F Q E v B MAGNETOSTATICS. Creation of magnetic field B. Effect of B on a moving charge. On moving charges only. Stationary and moving charges

Similar documents
Electrostatics (Electric Charges and Field) #2 2010

MAGNETIC FIELD INTRODUCTION

Review: Electrostatics and Magnetostatics

TUTORIAL 9. Static magnetic field

Ch 30 - Sources of Magnetic Field! The Biot-Savart Law! = k m. r 2. Example 1! Example 2!

2. Electrostatics. Dr. Rakhesh Singh Kshetrimayum 8/11/ Electromagnetic Field Theory by R. S. Kshetrimayum

16.1 Permanent magnets

3. Magnetostatic fields

( ) Make-up Tests. From Last Time. Electric Field Flux. o The Electric Field Flux through a bit of area is

Magnetic Fields Due to Currents

A moving charged particle creates a magnetic field vector at every point in space except at its position.

Unit 7: Sources of magnetic field

B da = 0. Q E da = ε. E da = E dv

Magnetic fields (origins) CHAPTER 27 SOURCES OF MAGNETIC FIELD. Permanent magnets. Electric currents. Magnetic field due to a moving charge.

PHYS 110B - HW #7 Spring 2004, Solutions by David Pace Any referenced equations are from Griffiths Problem statements are paraphrased

Gauss s Law: Circuits

EM-2. 1 Coulomb s law, electric field, potential field, superposition q. Electric field of a point charge (1)

Physics Spring 2012 Announcements: Mar 07, 2012

21 MAGNETIC FORCES AND MAGNETIC FIELDS

Magnetic Field. Conference 6. Physics 102 General Physics II

Magnetostatics. Magnetic Forces. = qu. Biot-Savart Law H = Gauss s Law for Magnetism. Ampere s Law. Magnetic Properties of Materials. Inductance M.

Section 1: Main results of Electrostatics and Magnetostatics. Electrostatics

Physics 2212 GH Quiz #2 Solutions Spring 2016

Sources of Magnetic Fields (chap 28)

EELE 3331 Electromagnetic I Chapter 4. Electrostatic fields. Islamic University of Gaza Electrical Engineering Department Dr.

Chapter 22 The Electric Field II: Continuous Charge Distributions

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

Physics 122, Fall October 2012

Calculate the electric potential at B d2=4 m Calculate the electric potential at A d1=3 m 3 m 3 m

Physics 2B Chapter 22 Notes - Magnetic Field Spring 2018

CHAPTER 25 ELECTRIC POTENTIAL

FARADAY'S LAW. dates : No. of lectures allocated. Actual No. of lectures 3 9/5/09-14 /5/09

Phys-272 Lecture 17. Motional Electromotive Force (emf) Induced Electric Fields Displacement Currents Maxwell s Equations

Flux. Area Vector. Flux of Electric Field. Gauss s Law

Phys102 Second Major-182 Zero Version Monday, March 25, 2019 Page: 1

The Law of Biot-Savart & RHR P θ

Conventional Current B = In some materials current moving charges are positive: Ionic solution Holes in some materials (same charge as electron but +)

University Physics (PHY 2326)

Physics 235 Chapter 5. Chapter 5 Gravitation

EM Boundary Value Problems

Physics 2A Chapter 10 - Moment of Inertia Fall 2018

( ) ( )( ) ˆ. Homework #8. Chapter 27 Magnetic Fields II.

Review. Electrostatic. Dr. Ray Kwok SJSU

Course Updates. Reminders: 1) Assignment #8 will be able to do after today. 2) Finish Chapter 28 today. 3) Quiz next Friday

FI 2201 Electromagnetism

1 Spherical multipole moments

Gauss Law. Physics 231 Lecture 2-1

PHY 114 A General Physics II 11 AM-12:15 PM TR Olin 101

7.2. Coulomb s Law. The Electric Force

Sources of the Magnetic Field. Moving charges currents Ampere s Law Gauss Law in magnetism Magnetic materials

Physics 122, Fall October 2012

$ i. !((( dv vol. Physics 8.02 Quiz One Equations Fall q 1 q 2 r 2 C = 2 C! V 2 = Q 2 2C F = 4!" or. r ˆ = points from source q to observer

DEVIL PHYSICS THE BADDEST CLASS ON CAMPUS IB PHYSICS

7.2.1 Basic relations for Torsion of Circular Members

Physics 2112 Unit 14

? this lecture. ? next lecture. What we have learned so far. a Q E F = q E a. F = q v B a. a Q in motion B. db/dt E. de/dt B.

