Chapter 22 Magnetism - California State University,

Chapter 22MagnetismOutline22-1The Magnetic Field22-222-3The Magnetic Force on Moving ChargesThe Motion of Charged Particles in Magnetic Field22-4The Magnetic Force Exert on a Current-CarryingWire22-5Loops of Current and Magnetic Torque22-6Electric Current, Magnetic Fields, and Ampère’sLaw22-7Electric Loops and Solenoid

22-3The Motion of Charges Particles in Magnetic FieldElectric versus Magnetic ForcesLet us compare two cases in Fig (a) and (b).The work done W F S(a) Electric force does work on the particle:(b) No work done by the magnetic force.Figure 22-10Differences Between Motionin Electric and Magnetic Fields

Circular MotionAssume a particle with a velocity that is perpendicular to themagnetic field, as shown in Fig. 22-12.At every points, the force is vertical to the velocity and point to acommon center.V is perpendicular to the B !Figure 22-12Circular Motion in aMagnetic Field

For centripetal motion, one hasv2m q vBrTherefore, we have the radiusmvr qB22 3

Problem 22-17An electron accelerated from rest through a voltage of 410 V entersa region of constant magnetic field. If the electron follows a circularpath with radius of 17 cm, what is the magnitude of the magneticfield ?

Solution:(1) Apply energy conservation:e V 12 mv 2 v 2 e V m2 1.60 10 19 C 410 V 9.11 10 31kg 1.2 107 m/s(2) Solve Eq(22-3) for B:mvr qBmv 9.11 10 kg 1.2 10 m/s 4.0 10B er 1.60 10 C 0.17 m 317 19 4T 0.40 mT

Helical MotionFigure 22-14Helical Motionin a Magnetic FieldWhy ?

22-4 The Magnetic Force Exert on a Current-Carrying WireHow to derive the force?Assume a wire with current is locatedin a magnetic field. The wire has alength L, and the charge is moving at aspeed of v.The time require the charge to gothrough the wire L is t L / vTherefore, the amount of charge isq I t IL / vThus, the force exerted on the wire isF qvB sin (IL)vB sin ILB sin vFigure 22-15The Magnetic Force on aCurrent-Carrying Wire

Magnetic Force on a Current-Carrying WireF ILB sin (22 4)SI unit: Newton, NThe direction of the force is determined by “Right-hand-Rule”.

Problem 22-25The magnetic force exerted on a 1.2-m straight wire is 1.6 N. Thewire carries a current of 3.0 A in an region with constant magneticfield of 0.50 T. what is the angle between the wire and the magneticfield?

Solution:Solve Eq (22-4) for ɵF ILB sin Fsin I LB sin 1F1.6 N sin 1 63 I LB 3.0 A 1.2 m 0.50 T

SummaryMagnetic Force on a Current-Carrying WireF ILB sin (22 4)SI unit: Newton, NThe direction of the magnetic force is determined by Right-HandRule

Exercise 22-1An electron moving perpendicular to a magnetic field of 4.60 x10-3 T followsa circular path of radius 2.80 mm. What is the electron’s speed?Solution:Sincemvr qBr q B (2.80 10 3 m)(1.60 10 19 C )(4.60 10 3 T )v 9.11 10 31 kgm 2.26 106 m / sFigure 22-13The Operating Principle of aMass Spectrometer

Active Example 22-1 Find the time for one orbitCalculate the time T required for a particle of mass m with charge qto complete a circular orbit in a magnetic field.SolutionSince for a circular motion, one hasT 2 r / v(1)Also, we havemvr qBSolving (1) and (2) for T:2 mT qB(2)

Example 22-4 Magnetic LevityA copper rod 0.150m long and with a mass 0.0500kg is suspendedfrom two thin wire. At right angle to the rod is a uniform magneticfield of 0.550 T pointing into the page. Find(a) The direction and (b) magnitude of the electric current to levitatethe copper rod’s gravitation force.Example 22-4Magnetic Levity

SolutionPart (b)The magnetic force must cancel the force of the gravity,ILB mgmg (0.0500kg )(9.81m / s 2 )I 5.95 A(0.150m)(0.550T )LB

Chapter 22 Magnetism Outline 22-1 The Magnetic Field 22-2 The Magnetic Force on Moving Charges 22-3 The Motion of Charged Particles in Magnetic Field 22-4 The Magnetic Force Exert on a Current-Carrying Wire 22-5 Loops of Current and Magnetic Torque 22-6 Electric Cu