Section 24.5 Magnetic Fields Exert Forces on Moving...

Section 24.5 Magnetic Fields Exert Forces on Moving Charges

© 2015 Pearson Education, Inc.

Magnetic Fields

• Sources of Magnetic Fields

• You already know that a moving charge is the “creator” of a

magnetic field.

• Effects of Magnetic Fields

• If a moving charge “experiences” a magnetic field, it there will be a

force on the charge.

• Sound familiar?!?!

3

Magnetic Fields Exert Forces on Moving Charges

There is no magnetic

force on a charged

particle at rest.

There is no magnetic force on

a charged particle moving

parallel to a magnetic field.


• Magnetic fields also exert forces on moving charged

particles and on electric currents in wires.

4

As the angle α between the velocity and the magnetic field

increases, the magnetic force also increases. The force is

greatest when the angle is 90°. The magnetic force is always

perpendicular to the plane containing and .



5

Magnetic Force on Particles

• A magnetic force is exerted on a particle within a magnetic

field only if

• the particle has a charge.

• the charged particle is moving….

• …with at least a portion of its velocity perpendicular to the

magnetic field.

Magnetic Force on a Charged Particle

• magnitude:

• q: charge in Coulombs

• v: speed in meters/second

• B: magnetic field in Tesla

• Θ: angle between v and B

• direction: Right Hand Rule

θsinBvqFB

vv=

charge

movingin a

magnetic

field

with an angle not equal to 0

or 180 degrees

7


• We determine the correct direction of the force using the

right-hand rule for forces.


8

The right hand rule to determine magnetic force

This right hand rule is a little different than what your book uses, but I think it’s easier to remember. You must keep your hand in this configuration and turn your whole wrist!!

• Point in the direction of the velocity.

• Turn your hand so that your middle finger (or three remaining fingers) point in the direction of the field.

• Your thumb gives you the direction of the force.

v

B

F

9



10

QuickCheck 24.15

The direction of the magnetic force on the proton is

A. To the right.

B. To the left.

C. Into the screen.

D. Out of the screen.

E. The magnetic force is zero.


11

QuickCheck 24.15

The direction of the magnetic force on the proton is

A. To the right.

B. To the left.

C. Into the screen.




12

QuickCheck 24.16

The diagram shows a top view of an electron beam passing

between the poles of a magnet. The beam will be deflected

A. Toward the north pole of the magnet.

B. Toward the south pole of the magnet.

C. Out of the plane of the figure

D. Into the plane of the figure.


13

QuickCheck 24.16

The diagram shows a top view of an electron beam passing

between the poles of a magnet. The beam will be deflected

A. Toward the north pole of the magnet.

B. Toward the south pole of the magnet.

C. Out of the plane of the figure

D. Into the plane of the figure.


14

QuickCheck 24.17

A beam of positively charged particles passes between the

poles of a magnet as shown in the figure; the force on the

particles is noted in the figure. The magnet’s north pole is

on the _____, the south pole on the _____.

A. Left, right

B. Right, left

C. There’s not enough

information to tell.


15

QuickCheck 24.17

A beam of positively charged particles passes between the

poles of a magnet as shown in the figure; the force on the

particles is noted in the figure. The magnet’s north pole is

on the _____, the south pole on the _____.

A. Left, right

B. Right, left

C. There’s not enough

information to tell.


16

QuickCheck 24.18

The direction of the magnetic force on the electron is

A. Upward.

B. Downward.

C. Into the screen.




17

QuickCheck 24.18

The direction of the magnetic force on the electron is

A. Upward.

B. Downward.

C. Into the screen.




18

QuickCheck 24.19

Which magnetic field causes the observed force?


19

QuickCheck 24.19

Which magnetic field causes the observed force?


C.

20

Sample Problem

• What is the magnetic force exerted on a 3.0 µC charge

moving north at 300,000 m/s in a magnetic field of 200

mT if the field is directed

a) North.

b) South.

c) East.

d) West.

Sample Problem

• What is the magnetic force exerted on a 3.0 µC charge

moving north at 300,000 m/s in a magnetic field of 200

mT if the field is directed

a) North.

b) South.

c) East.

d) West.