Magnetic Field of a Wire

How Electric Currents Interact with Magnetic Fields

Algebra-based Physics II

ELECTRIC FIELD. decos. 1 dq x.. Example:

4. Electrodynamic fields

F = net force on the system (newton) F,F and F. = different forces working. E = Electric field strength (volt / meter)

! E da = 4πkQ enc, has E under the integral sign, so it is not ordinarily an

Chapter 31 Faraday s Law

Physics 2020, Spring 2005 Lab 5 page 1 of 8. Lab 5. Magnetism

ELECTROSTATICS::BHSEC MCQ 1. A. B. C. D.

FARADAY'S LAW dt

Faraday s Law (continued)

ELECTROMAGNETISM (CP2)

Prepared by: M. S. KumarSwamy, TGT(Maths) Page - 1 -

Introduction: Vectors and Integrals

Chapter 13 Gravitation

1 Fundamental Solutions to the Wave Equation

PHY2061 Enriched Physics 2 Lecture Notes. Gauss Law

(read nabla or del) is defined by, k. (9.7.1*)

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

Antennas & Propagation

Electrostatics. 1. Show does the force between two point charges change if the dielectric constant of the medium in which they are kept increase?

Exam 3, vers Physics Spring, 2003

Math 2263 Solutions for Spring 2003 Final Exam

ECE 3318 Applied Electricity and Magnetism. Spring Prof. David R. Jackson ECE Dept. Notes 13

Force between two parallel current wires and Newton s. third law

AE301 Aerodynamics I UNIT B: Theory of Aerodynamics

Qualifying Examination Electricity and Magnetism Solutions January 12, 2006

Module 05: Gauss s s Law a

Magnetic Dipoles Challenge Problem Solutions

Chapter 26: Magnetism: Force and Field

PHYS 1444 Lecture #5

Lecture 8 - Gauss s Law

Chapter 22: Electric Fields. 22-1: What is physics? General physics II (22102) Dr. Iyad SAADEDDIN. 22-2: The Electric Field (E)

Appendix B The Relativistic Transformation of Forces

AST 121S: The origin and evolution of the Universe. Introduction to Mathematical Handout 1

Current, Resistance and

q r 1 4πε Review: Two ways to find V at any point in space: Integrate E dl: Sum or Integrate over charges: q 1 r 1 q 2 r 2 r 3 q 3

Module 19: Sections 9.1 through 9.2 Module 20: Sections 9.3 through Table of Contents

Welcome to Physics 272

arxiv: v1 [physics.pop-ph] 3 Jun 2013

Hopefully Helpful Hints for Gauss s Law

(Sample 3) Exam 1 - Physics Patel SPRING 1998 FORM CODE - A (solution key at end of exam)

Electromagnetic scattering. Graduate Course Electrical Engineering (Communications) 1 st Semester, Sharif University of Technology

Transcription:

MAGNETOSTATICS Ceation of magnetic field. Effect of on a moving chage. Take the second case: F Q v mag On moving chages only F QE v Stationay and moving chages dw F dl Analysis on F mag : mag mag Qv. vdt Magnetic foces do no wok!

Cuent-caying wie v v t I v F mag vdq vdl I dl K di Kˆ K v dl Fmag v da K da Chage-flow ove a suface with a suface cuent density K dl K V l t d it Volume cuent density di da da v Flow F mag v d mag d vˆ

The amount of chage passing though aea A in time Δt is nq(avt). Amount flowing pe unit aea pe unit time is nqv, giving you the cuent density. j = nqv. dq I jds (ate of flow of chage) dt s Conside a closed suface enclosing volume V. If be the chage density fo an infinitesimal volume dv, then dv epesents the total chage inside the volume V. Accoding to the law of consevation o of chage, the ate of flow of chage though the enclosed suface is equal to the ate of decease of chage in it. s j ds - t V dv V dv t

Accoding to divegence theoem s --> j dv dv t j ds jdv V V V j t This is called the equation of continuity it and epesents the physical facts of consevation of chages.

