Chapter 28 Magnetic Field and Magnetic Forces Iron

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Chapter 28: Magnetic Field and Magnetic Forces Iron ore found near Magnesia Compass needles

Chapter 28: 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. p 212 c 28: 1

Magnetic Fields in analogy with Electric Fields Electric Field: – Distribution of charge creates

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: – Moving charge or current creates a magnetic field B(r) in the surrounding space. – Field exerts a force F on a charge moving q at r – (emphasis this chapter is on force law) p 212 c 28: 2

Magnetic Fields and Magnetic Forces Magnetic Force on a moving charge – proportional to

Magnetic Fields and Magnetic Forces Magnetic Force on a moving charge – proportional to electric charge – perpendicular to velocity v – proportional to speed v (for a given geometry) – perpendicular to Magnetic Field B – proportional to field strength B (for a given geometry) F=qv´B p 212 c 28: 3

F=qv´B F = |q| v B sinq = |q| v B (v ^ B)

F=qv´B F = |q| v B sinq = |q| v B (v ^ B) F B + v F F=qv´B F = |q| v^ B B + v^ v F F=qv´B F = |q| v B^ v + B^ B p 212 c 28: 4

Magnetic Fields Units of Magnetic Field Strength: [B] = [F]/([q][v]) = N/(C m s-1)

Magnetic Fields Units of Magnetic Field Strength: [B] = [F]/([q][v]) = 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 velocity which generates no force Electromagnetic Force: F=q(E+ v´B) = Lorentz Force Law p 212 c 28: 5

Magnetic Field Lines and Magnetic Flux Magnetic Field Lines Mapped out with compass Are

Magnetic Field Lines and Magnetic Flux Magnetic Field Lines Mapped out with compass Are not lines of force (F is not parallel to B) Field Lines never intersect Magnetic Flux d. FB = B. d. A p 212 c 28: 6

 • SI Unit of Flux: – 1 Weber = 1 Tesla x 1

• SI Unit of Flux: – 1 Weber = 1 Tesla x 1 m 2 – for a small area B = d. FB /d. A^ – B = “Magnetic Flux Density” Flux through an open surface will play an important role p 212 c 28: 7

Motion of Charged Particles in a Magnetic Field Charged Particle moving perpendicular to the

Motion of Charged Particles in a Magnetic Field Charged Particle moving perpendicular to the Magnetic Field – Circular Motion! – (simulations) F + v p 212 c 28: 8

Charged Particle moving perpendicular to a uniform Magnetic Field v + R + v

Charged Particle moving perpendicular to a uniform Magnetic Field v + R + v p 212 c 28: 9

In a non-uniform field: Magnetic Mirror v F B Net component of force away

In a non-uniform field: Magnetic Mirror v F B Net component of force away from concentration of field lines. Magnetic Bottle Van Allen Radiation Belts p 212 c 28: 10

Work done by the Magnetic Field on a free particle: => no change in

Work done by the Magnetic Field on a free particle: => no change in Kinetic Energy! Motion of a free charged particle in any magnetic field has constant speed. p 212 c 28: 11

Applications of Charged Particle Motion in a Magnetic Field Recall: Charged Particle moving perpendicular

Applications of Charged Particle Motion in a Magnetic Field Recall: Charged Particle moving perpendicular to a uniform Magnetic Field v + R + v p 212 c 28: 12

Velocity Selector makes use of crossed E and B to provide opposing forces +

Velocity Selector makes use of crossed E and B to provide opposing forces + - + - E upwards F=qv´B downwards F = q. E No net deflection => forces exactly cancel: |q| v B=|q| E v = E/B p 212 c 28: 13

J. J. Thomson’s Measurement of e/m Electron Gun + - + - + -

J. J. Thomson’s Measurement of e/m Electron Gun + - + - + - + + - - E V and velocity selector: e/m = 1. 76 x 1011 C/kg with Millikan’s measurement of e => mass of electron p 212 c 28: 14

Example: Using an accelerating Potential of 150 V and a transverse Electric Field of

Example: Using an accelerating Potential of 150 V and a transverse Electric Field of 6 x 106 N/C. Determine a) the speed of the electrons, b) the magnetic field magnitude required for no net deflection p 212 c 28: 15

Mass Spectrometer R 2 + + + R 1 - - - E One

Mass Spectrometer R 2 + + + R 1 - - - E One method velocity selector + circular trajectory p 212 c 28: 16

Example: Vacuum System Leak Detector uses Helium atoms. Ionized helium atoms (He +) are

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 velocity selector tuned to 1 x 105 m/s, where must the detector be placed to detect 4 He + ions? p 212 c 28: 17

Magnetic Force on a Current Carrying Wire B I A vd dl Fi p

Magnetic Force on a Current Carrying Wire B I A vd dl Fi p 212 c 28: 18

Example: A 1 -m bar carries 50 A from west to east in a

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 B magnetic force on the bar? I dl Force will be directed upwards (out of the plane of the page) p 212 c 28: 19

Torque on a Current Loop (from F = I l x B ) Rectangular

Torque on a Current Loop (from F = I l x B ) 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 B, long sides still perpendicular to B. B Fb Fa 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. p 212 c 28: 20

Torque calculation: Side view moment arm a/2 sin q Fb = IBb q Fb

Torque calculation: Side view moment arm a/2 sin q Fb = IBb q Fb t = Fb a/2 sin q + Fb a/2 sin q = Iab B sin q = I A ´B = m ´B magnetic moment m q B Magnetic Dipole ~ Electric Dipole U = - m. B Switch current direction every 1/2 rotation => DC motor p 212 c 28: 21

y Hall Effect Bz Conductor in a uniform magnetic field x z + Jx

y Hall Effect Bz Conductor in a uniform magnetic field x z + Jx Magnetic force on charge carriers F = q vd ´ B Fz = qvd. B Charge accumulates on edges + + - - - - - + + + + + - - Jx p 212 c 28: 22

Equilibrium: Magnetic Force = Electric Force on bulk charge carriers Bz E w t

Equilibrium: Magnetic Force = Electric Force on bulk charge carriers Bz E w t + + - - - + + + y - - + + - - Jx Charge accumulates on edges Fz = 0 = qvd. By + q Ez p 212 c 28: 23

Negative Charge carriers: velocity in negative x direction magnetic force in positive z direction

Negative Charge carriers: velocity in negative x direction magnetic force in positive z direction => resulting electric field has reversed polarity + - - + - + -Ey + + - + - + - + Bz Jx p 212 c 28: 24

Example: A ribbon of copper 2. 0 mm thick and 1. 5 cm wide

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 m. V. What is the density of charge carrying electrons? p 212 c 28: 25

p 212 c 28: 26

p 212 c 28: 26