( )( )( ) ( )

0

0sinT 10200m/s 000,300C 103

sin

36

=

°××=

=

−−

B

B

B

F

F

qvBF θ

( )( )( ) ( )

0

180sinT 10200m/s 000,300C 103

sin

36

=

°××=

=

−−

B

B

B

F

F

qvBF θ

( )( )( ) ( )

Earth) (towarddown N, 18.0

90sinT 10200m/s 000,300C 103

sin

36

=

°××=

=

−−

B

B

B

F

F

qvBF θ

( )( )( ) ( )

sky) (toward up N, 18.0

90sinT 10200m/s 000,300C 103

sin

36

=

°××=

=

−−

B

B

B

F

F

qvBF θ

Sample Problem

• Calculate the magnitude and direction of the magnetic

force in the situation below.

( )( )( ) ( )

page theofout N, 101.0

34sinT 10200m/s 000,300C 103

sin

36

=

°××=

=

−−

B

B

B

F

F

qvBF θ

23

Conceptual Example 24.6 Determining the force on a moving electron

An electron is moving to the right in a magnetic field that

points upward, as in FIGURE 24.26. What is the direction

of the magnetic force?


24

Magnetic forces…

• are always orthogonal (at right angles) to the plane

established by the velocity and magnetic field vectors.

• can accelerate charged particles by changing their

direction.

• can cause charged particles to move in circular or helical

paths.

25

Magnetic forces cannot…

• do work on charged particles. Why?

• The force is always perpendicular to the motion.

• What are the implications of this?

• They cannot change the speed or kinetic energy of charged

particles.

26

This means magnetic forces…

• …are centripetal!

• Remember that centripetal acceleration is

• Therefore, centripetal force isr

vac

2

=

r

mvmaF cc

2

==Σ

27

Paths of Charged Particles in Magnetic Fields

• When we studied the motion of objects subject to a force

that was always perpendicular to the velocity, the result

was circular motion at a constant speed. For example, a

ball moved at the end of a string moved in a circle due to

the perpendicular force of tension in the string.

• For a charged particle moving in a magnetic field, the

magnetic force is always perpendicular to and so it

causes the particle to move in a circle.


28


• A particle moving

perpendicular to a

uniform magnetic field

undergoes uniform

circular motion at

constant speed.


29


• Derive an equation for the radius of orbit for a charged

particle in a magnetic field.


30

Sample Problem

• What is the orbital radius of a proton moving at 20,000 m/s perpendicular to a 40 T magnetic field?

Sample Problem

• What is the orbital radius of a proton moving at 20,000 m/s perpendicular to a 40 T magnetic field?

( )( )( )( ) ( )

m 1022.590sinT 04C 106.1

m/s 000,20kg 1067.1

sin

sin

sin

6

19

27

2

2

−

−

−

×=°×

×==

=

=

=

Σ=

θ

θ

θ

qB

mvr

qBr

vm

qvBr

vm

Fr

vm

Fma

B

cc

32


• The motion of a charged particle when its velocity is

neither parallel nor perpendicular to the magnetic field:


33



34



35


• High-energy particles stream out from the sun in the solar

wind, some of which becomes trapped in the earth’s

magnetic field.

• The particles spiral in helical trajectories along the earth’s

magnetic field lines. When they enter the atmosphere at

the poles, they ionize gas, creating the aurora.


36

Sample Problem

• An electric field of 2000 N/C is directed to the south. A proton is traveling at 300,000 m/s to the west. What is the magnitude and direction of the force on the proton? Describe the path of the proton. Ignore gravitational effects.

E

v

( )( )

N 102.3

2000C 106.1

16

CN19

−

−

×=

×=

=

F

F

qEF

The force is south. (Since the charge is positive,

it will experience a force in the direction of the

field.) Since the horizontal velocity is

unchanged, the proton will follow a parabolic

path downward. E

v

( )( )

N 102.3

2000C 106.1

16

CN19

−

−

×=

×=

=

F

F

qEF

37

Sample Problem

• A magnetic field of 2000 mT is directed to the south. A proton is traveling at 300,000 m/s to the west. What is the magnitude and direction of the force on the proton? Describe the path of the proton. Ignore gravitational effects.

