Magnetic fieldMagnetic field

Chapter 28Chapter 28

Magnetism 2

Magnetism

• Refrigerators are attracted to magnets!

Magnetism 3

Where is Magnetism Used??Where is Magnetism Used??

• Motors• Navigation – Compass• Magnetic Tapes

– Music, Data

• Television– Beam deflection Coil

• Magnetic Resonance Imaging (MRI)

• High Energy Physics Research

Magnetism 4

EF

BF

Cathode

Anode

EF qE

BF qv B

(28 – 8)

Magnetism 5

Consider a Permanent Magnet

N S

B

The magnetic Field B goes from North to South.

Magnetism 6

Units

N/(A.m) 1 T 1 tesla1

m.Amp

N

/

:

)

sCm

N

qv

FB

Units

Bqv Sin(θF

Magnetism 7

Typical Representation

Magnetism 8

A Look at the Physics

B

q

There is NO force ona charge placed into amagnetic field if thecharge is NOT moving.

Bq

v

• If the charge is moving, thereis a force on the charge,perpendicular to both v and B.

F = q v x B

There is no force if the chargemoves parallel to the field.

Magnetism 9

The Lorentz Force

This can be summarized as: F q B v

vF

Bqm

or:

F qvBsin

is the angle between B and V

Magnetism 10

Nicer Picture

Magnetism 11

The Wire in More Detail

B out of plane of the paper

Assume all electrons are moving with the same velocity vd.

(i). charge POSITIVE ofmotion

theofdirection in the

:

L

BLF

Vector

i

vector

iLBBvv

LiBqvF

v

Litiq

dd

d

d

L

Magnetism 12

. dL

B

dF

i

If we assume the more general case for which the

magnetic field froms and angle with the wire

the magnetic force equation can be writ

B

Magnetic force on a straight wire in a uniform

magnetic field.

ten in vector

form as: Here is a vector whose

magnitude is equal to the wire length and

has a direction that coincides with that of the current.

The magnetic force magnitude

B

B

F iL B L

L

F

sin

In this case we divide the wire into elements of

length which can be considered as straight.

The magnetic f

iLB

dL

Magnetic force on a wire of arbitrary shape

placed in a non - uniform magnetic field.

orce on each element is:

The net magnetic force on the

wire is given by the integral:

B

B

dF idL B

F i dL B

=

BF iL B

BdF idL B=

BF i dL B

(28 – 12)

Magnetism 13

Current Loop

Loop will tend to rotate due to the torque the field applies to the loop.

What is forceon the ends??

Magnetism 14

Magnetic Force on a Current Loop

I

N S

F=BIL

F=BIL

L

F

F

B

Magnetism 15

Magnetic Force on a Current Loop

IF=BIL

Simplified view:

dL

B

F=BIL

Magnetism 16

Magnetic Force on a Current Loop Torque & Electric Motor

IF=BIL

Simplified view:

dL

Torque BILd

IAB

2

2sin

sin

B

F=BIL

Magnetism 17

Magnetic Force on a Current LoopTorque & Electric Motor

I F=BIL

dL

IA B

B

for a current loop

F=BIL

Magnetism 18

C

C

Top viewSide view

sinnet iAB

0netF

(28 – 13)

Magnetism 19

By analogy with electric dipoles, for which:

The expression,

implies that a current loop acts as a magnetic dipole!Here is the magnetic dipole moment, and

Magnetic Force on a Current Loop Torque & Magnetic Dipole

IA B

p E

IA

B

(Torque on acurrent loop)

Magnetism 20

Dipole Moment DefinitionDefine the magneticdipole moment ofthe coil as:

=NiA= x B

We can convert thisto a vector with Aas defined as being normal to the area asin the previous slide.

Magnetism 21

U B U B

The torque of a coil that has loops exerted

by a uniform magnetic field and carrries a

current is given by the equation:

We define a new vector associated wi

N

B

i NiAB

Magnetic dipole moment :

th the coil

which is known as the magnetic dipole moment of

the coil.

