Physics 202, Lecture 13. Today s Topics

Similar documents
Physics 202, Lecture 14

Physics 202, Lecture 13. Today s Topics. Magnetic Forces: Hall Effect (Ch. 27.8)

Chapter 7 Steady Magnetic Field. september 2016 Microwave Laboratory Sogang University

Physics 2135 Exam 3 April 21, 2015

Handout 8: Sources of magnetic field. Magnetic field of moving charge

in a uniform magnetic flux density B = Boa z. (a) Show that the electron moves in a circular path. (b) Find the radius r o

F is on a moving charged particle. F = 0, if B v. (sin " = 0)

Physics 3323, Fall 2016 Problem Set 7 due Oct 14, 2016

Unique Solutions R. All about Electromagnetism. C h a p t e r. G l a n c e

This final is a three hour open book, open notes exam. Do all four problems.

Reference. Vector Analysis Chapter 2

Lecture 13 - Linking E, ϕ, and ρ

Homework Assignment 9 Solution Set

Ch 30 - Sources of Magnetic Field

Physics 1402: Lecture 7 Today s Agenda

PHYS152 Lecture 8. Eunil Won Korea University. Ch 30 Magnetic Fields Due to Currents. Fundamentals of Physics by Eunil Won, Korea University

- 5 - TEST 2. This test is on the final sections of this session's syllabus and. should be attempted by all students.

Chapter 29. Magnetic Fields due to Currentss

Reading from Young & Freedman: For this topic, read the introduction to chapter 24 and sections 24.1 to 24.5.

Homework Assignment 6 Solution Set

Physics 202, Lecture 14

A = Qinside. E d. Today: fundamentals of how currents generate magnetic fields 10/7/15 2 LECTURE 14. Our Study of Magnetism

Energy creation in a moving solenoid? Abstract

PHYSICS ASSIGNMENT-9

μ 0 I enclosed = B ds

Ampere s law. Lecture 15. Chapter 32. Physics II. Course website:

#6A&B Magnetic Field Mapping

Version 001 HW#6 - Electromagnetism arts (00224) 1

Candidates must show on each answer book the type of calculator used.

ECE 3318 Applied Electricity and Magnetism. Spring Prof. David R. Jackson Dept. of ECE. Notes 31 Inductance

Summary of equations chapters 7. To make current flow you have to push on the charges. For most materials:

Problem Set 4: Mostly Magnetic

Magnetic Fields due to Currents

Lecture 1: Electrostatic Fields

ragsdale (zdr82) HW2 ditmire (58335) 1

Physics 202, Lecture 10. Basic Circuit Components

Week 10: Line Integrals

CAPACITORS AND DIELECTRICS

Sources of the Magnetic Field

Problem 1. Solution: a) The coordinate of a point on the disc is given by r r cos,sin,0. The potential at P is then given by. r z 2 rcos 2 rsin 2

Physics 1402: Lecture 17 Today s Agenda

Magnetic forces on a moving charge. EE Lecture 26. Lorentz Force Law and forces on currents. Laws of magnetostatics

Physics Jonathan Dowling. Lecture 9 FIRST MIDTERM REVIEW

The Steady Magnetic Field LECTURE 7

200 points 5 Problems on 4 Pages and 20 Multiple Choice/Short Answer Questions on 5 pages 1 hour, 48 minutes

Chapter 4 Syafruddin Hasan

Mathematics. Area under Curve.

Physics 227: Lecture 16 Ampere s Law

Potential Formulation Lunch with UCR Engr 12:20 1:00

Problem Solving 7: Faraday s Law Solution

Chapter 27 Sources of Magnetic Field

Magnetic Fields Part 2: Sources of Magnetic Fields

Motion of Electrons in Electric and Magnetic Fields & Measurement of the Charge to Mass Ratio of Electrons

Lecture 27: MON 26 OCT Magnetic Fields Due to Currents II

Homework Assignment 3 Solution Set

Chapter 30. Sources of the Magnetic Field Amperes and Biot-Savart Laws

Elements of Physics II. Agenda for Today. Physics 201: Lecture 1, Pg 1

Electricity and Magnetism

CHAPTER 30: Sources of Magnetic Fields

Magnetic Fields due to Currents

Physics 207 Lecture 5

CHETTINAD COLLEGE OF ENGINEERING & TECHNOLOGY NH-67, TRICHY MAIN ROAD, PULIYUR, C.F , KARUR DT.

