1

The Magnetic Field

Chapter 30

2

Magnetic Forces

• Magnetic Force - A force present when an electric charge is in motion.

• A moving charge is said to produce a magnetic field .

• Magnetic fields exert forces on moving charges.

3

Magnetic Fields

• Represented by field lines .• By definition:

• or more commonly:

• Where q is the angle between v and B.

B =F

Qvsinq

F = qvBsinq

F = qv B

4

Magnetic Field Units

• Standard Unit = Tesla (T)

• 1 T = 1 N/A•m

• 1 T = 104 gauss

5

Force on Moving Charges• The diagram below shows a uniform

magnetic field with several charges in motion.

+ +

+ – x

vv

vv

6

Force on Moving Charges

• The magnitude of the force on each charge can be found by qv X B or qvBsinq.

• The direction of the force is found by a right hand rule.

7

Right Hand Rule

• 1) Place your fingers in the direction of the velocity.

• 2) Curl your fingers toward the direction of the field. You might need to turn your hand.

• 3) Your thumb points in the direction of the force.

8

Direction of Force

+ +

+ – x

vv

vv

F = 0

F

F

F

9

Magnetic Field Lines

• NOT lines of force.• Force on charges is not in the direction of

the magnetic field.• Force is always perpendicular to the

velocity of the charge.• Force is always perpendicular to the

magnetic field.• RHR & LHR

10

Permanent Magnets• Magnetic field lines point away from north

poles• and toward south poles.

11

Magnetic Flux

• The amount of a magnetic field passing through a given area.

• Proportional to the number of magnetic field lines which pass through an area.

B = BA cosq = B • AB =

B d A

12

Magnetic Flux

Maximum Flux

No Flux

A

A

A

13

Flux Units

• Weber• 1 Wb = 1 T/m2

14

Gauss's Law for Magnetism

• The magnetic flux through any closed surface must be zero.

N S BdA 0

15

Example

• Exercise 4

16

homework

• E 1, 2, 7

17

Motion of Charges in a Magnetic Field

• Two possible paths can result for the motion of the charge:

• 1) If vo is perpendicular to B, a circular path will result.

• 2) If vo is not perpendicular to B, the charge will travel in a spiral path.

18

vo perpendicular to B

19

vo at an angle to B

20

Magnetic Bottle

21

Van Allen Belts

22

Motion of Charges• As a charge circles or spirals in a magnetic

field, the radius of its path is dependent on the perpendicular component of its velocity.

FB = Fc

QvB =mv2

r

r =mv

QB

T =2rv

=2mqB

vr

qBm

23

Mass Spectrometer

r =mv

QB

24

Velocity Selector

• Only allows charges with a specific velocity to pass through undeflected.

• FB is opposite of FE

• E is perpendicular to B

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Velocity Selector

+ v

–  –  –  –  –  –  –  –  –  –

+ + + + + + + + + +

•

•

•

••

•FE

FB

26

Velocity Selector

• For a specific value of v, the electric force and the magnetic force will be equal to each other and opposite in direction.

• FB = FE

• qvB = qE• vB = E

27

Current-Carrying Wire• Since a current is moving charges, a

current-carrying wire experiences a force in a magnetic field. (B into screen)

X X X X X X X X

X X X X X X X X

F

28

Magnitude of Force

F QvBsinq Q Itv Lt

Qv ItLt

IL

F ILBsinq F IL B

29

Example

• Exercise 14

30

homework

• E 19, 20

31

Sources of Magnetic Fields

Chapter 31

32

Long, straight wire

B =oI2r

o is equal to 4 x 10–7 T•m/A.

33

Current Carrying Wire

• Shape of the field is circular.• Concentric circles• The direction is given a Right Hand Rule:

– Thumb in the direction of the current.– Curl your fingers and they give the direction of

the field.

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Moving Charge

+ • v

35

Wire

I

• • • • • • • • •

x x x x x x x x x

B

I

36

Parallel conductors

• Each creates a magnetic field that produces a force on the other

• Can calculate force per unit length

• To find direction, use both right hand rules

rII

lF

2

0

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Definition of Ampere

• Comes from force exerted by two parallel conductors

• 1 A is the current necessary in each conductor (if 1 m apart) to produce a force of 2 x 10-7 N.

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Field of a circular loop or coil

• At center of loop

• Direction found with right hand rule – like current in straight wire

RNIB

20

39

Field of a Solenoid• Long Spring-like Coil• Uniform field in the interior:

B = onI

B oNL

I

40

Examples

• Exercises 1 and 7

41

homework

• E 2, 6, 10, 12

42

Ampere’s Law

• Like Gauss’s law

encId 0sB

Iencparallel IsB 0

43

Long straight wire

I

rIB

IrB

2

2

0

0

encparallel IsB 0

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Example• A wire has a radius of R and carries a current I that

is uniformly distributed across its area.• Determine how to calculate the magnitude of the

magnetic field inside and outside the conductor.

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Inside• The current inside a circle of radius r would

be a fraction of the total current.• Same ratio as areas.• With total current, I:

Rr

B2r oI r2

R2

B oIr

2R2

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Outside• A circle of radius r, where r > R, encloses

all the current.

B oI2rR

r

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Example

• Determine the field inside a solenoid

lengthturns

n

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Solenoid• Vertical sides – zero

because B is perpendicular to sides

• Side outside solenoid – if it is far away from the solenoid, B is zero

nLIBL 0

nIB 0

49

Paramagnetic materials

• Can become magnetized• An external magnetic field causes atoms to

line up so their currents add to the external field

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Ferromagnetic materials

• Atomic currents line up even when no external field is present

• Permanent magnets

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Electromagnets

52

homework

• E 24-26