IOT SAVAT LAW Ceation of magnetic field with the movement of chages. Unde steady state movement of chages (steady cuent), themagnetic field poduced is given by the iot-savat Law: (i) d I (ii) d dl 1 (iii) d (iv) d s in with k as constant of popotionality, p in SI units, k 1 4 7 NA Idl sin d 4 dl Idl p

Idl ( l ) d 4 3 ( l ) I 1 dl ( l) 3 4 ( l ) P l dl

EXAMPLE - 1 I dl (yeˆ ˆ z le x) 3/ 4 y l P y 1 Using the tan = l /y /y, we get the final esult as I sin sin eˆ 4y 1 z I sin eˆ 4y If - 1 = : z If = / Infinite line: y O l dl I e ˆ y z x I

EXAMPLE - I adeˆ zeˆ ˆ z ae 4 3/ z a azd e a d e I z 4 3/ z a ˆ ˆ I P O dl deˆ I a I a eˆ ˆ e 3/ z 3/ 4 z a z a z

If thee ae n tuns, then ni a z a 3/ eˆ z At the cente of the loop: I e ˆ a z

EXAMPLE - 3 P z 1 dz Use the evaluation of the magnetic field fo a loop and follow supeposition pinciple to evaluate fo a solenoid. The vaiable is now z. What about cuent? If thee ae n no. of tuns pe unit length and each tun having cuent I, then I a ni a dz d (ndz)e ˆ ˆ z e 3/ 3/ z z a z a

z a cot dz a cosec d z a a cot a a cosec ni ni sindeˆ cos cos eˆ 1 z 1 z Case 1 If we take a long solenoid (adius of the solenoid vey small compaed to its length) and obsevation point p is well with in the solenoid, then 1 =and = nieˆ z

Case When the obsevation point P is taken one end of the solenoid 1 = and = / ni Hence in case of semi-infinitely long solenoid, the magnetic field at y g, g a point at the end of the solenoid is half the magnetic field at a point well inside the solenoid.

Fo suface and volume cuents, iot-savat law becomes da K ˆ ) ( 4 ) ( d ˆ ) ( 4 ) ( 4 4

Infinite plane of unifom cuent sheet Constant cuent density K / y K / Cuent sheet lies in the xy plane and K Keˆ y xeˆ x yeˆ y x eˆ yeˆ x e y zeˆ xx eˆ y y eˆ zeˆ da z x y z x x y y z dxdy K Kzeˆ O da / / x x eˆ x z x P (x,y,z) z

Use iot-savat law fo a suface cuent density and integate ˆ ˆ d d 3 ˆ ˆ ' 4 x z ze x x e dx dy K x x y y x x y y z ˆ ˆ K ' ze X e dx dy 3 4 x z ze X e dx dy K X Y z x x X y y Y z-component will vanish because the integand is an odd function of X / x e x z z K K e ˆ 1 ˆ 1

The Divegence and cul of Nonzeo cul?? dl I dl s I I s dl If we use cylindical coodinates (s,, z) with cuent along z axis dl dseˆ I ˆ e s sdeˆ dzˆ s e z I 1 sd I d s I

undle of staight wies I 5 I 1 I Each wie that passes though the loop contibutes I dl I enc If the flow of chage is epesented by a volume chage density I enc da Applying stokes theoem Integal taken ove the suface bounded by the loop da da Above deivation is esticted by the condition that we need infinitely staight line cuents I 4 I 3

Divegence and cul of iot-savat law fo a volume cuent distibution is ( ) ˆ ( ) dv ' 4 is a function of is a function of x, y, z x, y, z x xeˆ x y y eˆ y z z eˆ z dv ' dxdydz d x, y, z x, y, z Integation is done ove the pimed coodinates Divegence and cul is done ove the unpimed coodinates