12:40-2:40 3:00-4:00 PM

Sources of Magnetic Field I

THREE-DIMENSIONAL KINEMATICS OF RIGID BODIES

Announcements This week:

PHYS 1444 Section 501 Lecture #17

Ch. 28: Sources of Magnetic Fields

Physics 2212 G Quiz #4 Solutions Spring 2018 = E

Ampere s Law. Outline. Objectives. BEE-Lecture Notes Anurag Srivastava 1

We partition C into n small arcs by forming a partition of [a, b] by picking s i as follows: a = s 0 < s 1 < < s n = b.

Phys 4321 Final Exam December 14, 2009

March 11. Physics 272. Spring Prof. Philip von Doetinchem

Physics 212. Faraday s Law

Version 001 HW#6 - Electromagnetic Induction arts (00224) 1 3 T

Every magnet has a north pole and south pole.

ECE 470 Electric Machines Review of Maxwell s Equations in Integral Form. 1. To discuss a classification of materials

Physics 202 Review Lectures

18. Ampere s law and Gauss s law (for B) Announcements: This Friday, Quiz 1 in-class and during class (training exam)

Agenda for Today. Elements of Physics II. Forces on currents

n Higher Physics 1B (Special) (PHYS1241) (6UOC) n Advanced Science n Double Degree (Science/Engineering) n Credit or higher in Physics 1A

Physics 202, Lecture 12. Today s Topics

That reminds me must download the test prep HW. adapted from (nz118.jpg)

DIRECT CURRENT CIRCUITS

INTRODUCTION MAGNETIC FIELD OF A MOVING POINT CHARGE. Introduction. Magnetic field due to a moving point charge. Units.

Experiment No: EM 4 Experiment Name: Biot-Savart Law Objectives:

Math 8 Winter 2015 Applications of Integration

Physics 2212 GH Quiz #4 Solutions Spring 2016

Physics 202, Lecture 14

Physics 2135 Exam 1 February 14, 2017

For the flux through a surface: Ch.24 Gauss s Law In last chapter, to calculate electric filede at a give location: q For point charges: K i r 2 ˆr

Physics 1402: Lecture 18 Today s Agenda

Magnetostatics. Lecture 23: Electromagnetic Theory. Professor D. K. Ghosh, Physics Department, I.I.T., Bombay

B for a Long, Straight Conductor, Special Case. If the conductor is an infinitely long, straight wire, θ 1 = 0 and θ 2 = π The field becomes

III.Sources of Magnetic Fields - Ampere s Law - solenoids

Chapter 5. Magnetostatics

Today in Physics 122: work, energy and potential in electrostatics

Chapter 28 Sources of Magnetic Field

Chapter 28 Sources of Magnetic Field

Magnetic Fields! Ch 29 - Magnetic Fields & Sources! Magnets...! Earth s Magnetic Field!

Transcription:

Physics 202, Lecture 13 Tody s Topics Sources of the Mgnetic Field (Ch. 30) Clculting the B field due to currents Biot-Svrt Lw Emples: ring, stright wire Force between prllel wires Ampere s Lw: infinite wire, solenoid, toroid Displcement Current: Ampere-Mwell Mgnetism in Mtter On WebAssign Tonight: Homework #6: due 10/22,10 PM. Optionl reding quiz: due 10/19, 7 PM Mgnetic Fields of Current Distributions First, review: bck to electrosttics: Two wys to clculte the electric field directly: Coulomb s Lw d E = 1 4" 0 dq r 2 Guss Lw E i d A = q in " 0 ˆr "Brute force" "High symmetry" ε 0 = 8.8510-12 C 2 /(N m 2 ): permittivity of free spce 1

