Chapter 22 Magnetism - California State University, Northridgerd436460/100B/lectures/chapter22-3-4.pdfChapter 22 Magnetism Outline 22-1 The Magnetic Field 22-2 The Magnetic Force on

Chapter 22 Magnetism

Outline

22-1 The Magnetic Field

22-2 The Magnetic Force on Moving Charges22-3 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 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=FS

(a)Electric force does work on the particle:

(b)No work done by the magnetic force.

Figure 22-10Differences 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.

Figure 22-12Circular Motion in a

Magnetic Field

V is perpendicular to the B !

Therefore, we have the radius

2vm q vBr

For centripetal motion, one has

22 3mvrq B

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 ?

212

19

31

7

2 1.60 10 C 410 V29.11 10 kg

1.2 10 m/s

e V mv

e Vvm

Solution:

(1) Apply energy conservation:

31 74

19

9.11 10 kg 1.2 10 m/s4.0 10 T 0.40 mT

1.60 10 C 0.17 m

mvrq B

mvBer

(2) Solve Eq(22-3) for B:

Helical Motion

Figure 22-14Helical Motion

in a Magnetic Field

Why ?

22-4 The Magnetic Force Exert on a Current-Carrying Wire

Figure 22-15The Magnetic Force 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

sin ( ) sin sinILF qvB vB ILBv

sin (22 4)F ILB

Magnetic Force on a Current-Carrying Wire

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

1 1

sin

sin

1.6 Nsin sin 633.0 A 1.2 m 0.50 T

F ILBF

I L BF

I L B

Summary

Magnetic Force on a Current-Carrying Wire

SI unit: Newton, N

sin (22 4)F ILB

The direction of the magnetic force is determined by Right-Hand Rule

Figure 22-13The Operating Principle of a

Mass Spectrometer

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:

3 19 3

31

6

(2.80 10 )(1.60 10 )(4.60 10 )9.11 10

2.26 10 /

r q B m C Tvm kg

m s

mvSince rq B

Solution

Since for a circular motion, one has

Also, we have

(2)mvrq B

2 / (1)T r v

Solving (1) and (2) for T:

2 mTq B

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.

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 22-4Magnetic Levity

Solution

Part (b)

The magnetic force must cancel the force of the gravity,

ILB mg

2(0.0500 )(9.81 / ) 5.95(0.150 )(0.550 )

mg kg m sI ALB m T