Applying divegence to the magnetic field due to a volume chage distibution d ˆ 4 4 A A A A A A ˆ ˆ ˆ because does not depend on unpimed coodinates ˆ and

Applying cul to the magnetic field due to a volume chage distibution d ˆ 4 ) ( ) ( A A A A A ˆ ˆ ˆ ˆ ˆ ˆ (a) The tems involving deivatives of is dopped since does not depend on (x,y,z) ) ( 4 ˆ 3 The second tem in (a) ) ( 4 ( ) integates to zeo 3 ( )4 ( )d ( ) 3 ( )4 ( )d () 4

How does the othe tem vanish? th d i ti t l t it h t ecause the deivative acts only on tem, we can switch to ˆ ˆ Conside the x-component x x x x x x ) ( ) ( 3 3 3 ) ( ) ( We ae dealing with steady cuents, hence second tems in zeo x x 3 ) ( ˆ Contibution to the integal fom this tem is da x x d x x 3 3 ) ( ) ( a d d S V 3 3 We ae integating ove the souce egion that include all the cuent. On the bounday the cuent is zeo and hence the suface integal vanishes

FUTHE EXAMPLES FO IOT SAVAT LAW b P a b I P a

AMPEE S LAW in diffeential fom Using Stokes theoem da da dl da is the total cuent passing though the suface-- I enc dl I enc in integal fom

EXAMPLES Example 1 s Ampeian loop I y symmety, the magnitude of is constant aound an ampeian loop of adius s dl dl s I enc I eˆ s I

Example Magnetic field of an infinite unifom suface cuent flowing ove the xy plane K Keˆ x can only have a y-component It points towads the left above the plane and towads the ight in the plane below K dl l I enc Kl Ampeian loop z K K z l

Example 3 Magnetic field of a vey long solenoid consisting of n closely wound tuns pe unit length on a cylinde of adius and caying a steady cuent I Loop1 dl Loop ( a) ( b) L I (a) (b) enc dl L I enc nil nieˆ Keˆ Inside the solenoid z z Outside the solenoid K ni a b 1 Ampeian loops L

EXAMPLE 4 A long coppe pipe with thick walls has an inne adius and an oute adius. What is the cuent density fo all? Use Ampee's Law to find the magnetic fields as a function of adial distance fom the cente of the pipe. <<: < <: () I 3 eˆ z I > :

EXAMPLE 5 Along coppe wie of coss-sectional adius caies a k cuent density ( ) e eˆ. Use Ampee's Law to z detemine as a function of the distance fom the cente of the wie. inside the wie: 1 ke e e k k k e e k k k outside the wie: k k k e e k k 1ke e e k k k

() A( ) POISSON S EQN. IN MAGNETOSTATICS da S Flux though any closed suface is always zeo. No monopole fo the magnetic chage A() is the vecto potential Divegence of a cul is always zeo A A A We can add to magnetic potential any function whose cul vanishes, with no effect on

We use this feedom to eliminate the divegence of A A If the oiginal potential A is not divegenceless, then add to it gadient of function such that, A becomes divegenceless. Gauge Tansfomation: A A A A 1 A d 4 A μ P i In the cuent fee egion, Poisson s eqn. Coulomb gauge A A Laplace Laplace eqn.

dl ˆ ( ) I 4 dl ˆ 1 dl dl dl I dl I dl 4 C 4 C Idl K( )da A( ) 4 4 4 C S dl A ( )d 4 V

DIVEGENCE OF A Idl I A( ) 4 C 4 C dl 1 1 dl dl A( I 1 ) dl 4 C oi 1 da 4 S dl 1 = dl

MAGNETIC SCALA POTENTIAL In the cuent fee egion, Theefoe, can be expessed as: m We call m as magnetic scala potential. We see that m satisfies the Laplace s equation. m m

