Chapter 22 MagnetismOutline22-1 The Magnetic Fieldeagetced22-2 The Magnetic Force on Moving Charges22-3 The Motion of Charges Particles in Magnetic Field22-4 The Magnetic Force Exert on a Current-Carrying WireWire22-5 Loops of Current and Magnetic Torque 226EltiC tM tiFild dAè’22-6 Electric Current, Magnetic Fields, and Ampère’s Law22-7 Electric Loops and Solenoid22-1 The Magnetic FieldPermanent MagnetsPermanent magnet: the magnetism is permanently exist, such as bar magnets, the Earth (two poles)• A magnet always consists of two poles: North Pole (N) and South Pole (S).• Opposite poles attract; Like poles repel.Magnetic Field LinesMagnetic field is Very similar with Electric field of point charge.1) Direction: The direction of the magnetic field, at a given location, is theThe direction of the magnetic field, at a given location, is the direction in which the north pole of a compass points.2) Magnetic field lines begin from the north pole and end at the south pole (close loop)south pole (close loop).3) Dense lines have higher magnitude.224Magnetic Field LinesFigure 22-4Magnetic Field Lines for a Bar MagnetFigure 22-6Magnetic Field of the Earthg22-2 The Magnetic Force on Moving ChargesMagnitude of the magnetic ForceCid tifild AtilfhBJKConsider a magnetic filed . A particle of charge q moves through with a velocity . The velocity and the magnetic field has an angle of ,BvKvKBJKθFigure 22-7The Magnetic Force on a Moving Charged ParticleMagnitude of the magnetic force , F, issin (22 1)FqvBθ=−SI unit: newtonDefinition of the magnitude of the magnetic field. BF(22 2)sinFBqvθ=−SI unit: 1 tesla = 1 T =1 N / (A·m)1 Gauss = 1 G = 10-4TProblem 22-5 Electromagnetic ForceA 0.32-µC particle moves with a speed of 16m/s through a region where the magnetic field has a strength of 0.95 T. At what angle to the field is the particle moving if the force applied on it is(a) 4.8x10-6N , (b) 3.0x10-6N.Magnetic Force Right-Hand Rule (RHR)Magnetic Force Right-Hand Ruleag et c o ce g tad ueTo find the direction of the magnetic force on positive charge, start by pointing the fingers of your right hand in the direction of the velocity . Now curl your fingers toward the direction of as shown in FigurevGNow curl your fingers toward the direction of , as shown in Figure. Your thumb will point in the direction of ; If the charge is negative, the force points opposite to the direction of th bBGFGyour thumb.Figure 22-8The Magnetic Force Right-Hand RuleThe relationship can be mathematically expressed in terms of the tdtvector cross product, Fqv B=×JGGGWe use the symbols to indicate a vector that point into or out of page: • Circle with point: a vector points out of the page.• Circle with cross: a vector points into the page.Figure 22-9Th M ti F fThe Magnetic Force for Positive and Negative ChargesProblem 22-11 Electromagnetic forceWhen at rest, a proton experiences a net electromagnetic force of magnitude 8.0x10-13N pointing in the positive x direction. When the proton moves with a speed of 1.5x106m/s in the positive y direction, the net electromagnetic force on it decreases in magnitude to 7.5x10-13N, still pointing in the positive x direction.Find the magnitude and direction ofFind the magnitude and direction of (a) The electric field and (b) the magnetic field.SSummary1) MagnetismOpposite poles attract; Like poles repelOpposite poles attract; Like poles repel.1)Magnitude of magnetic force)ggsin (22 1)FqvBθ=−3) Right-Hand Rule (RHR): direction of magnetic force JGGGFqv B=×GExample 22-1 Two ChargesParticle 1, with a charge q1=3.6 uC and a speed v1= 862 m/s, travel at right angle to a uniform magnetic field. The magnetic force on particle 1 is 4.25x10-3N. Particle 2, with a charge q1= 53.0 uC and a speed v2= 1.3x103m/s, move at a angle of 55.0˚ to the same magnetic field.Find (a) the strength of the magnetic field; (b) magnitude of the magnetic force exerted on particle 2.force exerted on particle 2.Example 22-1A Tale of Two ChargesPart (a)36sin4.25 10(3.60 10)(862 /)(sin 90)FqvBNCmsBθ−−=×=× °()()()1.37BT=Part (b)63sinFqvBθ=63(53.0 10 )(1.30 10 / )(1.37 )sin 55.00.0773CmsTN−=× × °=Example 22-2 Electric and Magnetic FieldA particle with a charge of 7 70μC and speed 435 m/s is actedA particle with a charge of 7.70 μC and speed 435 m/s is acted on by both en electric and a magnetic field. The particle moves along x axis in positive direction. The magnetic field has a strength of 3.2 T and points in the positive y direction, and thestrength of 3.2 T and points in the positive y direction, and the electric field points in the positive z direction with a magnitude of 8.10x103N/C. Find the magnitude and direction of the net force acting on the particle. Example 22-2Electric and Magnetic FieldsSolution1)Calculate the magnitude of the electric force on the particle63 2(7 70 10 )(8 10 10 / ) 6 24 10EFqECNC N=1)Calculate the magnitude of the electric force on the particle63 2(7.70 10 )(8.10 10 / ) 6.24 10CNC N−−=× × =×2) Calculate the magnitude of the magnetic force6 2sin(7.7 10)(435 /)(3.20)sin 90 1.07 10FqvBCms TNθ− −==× °=×()()()CN3) Both forces are in positive z direction, and can be added directlyy222(6.24 10 ) (1.07 10 )7.31 10net E BFFF N NN−−−=+= × +