University:
California State University, Northridge
Course:
PHYS 100B | General Physics II
Academic year:

2008
Views:

224

Pages:

19
Author:

Celeste Austin

Chapter 22 Magnetism Outline 22-1 The e Magnetic ag et c Field ed 22-2 22-3 The Magnetic Force on Moving Charges The Motion of Charges 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 22-6 El Electric ti C Current, t M Magnetic ti Fi Fields, ld and dA Ampère’s è ’ Law 22-7 Electric Loops and Solenoid 22-1 The Magnetic Field Permanent Magnets Permanent 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 Lines Magnetic field is Very similar with Electric field of point charge. 1) Direction: The 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) loop). 3) Dense lines have higher magnitude. Magnetic Field Lines Figure 22-4 22 4 Magnetic Field Lines for a Bar Magnet Figure 22-6 Magnetic g Field of the Earth 22-2 The Magnetic Force on Moving Charges Magnitude of the magnetic Force JK C Consider id a magnetic ti fil filed Kd B . A particle ti l Koff charge h q moves through with a velocity v . The velocity v and the magnetic field has an angle of θ , Figure 22-7 The Magnetic Force on a Moving Charged Particle JK B Magnitude of the magnetic force , F, is F = q vB sin θ (22 − 1) SI unit: newton Definition of the magnitude of the magnetic field. B F B = q v sin θ SI unit: 1 tesla = 1 T =1 N / (A·m) 1 Gauss = 1 G = 10-4 T (22 − 2) Problem 22-5 Electromagnetic Force A 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) 3.0x10-6 N. Magnetic Force Right-Hand Rule (RHR) Magnetic ag et c Force o ce Right-Hand g t a d Rule ue To 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 G . Now curl your fingers toward the direction of G, as shown in Figure Figure. v B Your thumb will point in the direction of G; F If the charge is negative, the force points opposite to the direction of your thumb. th b Figure 22-8 The Magnetic Force Right-Hand Rule The relationship can be mathematically expressed in terms of the vector t cross product, d t JG G G F = qv × B We 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 22-9 Th Magnetic The M ti Force F for f Positive and Negative Charges Problem 22-11 Electromagnetic force When at rest, a proton experiences a net electromagnetic force of magnitude 8.0x10-13 N pointing in the positive x direction. When the proton moves with a speed of 1.5x106 m/s in the positive y direction, the net electromagnetic force on it decreases in magnitude to 7.5x10-13 N, still pointing in the positive x direction. Find the magnitude and direction of (a) The electric field and (b) the magnetic field. Summary 1) Magnetism Opposite poles attract; Like poles repel repel. 1)) Magnitude g of magnetic g force F = q vB sin θ (2 2 − 1) 3) Right-Hand Rule (RHR): direction of magnetic force JG G G F = qv × B Example 22-1 Two Charges Particle 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 4.25x10-3 N. Particle 2, with a charge q1= 53.0 uC and a speed v2= 1.3x103 m/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. Example 22-1 A Tale of Two Charges Part (a) F = q vB sin θ 4.25 × 10 − 3 N = ((3.60 × 10 − 6 C )(862 m / s ) B (sin )( ( 90 ° ) B = 1.37 T Part (b) F = q vB sin θ = (5 3 .0 × 1 0 − 6 C ) (1.30 × 1 0 3 m / s )(1 .3 7 T ) sin 5 5 .0 ° = 0.0773 N Example 22-2 Electric and Magnetic Field A particle with a charge of 7 7.70 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 the electric field points in the positive z direction with a magnitude of 8.10x103 N/C. Find the magnitude and direction of the net force acting on the particle. Example 22-2 Electric and Magnetic Fields Solution 1) Calculate the magnitude of the electric force on the particle F E = qE = (7.70 (7 70 × 10 − 6 C ) (8 (8.10 10 × 10 3 N / C ) = 66.24 24 × 10 − 2 N 2) Calculate the magnitude of the magnetic force F = q vB sin θ ( m / s )(3.20 )( T ) sin 90 ° = 1 .0 7 × 10 − 2 = ((7.7 × 10 − 6 C ) (435 3) Both forces are in positive z direction, and can be added directly y Fnet = F E + F B = (6.24 × 10 − 2 = 7.31 × 10 − 2 N N ) + (1.07 × 10 − 2 N) N