ELECTOSTATICS AND MAGNETOSTATICS E E q enc Eds dl I V A V A I enc

MULTIPOLE EXPANSION FO MAGNETIC VECTO POTENTIAL Simila to the electic scala potential, one can use multipole expansion to find out the magnetic vecto potentialatafa away place due to a cuent distibution. p d d = dl

cos 1 cos d This expession can be ewitten as 1 1 1 d cos 1 d Using bionomial expansion 3 16 5 8 3 1 1 1 1 x x x x cos 8 3 cos 1 1 1 1 d 8 d Vecto potential due to the cuent loop is I dl A 4 d I 1 1 1 3 1 A d l cos d l ( ) cos d l Vecto potential due to the cuent loop is 4 d 3 A dl cos dl ( ) cos dl 4

As in multipole expession of V, The fist tem, goes like 1/, is monopole tem. The second tem, which goes like 1/,isdipole tem. The thid tem, goes like 1/ 3,isquadupole tem. The magnetic monopole tem is always zeo as the total vecto displacement aound close loop is dl= Hence no magnetic monopole exists in natue. In the absence of the monopole tem, the dominant tem is the dipole-except p in the ae case whee it, too vanishes. A dipole I 4 I 4 cos dl ˆ The integal in the above expession afte some manipulations can be witten as 1 ˆ dl ˆ dl μi 1 A = - ˆ dl dipole 4π dl

μ ˆ m A dip= 4π 1 m= I d l Whee m is the magnetic dipole moment of the loop, defined as If the cuent loop is a plane loop (cuent located on the suface of a plane, then dl Is the aea of the shaded tiangle as shown in figue. So the integal is the aea of whole loop line 1 dl Aea,a

In this case, the dipole moment of the cuent loop is equal to m I a Whee the diection of a must be consistent with the diection of the cuent loop (ight hand ule) Since the magnetic monopole tem is always zeo, the magnetic dipole moment always independent d of oigin. i is Assuming that the magnetic dipole is located at the oigin of ou coodinate system and that m is pointing along the positive z axis, μ m ˆ μ msinθ A dipole= = e ˆ 4π 4π Coesponding magnetic field is equal to dipole A dipole 1 m sin 1 m sin sin ˆ ˆ sin 4 4 m 4 dipole cos e 3 sin e ˆ ˆ 1 ˆ ˆ 4 3 me 3 e m

Poblem 5.38: A phonogaph ecod of adius, caying a unifom suface chage, is otating with a constant angula velocity. Find the magnetic dipole moment. Poblem 5.39: Find the magnetic dipole moment of a spinning spheical shell of adius, caies a unifom suface chage. Show that fo > the potential is that of pefect dipole. Poblem 5.41: Show that magnetic dipole moment of an abitay localized cuent loop is independent of the location of efeence point.

Magnetic dipole and magnetic dipole moment: A magnetic dipole consists of pai of magnetic dipole of equal and opposite stength sepaated by small distance. Examples of magnet dipoles ae Magnetic needle, a magnet, Cuent caying solenoid, A cuent loop etc. Atom is also consideed to behave like a dipole - so the fundamental magnetic dipole in natue is associated with the electons.

The poduct of pole stength of eithe poles and distance between them is called as magnetic dipole moment. The distance between two poles is called magnetic length. o m M l m M ( l ) m N l S The vecto l is diected fom south to noth. Thus the diection of magnetic dipole moment is fom south to noth S.I. units of M is ampee-mete (Am) and that of m is ampee-mete (Am ) o joule tesla -1

Field at a point due to magnetic dipole in the end on-position (on the axis) Let NS be a magnetic dipole of pole stength M and length l. Let P be the point on its axis at a distance d fom the cente of the dipole. l S N S p N d The magnetic field at point P due to noth pole is M 4 l d And will be diected away fom the magnet. The magnetic field at point P due to South pole pole is

M 4 l d And will be diected towads fom the magnet. Theefoe esultant field is = N - S 4Mld 4 d l md 4 l d Hee m = Ml is dipole moment of the magnet If d>>l mm 4 d 3

