P212c28: 1 Chapter 28: Magnetic Field and Magnetic Forces Iron ore found near Magnesia Compass needles align N-S: magnetic Poles North (South) Poles attracted.

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Presentation transcript:

p212c28: 1 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.

p212c28: 2 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)

p212c28: 3 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 = q v  B

p212c28: 4 F = q v  B F = |q| v B sin  = |q| v B (v  B) v B F + F = q v  B F = |q| v  B vv B F + v F = q v  B F = |q| v B  v B F + BB

p212c28: 5 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 = 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 Magnetic Fields

p212c28: 6 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  B = B. dA

p212c28: 7 SI Unit of Flux: –1Weber = 1Tesla x 1 m 2 –for a small area B = d  B /dA  –B = “Magnetic Flux Density” Flux through an open surface will play an important role

p212c28: 8 Motion of Charged Particles in a Magnetic Field Charged Particle moving perpendicular to the Magnetic Field –Circular Motion! –(simulations) + + v F F v

p212c28: 9 Charged Particle moving perpendicular to a uniform Magnetic Field + + v v R

p212c28: 10 In a non-uniform field: Magnetic Mirror Net component of force away from concentration of field lines. B v F Magnetic Bottle Van Allen Radiation Belts

p212c28: 11 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.

p212c28: 12 Recall: Charged Particle moving perpendicular to a uniform Magnetic Field + + v v R Applications of Charged Particle Motion in a Magnetic Field

p212c28: 13 Velocity Selector makes use of crossed E and B to provide opposing forces No net deflection => forces exactly cancel: |q| v B=|q| E v = E/B   E upwards F = q v  B downwards F = qE

p212c28: 14 J. J. Thomson’s Measurement of e/m Electron Gun and velocity selector:   E V e/m = 1.76x10 11 C/kg with Millikan’s measurement of e => mass of electron

p212c28: 15 Example: Using an accelerating Potential of 150 V and a transverse Electric Field of 6x10 6 N/C. Determine a) the speed of the electrons, b) the magnetic field magnitude required for no net deflection

p212c28: 16 Mass Spectrometer   E One method velocity selector + circular trajectory R1R1 R2R2

p212c28: 17 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 1x10 5 m/s, where must the detector be placed to detect 4 He + ions?

p212c28: 18 Magnetic Force on a Current Carrying Wire vdvd I dl B FiFi A

p212c28: 19 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? Force will be directed upwards (out of the plane of the page) I dl B

p212c28: 20 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 FaFa FaFa FbFb FbFb 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.

p212c28: 21 Torque calculation: Side view F b = IBb FbFb  moment arm a/2 sin   = F b a/2 sin  F b a/2 sin   Iab B sin  = I A B sin   Magnetic Dipole ~ Electric Dipole U =     Switch current direction every 1/2 rotation => DC motor  magnetic moment  B

p212c28: 22 Hall Effect Conductor in a uniform magnetic field x y z BzBz JxJx Magnetic force on charge carriers F = q v d  B F z = qv d B Charge accumulates on edges JxJx   

p212c28: 23 Equilibrium: Magnetic Force = Electric Force on bulk charge carriers BzBz Charge accumulates on edges F z = 0 = qv d B y + q E z EyEy w t JxJx  

p212c28: 24 Negative Charge carriers: velocity in negative x direction magnetic force in positive z direction => resulting electric field has reversed polarity  BzBz JxJx   EyEy

p212c28: 25 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  V. What is the density of charge carrying electrons?

p212c28: 26