiel oes no chnge wih ime. Does he lo eceies n cuen? luion: F sw Ψ Ψ ol ol wich is uning on n off he fequenc of 5H. o o chnges wih ime (Tnsfome nsfome onsie one insnce. Moing coil (Moionl moionl onsie one picul insnce. The mgneic flu insie he pink e is consn. No is inuce. Theefoe, hee is no cuen. AN 331 7 nsfome moionl 331 8

Poof: Tnsfome Ψ Ψ l ol Fom he efiniion ii of mgneic flu, No moemen, consn. 331 9 Poof: Moionl onsie consn so is inuce fom he chnge of coil s posiion (shpe. δ Ψol ( δ ( lim δ δ op boom sie lim lim δ δ δ δ δ sie lim δ δ sie ( 331 1 Moionl ( moionl Foce on he filmen cing inuce I in :, I F F I Foce esising he moemen. 331 11.1: Tnsfome I? Theoem: 1. F s lw:. Ohm s lw: V I 4 cm.cos5.75sin 4 1.cos314 T N5 uns, 1Ω nsfome moionl oluion: 1 Fin nsfome. nsfome 1loop 189.4sin 314 (1. 314 sin 314 331 1

.1: Tnsfome ( Fin fo 5 uns. nsfome 5 ( nsfome 1loop 3 No moion so moionl. Fin ol. 947 sin 314 nsfome moionl 947 sin 314 V 4 Fin cuen fom Ohm s lw. V 947sin 314 V I I 9.47sin314 A AN 1 5 Fin he iecion of he cuen. Posiie cuen is he iecion of he cuen hing he inuce in he sme iecion s. uen in iecion. AN 331 13 I.: uen n oss σ sin( T uen?, oss? Theoem: 1. F s lw: b nsfome moionl. Ohm s lw: V I 3. Powe: P VI [W] oluion: 1 Fin nsfome. nsfome No moion so moionl. Fin ol. nsfome moionl cos( ( ( π π cos( V 331 14.: uen n oss ( 3 Fin he iecion of he cuen. Negie cuen inices h he iecion is opposie. 4 Fin esisnce: uen in iecion. σ A π ( σ ( b 5 Fin cuen on ech hin esisnce: V π cos( σb cos( I π σ ( b 6 Inege fo he enie isc. σ b cos( I e A AN 4 331 15.: uen n oss (3 7 Fin powe loss in hin esisnce: σ b cos( P VI π cos( 8 Inege fo he enie isc: 4 πσb cos ( P 4 πσb 4 cos ( W AN 8 Quesion: Wh on we use P VI? 331 16

/s w.3: Moionl I? T l Theoem: 1. F s lw: nsfome moionl Ω Ohm s. lw: V I oluion: 1 oes no chnge wih ime nsfome. Fin onl moionl. moionl op ( sin cos cos boom cos 331 17 l.3: Moionl ( igh lef wcos Fin moionl (which is lso ol in his quesion. lef op wl cos wl cos V 3 Fin I b Ohm s lw. igh boom V wl cos V I I A AN 331 18.4: F s Disc Geneo I? /s /s T li oluion: Theoem: 1. F s lw: nsfome moionl. Ohm s lw: V I 1 oes no chnge wih ime nsfome. Fin onl moionl. moionl.4: F s Disc Geneo ( I Fin moionl (which is lso he ol in his quesion. moionl c lef 3 Fin I b Ohm s lw. op ( 1 igh V 1 V I I A AN boom 331 19 331

Noe: No-o n o Volge onsie he cse when he isc hs he esisnce. I? /s T oluion: The eluion fo is no ele o insie he isc so is he sme. onsie he equilen cicui. In he close loop,, V I No-lo olge: Open cicui ( V lo V o olge: V lo is less hn V becuse hee is he olge op in he isc 331 (geneo. 1 V A V V lo A w.5: Moionl A ighl woun ecngul coil hing N uns is oing in unifom mgneic fiel. Deemine he inuce in he coil. T N uns /s Theoem: 1. F s lw: nsfome moionl oluion: 1 ince oes no chnges wih ime. Fin onl moionl. moionl op ( sin cos cos boom cos 331 331.5: Moionl ( igh wcos lef w cos( 18 wcos Fin fo one loop. (ecll h nsfome so his is lso ol. 1loop /s T w N uns lef op igh boom w cos w cos wl cos AcosV 3 Fin fo N loop. loops ( N N 1loop N Acos V AN 3.5: Moionl (3 Meho : Using he oiginl fomul. Ψ Theoem: 1. F s lw: ol oluion: 1 Fin e of chnging flu in 1 loop Ψ 1loop ( ( ( cos sin Ae of ecngul coil A sin( A cos( ( ( 331 4

.5: Moionl (4 Fin chnging e of ol flu (N uns Ψ Ψ 1 N ol loop 3 Fin fom F s lw. Ψ ol ( N Acos N Acos V AN 4 Fin he iecion of he cuen b igh-hn ule: Posiie cuen is he iecion of he cuen hing he inuce in he sme iecion s. I AN.6: Moionl Tnsfome cos( T Theoem: 1. F s lw: nsfome moionl I?. Ohm s lw: V I oluion: 1 chnges wih ime. nsfome nsfome sin( ( sin( ( sin( ( 331 5 331 6.6: Moionl Tnsfome ( Thee is moion (fom he igh il. o fin moionl. cos( T moionl igh ih cos( ( cos( moionl cos( ( ( cos( 3 Fin ol nsfome moionl 4 Fin I b Ohm s lw. V V I I sin( cos( ( cos( sin( A AN 331 7 Diecion of he cuen is s shown b. AN.7: Moionl Tnsfome cos(cos(k T b (? oluion: Theoem: 1. F s lw: nsfome moionl 1 chnges wih ime. nsfome onsie when he lef sie of he loop is m. b sin( cos k ( nsfome sin( cos( k ( ( b sin( (sin( k( sin( k k 331 8

.7: Moionl Tnsfome ( Thee is moion in he loop. o fin moionl. moionl lef cos( cos( k ( cos( cos( k ( igh cos( cos( k( ( cos( cos( k( ( moionl b cos( cos( k b cos( cos( k( V o 3 Fin ol nsfome moionl b ( sin( i( k ( sin( i( k sin( i( k b cos( k cos( k( cos( ( Diecion of he cuen is s shown b. AN AN 331 9 Tnsfome : poin fom nsfome Fom oke s Theoem 331 3 Moionl : poin fom Moionl moionl Fom okes heoem ( Poin fom of moionl : ( 331 31 F s w: umm F s w: Inegl fom ( Poin fom ( onsie onl his p. 331 3

2024 © sciencedocbox.com Privacy Policy | Terms of Service | Feedback