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

2013
Views:

426

Pages:

20
Author:

Layla Bonilla

Chapter 22 Magnetism Outline 22-1 The Magnetic Field 22-2 22-3 The Magnetic Force on Moving Charges 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 Current, Magnetic Fields, and Ampère’s Law 22-7 Electric Loops and Solenoid 22-3 The Motion of Charges Particles in Magnetic Field Electric versus Magnetic Forces Let 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 22-10 Differences Between Motion in Electric and Magnetic Fields Circular Motion Assume a particle with a velocity that is perpendicular to the magnetic field, as shown in Fig. 22-12. At every points, the force is vertical to the velocity and point to a common center. V is perpendicular to the B ! Figure 22-12 Circular Motion in a Magnetic Field For centripetal motion, one has v2 m  q vB r Therefore, we have the radius mv r qB 22  3 Problem 22-17 An electron accelerated from rest through a voltage of 410 V enters a region of constant magnetic field. If the electron follows a circular path with radius of 17 cm, what is the magnitude of the magnetic field ? Solution: (1) Apply energy conservation: eV  12 mv 2  v 2 e V  m 2 1.60  1019 C   410 V   9.1110 31 kg   1.2  107 m/s (2) Solve Eq(22-3) for B: mv r  qB mv  9.11 10 kg 1.2  10 m/s    4.0  10 B er 1.60 10 C   0.17 m  31 7 19 4 T  0.40 mT Helical Motion Figure 22-14 Helical Motion in a Magnetic Field Why ? 22-4 The Magnetic Force Exert on a Current-Carrying Wire How to derive the force? Assume a wire with current is located in a magnetic field. The wire has a length L, and the charge is moving at a speed of v. The time require the charge to go through the wire L is ∆t = L / v Therefore, the amount of charge is q = I ∆t = IL / v Thus, the force exerted on the wire is F  qvB sin   ( IL )vB sin   ILB sin  v Figure 22-15 The Magnetic Force on a Current-Carrying Wire Magnetic Force on a Current-Carrying Wire F  ILB sin  (22  4) SI unit: Newton, N The direction of the force is determined by “Right-hand-Rule”. Problem 22-25 The magnetic force exerted on a 1.2-m straight wire is 1.6 N. The wire carries a current of 3.0 A in an region with constant magnetic field of 0.50 T. what is the angle between the wire and the magnetic field? Solution: Solve Eq (22-4) for ɵ F  ILB sin   F sin   I LB   sin 1 F 1.6 N  sin 1  63 I LB  3.0 A 1.2 m  0.50 T  Summary Magnetic Force on a Current-Carrying Wire F  ILB sin  (22  4) SI unit: Newton, N The direction of the magnetic force is determined by Right-Hand Rule Exercise 22-1 An electron moving perpendicular to a magnetic field of 4.60 x10-3 T follows a circular path of radius 2.80 mm. What is the electron’s speed? Solution: Since mv r qB r q B (2.80  103 m)(1.60  1019 C )(4.60 103 T ) v  9.111031 kg m  2.26 106 m / s Figure 22-13 The Operating Principle of a Mass Spectrometer Active Example 22-1 Find the time for one orbit Calculate the time T required for a particle of mass m with charge q to complete a circular orbit in a magnetic field. Solution Since for a circular motion, one has T  2 r / v (1) Also, we have mv r qB Solving (1) and (2) for T: 2 m T qB (2) Example 22-4 Magnetic Levity A copper rod 0.150m long and with a mass 0.0500kg is suspended from two thin wire. At right angle to the rod is a uniform magnetic field of 0.550 T pointing into the page. Find (a) The direction and (b) magnitude of the electric current to levitate the copper rod’s gravitation force. Example 22-4 Magnetic Levity Solution Part (b) The magnetic force must cancel the force of the gravity, ILB  mg mg (0.0500kg )(9.81m / s 2 ) I   5.95 A (0.150m)(0.550T ) LB