Mgnetic Fields of Current Distributions Two Wys to clculte the mgnetic field: Biot-Svrt Lw ( Brute force ) Ampere s Lw ( high symmetry ) d B = µ I 0 d l " ˆr 4 r 2 Bid s " = µ 0 I enclosed AMPERIAN LOOP I µ 0 = 4π10-7 T m/a: permebility of free spce Biot-Svrt Lw... dd up the pieces dl d B = µ 0 I 4 d l ˆr r 2 = µ 0 I 4 d l r r 3 θ r X db I B field circultes round the wire Use right-hnd rule: thumb long I, fingers curl in direction of B. 2

B Field of Circulr Current Loop on Ais (Tet emple 30.3) B field on is of current loop is: µ0 IR 2 B = 2( 2 + R 2 ) 3 / 2 Bcenter = µ0 I 2R See lso: center of rc (tet emple 30.2) B of Circulr Current Loop: Field Lines B 3

B Field of Stright Wire, length L (Tet emple 30.1) B field t point P is: B = µ 0 I (cos 1 # cos 2) 4" When the length of the wire is infinity: B = µ 0I 2 θ 1 θ 2 (return to this lter with Ampere s Lw) Superposition: emple Ch. 30, #3 Quick Question 1: Mgnetic Field nd Current Which figure represents the B field generted by the current I.................................................................. 4

Quick Question 2: Mgnetic Field nd Current Which figure represents the B field generted by the current I. Forces between Current-Crrying Wires A current-crrying wire eperiences force in n eternl B-field, but lso produces B-field of its own. Mgnetic Force: I b d F F I Prllel currents: ttrct d F F I b I Opposite currents: repel 5

Emple: Two Prllel Currents Force on length L of wire b due to field of wire : B = µ I 0 2d F b = I b d l B µ " = 0 I b I L 2#d d F F I b I Force on length L of wire due to field of wire b: B b = µ 0 I b 2d F = I d l B µ " b = 0 I I b L 2#d Ampere s Lw: " ny closed pth Ampere s Lw Bid s = µ 0 I enclosed I ds pplies to ny closed pth, ny sttic B field useful for prcticl purposes only for situtions with high symmetry Ampere s Lw cn be derived from Biot-Svrt Lw Generlized form: Ampere-Mwell (lter in lecture) 6

Ampere s Lw: B-field of Stright Wire Use symmetry (nd Ampere s Lw) I Bi d " s = µ 0 # I Choose loop to be circle of rdius R centered on the wire in plne to wire. Why? Mgnitude of B is constnt (function of R only) Direction of B is prllel to the pth. # Bi d s " = # Brd = 2"rB = µ 0 I B = µ I 0 2r Quicker nd esier method for B field of infinite wire (due to symmetry -- compre Guss Lw) B Field Inside Long Wire Totl current I flows through wire of rdius into the screen s shown. Wht is the B field inside the wire? By symmetry -- tke the pth to be circle of rdius r: r Bid s = B2r " Current pssing through circle: I enclosed = r 2 2 I Therefore: Ampere s Lw gives " Bid s = B2r = µ o I enclosed B = µ 0Ir 2 2 7

B Field of Long Wire Inside the wire: (r < ) B = µ 0 I 2 r 2 Outside the wire: ( r > ) B = µ 0 I 2 r B r Ampere s Lw: Toroid (Tet emple 30.5) Using Ampere s Lw, B field inside toroid is found to be: B = µ 0NI 2 r 8

Ampere s Lw: Toroid Toroid: N turns with current I. B φ =0 outside toroid (Consider integrting B on circle outside toroid: net current zero) B φ inside: consider circle of rdius r, centered t the center of the toroid. Bid s = B2r = µ o I enclosed " B = µ 0 NI 2r r B B Field of Solenoid Solenoid: source of uniform B field (inside of it) Solenoid: current I flows through wire wrpped n turns per unit length on cylinder of rdius nd length L. L To clculte the B-field, use Biot-Svrt nd dd up the field from the different loops. If << L, the B field is to first order contined within the solenoid, in the il direction, nd of constnt mgnitude. In this limit, cn clculte the field using Ampere's Lw 9

Ampere s Lw: Solenoid The B field inside n idel solenoid is: (see bord) B = µ 0nI n=n/l idel solenoid segment 3 t Solenoid: Field Lines Right-hnd rule: direction of field lines. In this emple, since the field lines leve the left end, the left end is the north pole. Like br mgnet (ecept it cn be turned on nd off) 10