Field due to a magnetic dipole in the boad side on position Let NS be a magnetic dipole of pole stength M and length l. Let P be a point on the boad side on position of the dipole at the distance d fom its cente. N P S The magnetic field at point P due to noth pole is d N M PN 3 4 PN S l N The magnetic field at point P due to noth pole is

S M PS 3 4 PS Now NP = PS = (d +l ) 1/ Theefoe esultant field at P 4 4 4 M d l 3 / M d l 3 / M d l 3 / m 4 l 3 / d ( NP NS l PS)

If d >> l Thus m 3 4 d axial = equatoial The same conclusion we had dawn fo electic field

When a pemanent magnet is placed in a field, Noth pole will expeience a foce in the diection of field and south pole has a foce opposite to the pole. If the field is unifom the net foce is zeo, but thee is a toque. Fo electic dipole, the toque is given by elation N p E p: electic dipole moment and E: unifom electic field The coesponding expession fo toque of magnetic dipole in magnetic field is N m In addition to the pemanent magnets being dipole, we see that cuent loops ae also magnetic dipole.

Toque on a dipole (ba magnet) in a magnetic field: If a magnetic dipole is placed in a unifom magnetic field as shown in figue, the Noth and South poles of the magnet will expeience equal and opposite foces. N l m M M S Z Let M be the pole stength of each pole and moment m and magnetic field, then be the angle between magnetic dipole Foce on Noth pole = M along Foce on South pole = M opposite to These foces will constitute a couple which tends to otate the magnet in the diection of. Thus a magnet expeience a toque

N = Foce distance between the foce = M ZN = M (SN sin) = M (l sin) because In tiangle SZN, sin = ZN/SN o ZN = SN sin o N = (M l ) sin = msin In vecto fom N m When = 1 unit and = 9, then N = m Thus the magnetic dipole moment can be defined as the toque acting on a magnetic dipole placed nomal to the unifom magnetic field of unit stength

Let us calculate the toque on a ectangula cuent loop in a unifom field. Conside a ectangula loop of sides a and b as shown in figue placed in a unifom magnetic field and let the diection of the field is along z-axis. The magnetic dipole moment is pependicula to the cuent loop and makes an angle with magnetic field. F 1 F 4 I 1 4 I 3 a F F 3 b

Since the cuents ae opposite on opposite sides of the loop, the foces ae also opposite, so thee is no net foce on the loop (when the field is unifom). The foces on the loop sides 3 and 4 tend to stetch the loop, but do not otate the loop. ecause of the foces on the two sides maked 1 and, tend to otate the loop about y-axis and geneates the toque. The magnitude of foces F 1 and F is F 1 F Ib And thei moment o leve am is asin So the toque N is equal to N Iabsin msin

Whee m = Iab is the magnetic dipole moment of the loop O N m The toque given by above equation is a special case, the esult is ight fo small cuent loop of any shape in the unifom magnetic field. Wok Done on a Magnetic Dipole. Since a magnetic dipole placed in an extenal magnetic field expeiences a toque, wok (positive o negative) must be done by an extenal agent in ode to change the oientation of the dipole. Let us calculate how much wok is done by the field when otating the dipole fom angle A to. W A Nd

W m sin d A m cos cos A If the dipole is initially at ight angle to the field i.e. A = 9 and finally makes an angle with field i.e. =, then W m cos cos9 -mcos This wok done is equal to the potential of the dipole U p -m cos - m

In case of non-unifom field the above discussion is exact only fo a pefect dipole of infinitesimal size. Now we will calculate foce of a infinitesimal loop of dipole moment m in the field. We have seen that the potential enegy of a magnetic dipole m in a magnetic field is U p -m cos - m We know foce is elated to potential enegy by the elation F U Theefoe F m Using poduct ule (ule 4)

A A A A A Theefoe F m m m m m Since m is not function of space co-odinate Theefoe m and m And Theefoe

F m Povided thee is no extenal cuent at the actual location of the dipole We must be vey caeful about the analogies between electic and magnetic dipoles. Fo example foce on a magnetic dipole in non unifom field is F m Whee as fo electic field F p E So one should be vey alet when solving the poblems.