Chapter 27: Magnetic Field and Magnetic Forces

Transcription

1 Chapter 27: Magnetic Field and Magnetic Forces Iron ore found near Magnesia Compass needles align N-S: magnetic Poles North (South) Poles attracted to geographic North (South) Like Poles repel, Opposites Attract No Magnetic Monopoles Magnetic Field Lines = direction of compass deflection. Electric Currents produce deflections in compass direction. =>Unification of Electricity and Magnetism in Maxwell s Equations. p212c27: 1 Magnetic Fields in analogy with Electric Fields Electric Field: Distribution of charge creates an electric field E(r) in the surrounding space. Field exerts a force F=q E(r) on a charge q at r Magnetic Field: Moing charge or current creates a magnetic field (r) in the surrounding space. Field exerts a force F on a charge moing q at r (emphasis this chapter is on force law) p212c27: 2

2 Magnetic Fields and Magnetic Forces Magnetic Force on a moing charge proportional to electric charge perpendicular to elocity proportional to speed (for a gien geometry) perpendicular to Magnetic Field proportional to field strength (for a gien geometry) F = q p212c27: 3 F = q F = q sinθ = q ( ) F F = q F = q F F = q F F = q p212c27: 4

3 TYU Which path will the charge take? The uniform magnetic field is going into the page, as indicated by the blue crosses p212c27: 5 Magnetic Fields Units of Magnetic Field Strength: [] = [F]/([q][]) = N/(C m s -1 ) = Tesla Defined in terms of force on standard current CGS Unit 1 Gauss = 10-4 Tesla Earth's field strength ~ 1 Gauss Direction = direction of elocity which generates no force Electromagnetic Force: F = q ( E ) = Lorentz Force Law p212c27: 6

4 Magnetic Field Lines and Magnetic Flux Magnetic Field Lines Mapped out with compass Are not lines of force (F is not parallel to ) Field Lines neer intersect Magnetic Flux r r dφ = da r r Φ = da r r da = 0 no magnetic charge! (no monopoles) p212c27: 7 Magnetic field lines of some sources circular current loop closely stacked current loops (solenoid) nearly uniform field between the poles of a permanent magnet field around long straight current p212c27: 8

5 TYU (A) Moing along the central axis of a current loop, starting from the left of the loop and ending to the right of the loop, the magnetic field strength would 1) remain the same at all points 2) increase then decrease 3) decrease then increase () Would the magnetic field direction change along this path? p212c27: 9 SI Unit of Flux: 1Weber = 1Tesla x 1 m 2 for a small area = dφ /da = Magnetic Flux Density Flux through an open surface will play an important role p212c27: 10

6 Motion of Charged Particles in a Magnetic Field Charged Particle moing perpendicular to the Magnetic Field Circular Motion! (simulations) F F p212c27: 11 Charged Particle moing perpendicular to a uniform Magnetic Field m F = q = R m R = q q ω = = R m = cyclotron frequency 2 R In 3-D, helical motion: p212c27: 12

7 In a non-uniform field: Magnetic Mirror F Net component of force away from concentration of field lines. Magnetic ottle Van Allen Radiation elts p212c27: 13 Work done by the Magnetic Field on a free particle: r r dw = F dx r r r = q dt ( ) = 0! => no change in Kinetic Energy! Motion of a free charged particle in any magnetic field has constant speed. p212c27: 14

8 TYU If the speed of a charge moing perpendicular to a uniform magnetic field is doubled, the radius of the resulting circular motion will 1) be quadrupled 2) be doubled 3) not change 4) be haled 5) be quartered p212c27: 15 Applications of Charged Particle Motion in a Magnetic Field Recall: Charged Particle moing perpendicular to a uniform Magnetic Field F = q = m R m R = q 2 R p212c27: 16

9 Velocity Selector makes use of crossed E and to proide opposing forces E upwards F = q downwards F = qe No net deflection => forces exactly cancel: q = q E = E/ p212c27: 17 J. J. Thomson s Measurement of e/m Electron Gun V and elocity selector: = 1 2 e m E m 2 = ev 2 E = 2V 2 E e/m = 1.76x10 11 C/kg with Millikan s measurement of e => mass of electron p212c27: 18

10 Example: Using an accelerating Potential of 150 V and a transerse Electric Field of 6x10 6 N/C. Determine a) the speed of the electrons, b) the magnetic field magnitude required for no net deflection p212c27: 19 Mass Spectrometer R 2 R 1 E One method elocity selector circular trajectory E = R = m q m = Rq = Rq E 2 p212c27: 20

11 Example: Vacuum System Leak Detector uses Helium atoms. Ionized helium atoms (He ) are detected with a mass spectrometer with a magnetic field strength of.1 T. With a elocity selector tuned to 1x10 5 m/s, where must the detector be placed to detect 4 He ions? p212c27: 21 TYU Mass spectrometer and similar instruments work by using singly ionized atoms (atoms with a single electron remoed). If an atom were doubly ionized, then the speed it would need to hae in order to trael straight through the elocity selector would 1) be quadrupled 2) be doubled 3) not change 4) be haled 5) be quartered p212c27: 22

12 Magnetic Force on a Current Carrying Wire I A d dl F i r r F = Fi = qii r r r r = Nqd = n olume qd r r r r = nadlqd = JAdl r r = Idl ( RHR) p212c27: 23 Example: A 1-m bar carries 50 A from west to east in a 1.2 T field directed 45 North of East. What is the magnetic force on the bar? I dl Force will be directed upwards (out of the plane of the page) F = IL sinθ = 50A 1m 12. T sin 45 = 42. 4N o p212c27: 24

13 Torque on a Current Loop (from F = I l x ) Rectangular loop in a magnetic field (directed along z axis) short side length a, long side length b, tilted with short sides at an angle with respect to, long sides still perpendicular to. F a F b F b F a Forces on short sides cancel: no net force or torque. Forces on long sides cancel for no net force but there is a net torque. p212c27: 25 Torque calculation: Side iew moment arm a/2 sin θ θ F b = Ib F b τ = F b a/2 sin θ F b a/2 sin θ = Iab sin θ = I A sin θ = Ι Α Β = µ Β Magnetic Dipole acts similarly to Electric Dipole U = µ. Β magnetic moment µ θ Switch current direction eery 1/2 rotation => DC motor p212c27: 26

14 TYU (A) What is the direction of the magnetic moment of the loop shown? 1) to the right 2) to the left () Which side of the loop is equialent to the north magnetic pole? 1) the right 2) the left p212c27: 27 Hall Effect y Conductor in a uniform magnetic field z z x J x Magnetic force on charge carriers F = q d F z = q d Charge accumulates on edges J x p212c27: 28

15 t Equilibrium: Magnetic Force = Electric Force on bulk charge carriers z Charge accumulates on edges F z = 0 = q d y q E z w d E = z y J nq nq E z x x = d = I y nq = J x y nq = VHt E z E y Hall EMF I = J tw y V H = E w z J x p212c27: 29 Negatie Charge carriers: elocity in negatie x direction magnetic force in positie z direction => resulting electric field has reersed polarity E y z J x p212c27: 30

16 Example: A ribbon of copper 2.0 mm thick and 1.5 cm wide carries a 75 A current in a.40 T magnetic Field. The resulting Hall emf is.81 µv. What is the density of charge carrying electrons? p212c27: 31 TYU A copper wire with a square cross section is oriented ertically, carrying a current downward. There is a uniform electric field from east to west. Which side of the wire is at the highest potential? 1) the north side 2) the south side 3) the east side 4) the west side p212c27: 32