PHYS 205 Chapter 21, Fall 2017

Chapter 21 Magnetic Forces and Magnetic Fields Outline 1. Magnetic Fields 2. The Force that a Magnetic Field Exerts on a Moving Charge 3. The Motion of a Charged Particle in a Magnetic Field 4. The Mass Spectrometer 5. The Force on a Current in a Magnetic Field 6. The Force on a Current-Carrying Coil 7. Magnetic Fields Produced by Currents 8. Ampère’s Law 9. Magnetic Materials 1- Magnetic Fields Like poles repel each other and unlike poles attract Magnetic poles can’t be isolated: they are always found in pairs. North Pole: under the influence of the earth’s magnetic field South Pole Magnetic field lines Magnetite Fe3O4 Surrounding a magnet there is a magnetic field. The direction of the magnetic field at any point in space is the direction indicated by the north pole of a small compass needle placed at that point. Earth’s Magnetic Field The difference between the true geographic north and south is indicated by a compass 2- The Force that a Magnetic Field Exerts on a Moving Charge = q v B sin  Magnetic field Speed Units of B: 1Wb/m2 = 1 Tesla (T) = 104 Gauss (G) Any charge moving in a magnetic field is subject to a magnetic force F. Note: • The force, F, has a maximum value when the charge moves perpendicularly to the magnetic field lines: = 90o or 270o • This force is zero when v and B are on the same line: = 0o or 180o The direction of F: the right hand rule Example 1 Magnetic Forces on Charged Particles A proton in a particle accelerator has a speed of 5.0x106 m/s. The proton encounters a magnetic field whose magnitude is 0.40 T and whose direction makes and angle of 30.0 degrees with respect to the proton’s velocity (see part (c) of the figure). Find (a) the magnitude and direction of the force on the proton and (b) the acceleration of the proton. (c) What would be the force and acceleration of the particle were an electron?       N 10 6.1 .0 40T sin 30 0. 10 m s 0.5 C .1 60 10 sin 13 6 19           q vB F o (a) (b) (c) 2 13 27 13 p 10 m s 6.9 kg 67 10 .1 N 10 6.1         m F a Magnitude is the same, but direction is opposite. 2 17 31 13 e m s 10 8.1 kg 11 10 .9 N 10 6.1         m F a 3- The Motion of a Charged Particle in a Magnetic Field The electrical force can do work on a charged particle. The magnetic force cannot do work on a charged particle. The magnetic force always remains perpendicular to the velocity and is directed toward the center of the circular path. r m v Fc 2  r m v qvB 2  qB r mv The trajectory of an electron moving in a magnetic field 4- The Mass Spectrometer eB mv qB mv r   mv eV 2 2 1 magnitude of electron charge KE=PE 2 2 2 B V er m       The mass spectrum of naturally occurring neon, showing three isotopes. 5- The Force on a Current in a Magnetic Field The magnetic force on the moving charges pushes the wire to the right. qvBsin F     v t Bsin t q F L I              F ILBsin Example 5 The Force and Acceleration in a Loudspeaker The voice coil of a speaker has a diameter of 0.0025 m, contains 55 turns of wire, and is placed in a 0.10-T magnetic field. The current in the voice coil is 2.0 A. (a) Determine the magnetic force that acts on the coil and the cone. (b) The voice coil and cone have a combined mass of 0.0200 kg. Find their acceleration. (a)        86 N .0 .0 10 T sin 90 .0 0025 m A 55 0.2 sin       ILB F (b) 2s 43m 020 kg .0 .0 86 N   m F a 6- The Torque on a Current-Carrying Coil Forces acting on each segment of the loop: 1 and 2 : I and B are parallel, thus F = 0 2 and 3 : F2 = I.B.b, into the page 3 and 4 : I and B are parallel, thus F = 0 4 and 1 : F1 = I.B.b, out of the page 1 4 3 2 Loop fixed so that it rotates about the axis O. Torque =   F1. a / 2 + F2. a / 2  = I.B.b.a  = I.B.A.sin With A being the area of the loop For N loops, the torque is: = N.I.B.A.sin Bottom view Example 6 The Torque Exerted on a Current-Carrying Coil A coil of wire has an area of 2.0x10-4m2, consists of 100 loops or turns, and contains a current of 0.045 A. The coil is placed in a uniform magnetic field of magnitude 0.15 T. (a) Determine the magnetic moment of the coil. (b)Find the maximum torque that the magnetic field can exert on the coil. (a)      2 4 2 4 moment magnetic A m 10 0.9 m 10 0.2 .0 045 A 100        NIA     N m 10 4.1 .0 15 T sin 90 A m 10 0.9 sin 4 2 4 moment magnetic             NIA B (b) 7- Magnetic Fields Produced by Currents A current-carrying wire produces a magnetic field Right-Hand Rule No. 2. Curl the fingers of the right hand into the shape of a half-circle. Point the thumb in the direction of the conventional current, and the tips of the fingers will point in the direction of the magnetic field. r I B o   2  T m A 10 4 7      o permeability of free space Magnetic Field Created by a Current in a Long, Strait Wire Example 7 A Current Exerts a Magnetic Force on a Moving Charge The long straight wire carries a current of 3.0 A. A particle has a charge of +6.5x10-6 C and is moving parallel to the wire at a distance of 0.050 m. The speed of the particle is 280 m/s. Determine the magnitude and direction of the magnetic force on the particle.     sin 2 sin         r I qv qvB F o r I B o   2 Magnetic Force Between Two Parallel Conductors 1,2 magnetic field created by wire 1 on wire 2 B  0 1 1,2 2 I B d    1,2 2 . 1,2 . Force due to magnetic field on wire 2 of lengh F I B l l   0 1 2 1,2 . 2 I I F l d    F2,1 has the same magnitude as F1,2 but it is downwards This results in attraction between the two wires Parallel conductors carrying currents in the same direction attract each other Parallel conductors carrying currents in the opposite directions repel each other Magnetic Field of a Current Loop center of circular loop R I B o 2  The field lines around the bar magnet resemble those around the loop. Example 10 Finding the Net Magnetic Field A long straight wire carries a current of 8.0 A and a circular loop of wire carries a current of 2.0 A and has a radius of 0.030 m. Find the magnitude and direction of the magnetic field at the center of the loop. Magnetic Field of a Solenoid Interior of a solenoid nI B o  number of turns per unit length 8- Ampère’s Law Ampère’s law states that the line integral around any closed path equals µo I where I is the total steady current passing through any surface bounded by the closed path. µo = 4  x 10-7 T m / A (permeability of free space) . . .(2 ) o o o o B ds I B ds I B ds I B r I                   2 oI B r    Simplified Ampere’s Law For Static Magnetic Fields For any current geometry that produces a magnetic field that does not change in time, I B  o   || net current passing through surface bounded by path Example 11 An Infinitely Long, Straight, Current-Carrying Wire Use Ampere’s law to obtain the magnetic field. I B  o   ||   I B  o   I r B o   2 r I B o   2 9- Magnetic Materials The intrinsic “spin” and orbital motion of electrons gives rise to the magnetic properties of materials. In ferromagnetic materials groups of neighboring atoms, forming magnetic domains, the spins of electrons are naturally aligned with each other. Soft magnetic materials, such as iron, are easily magnetized. They also tend to lose their magnetism easily Hard magnetic materials, such as cobalt and nickel, are difficult to magnetize. They tend to retain their magnetism