B

(28 – 14)

U B

Magnetism 22

L L

L

R R

R

In 1879 Edwin Hall carried out an experiment in which

he was able to determine that conduction in metals is due

to the motion of charges (electrons). He was also

able to determin

ne

The Hall

gat

e t

ive

ffec

e the concentration of the electrons.

He used a strip of copper of width and thickness . He passed

a current along the length of the strip and applied a magnetic

field perpendicular to the stri

n

d

i

B

p as shown in the figure. In the

presence of the electrons experience a magnetic force that

pushes them to the right (labeled "R") side of the strip. This

accumulates negative charge on the R-sid

BB F

e and leaves the left

side (labeled "L") of the strip positively charged. As a result

of the accumulated charge, an electric field is generated as

shown in the figure so that the electric force bala

E

nces the magnetic

force on the moving charges.

( ). From chapter 26 we have:

( )

E B d

d d

d

F F eE ev B

E v B J nev

J i iv

ne Ane dne

eqs.1

eqs.2 (28 – 15)

MagnetismMagnetism 2323

Motion of a chargedMotion of a chargedparticle in a magneticparticle in a magnetic

FieldField

Magnetism 24

Trajectory of Charged Particlesin a Magnetic Field

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

vB

F

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

vB

F

(B field points into plane of paper.)

Magnetism 25

Trajectory of Charged Particlesin a Magnetic Field

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

vvB B

FF

(B field points into plane of paper.)

Magnetic Force is a centripetal force

Magnetism 26

Review of Rotational Motion

atar

at = r tangential acceleration

ar = v2 / r radial acceleration

The radial acceleration changes the direction of motion,while the tangential acceleration changes the speed.

r s = s / r s = r ds/dt = d/dt r v = r

= angle, = angular speed, = angular acceleration

Uniform Circular Motion

= constant v and ar constant but direction changes

ar = v2/r = 2 rF = mar = mv2/r = m2r

KE = ½ mv2 = ½ mw2r2

v

ar

Magnetism 27

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

Radius of a Charged ParticleOrbit in a Magnetic Field

vB

F

r

mr

q B

r mqB

v v

v

2

Centripetal Magnetic Force Force =

Magnetism 28

Cyclotron Frequency

+ + + +

+ + + +

+ + + +

+ + + +

+ + + +

vB

F

r

The time taken to complete one orbit is:

qB

m

rT

v

v

2v

2

m

qBf

m

qB

Tf

c

2

2

1

Magnetism 29

Mass Spectrometer

Smaller Mass

Magnetism 30

Magnetism 31

An ExampleA beam of electrons whose kinetic energy is K emerges from a thin-foil “window” at the end of an accelerator tube. There is a metal plate a distance d from this window and perpendicular to the direction of the emerging beam. Show that we can prevent the beam from hitting the plate if we apply a uniform magnetic field B  such that

22

2

de

mKB

Magnetism 32

Problem Continued

r

22

22

2

e

2mKB

:Bfor Solvee

22

eB

m

2

2

1

qB

mvr

Before From

d

dB

mK

m

Kr

m

KvmvK

Magnetism 33

#14 Chapter 28 A metal strip 6.50 cm long, 0.850 cm wide, and 0.760 mm

thick moves with constant velocity through a uniform magnetic field B= 1.20mTdirected perpendicular to the strip, as shown in the Figure. A potential difference of 3.90 ηV is measured between points x and y across the strip. Calculate the speed v.

FIGURE 28-37   

Problem 14.

Magnetism 34

21.  (a) Find the frequency of revolution of an electron with an energy of 100 eV in a uniform magnetic field of magnitude 35.0 µT . (b) Calculate the radius of the path of this electron if its velocity is perpendicular to the magnetic field.

Magnetism 35

39.  A 13.0 g wire of length L = 62.0 cm is suspended by a pair of flexible leads in a uniform magnetic field of magnitude 0.440 T. What are the (a) magnitude and (b) direction (left or right) of the current required to remove the tension in the supporting leads?