B

v

( )( )( ) ( )

N 106.9

90sinT 102000 000,300C 106.1

sin

14

3-

sm19

−

−

×=

°××=

=

F

F

qvBF θ

The force is up (out of the page). (Since the

charge is positive, use the right hand rule.) It

will move in a circular path.

40

Sample Problem

• How would you arrange a magnetic field and an electric field so that a charged particle of velocity v would pass straight through without deflection?

41

QuickCheck 24.20

Which magnetic field (if it’s the correct strength) allows the

electron to pass through the charged electrodes without

being deflected?


42

QuickCheck 24.20

Which magnetic field (if it’s the correct strength) allows the

electron to pass through the charged electrodes without

being deflected?


E.

43

Electric and Magnetic Fields Together

e-

E

B

This electron will experience an upward force from the electric field (opposite the direction

of the field) and a downward force from the magnetic field (left hand rule).

44

Sample Problem

• It is found that protons traveling at 20,000 m/s pass

undeflected through the velocity filter below. What is

the magnitude and direction of the magnetic field

between the plates?

0.02 m20,000 m/s

e

400 V

Solution on next page.

45

Solution

( )( )

page theinto T, 190sin 000,20

000,20

sin

sin

sin

sin

sin

0

0ΣF that means dUndeflecte

sm

CN

=°

=

=

=

=

=

−==Σ

=

B

Bv

E

qv

qvB

qv

qE

qvBqE

FF

FFF

.

Be

eB

θ

θ

θ

θ

θ

mV 000,20

m 0.02

V 400===

=

d

VE

EdV

46

Electromagnetic Flowmeters

• An electromagnetic flowmeter is a device that can be used

to measure the blood flow in an artery.

• It applies a magnetic field across the artery, which

separates the positive and negative ions in the blood.

• The flowmeter measures the potential difference due to the

separation of the ions.

• The faster the blood’s ions are moving, the greater the

forces separating the ions become, therefore generating a

higher voltage.

• Therefore, the measured voltage is proportional to the

velocity of the blood.


47



48



57

Don’t Try It Yourself: Magnets and Television Screens

The image on a cathode-ray tube

television screen is drawn by an

electron beam that is steered by

magnetic fields from coils of wire.

Other magnetic fields can also

exert forces on the moving

electrons. If you place a strong

magnet near the TV screen, the

electrons will be forced along altered trajectories and will strike

different places on the screen than they are supposed to, producing an

array of bright colors. (The magnet can magnetize internal

components and permanently alter the image, so do not do this to your

television!)


Section 24.6 Magnetic Fields Exert Forces on Currents


59

The Form of the Magnetic Force on a Current

• We learned that the magnetic

field exerts no force on a

charged particle moving

parallel to a magnetic field.

• If a current-carrying wire is

parallel to a magnetic field,

we also find that the force on

it is zero.

• There is a force on a current-

carrying wire that is

perpendicular to a magnetic

field.


60

The right hand rule to determine magnetic force

This right hand rule is a little different than what your book uses, but I think it’s easier to remember. You must keep your hand in this configuration and turn your whole wrist!!

• Point in the direction of the CURRENT (velocity of the charges).

• Turn your hand so that your middle finger (or three remaining fingers) point in the direction of the field.

• Your thumb gives you the direction of the force.

v

B

F

61

Calculating Magnetic Force on a Current


• Derive an equation that can be used to calculate magnetic

force on a current-carrying wire.

62

Review

• What is the equation for a magnetic force on a moving

charge?

• What if there are many moving charges, like a current in a

wire?

θsinqvBF =

θθθ

θ

sinsinsin

sin

xBIxBt

q

t

xBqF

Bt

xqF

∆=∆

∆=

∆

∆=

∆

∆= ∆x represents the length of the wire (the

distance the charges move)

63

Magnetic Force on Current-Carrying Wire

• I: current in Amps

• ℓ: length in meters

• B: magnetic field in Tesla

• θ: angle between current and field

θsinBIFB l=

65

Sample Problem

• What is the force on a 100 m long wire bearing a 30

A current flowing north if the wire is in a downward-

directed magnetic field of 400 mT?

( )( )( )N 1200

90sinT 10400m 001 30

sin

3

sC

=

°×=

=

−

B

B

B

F

F

BIF θl

The force is west (use the right hand rule…remember we assume positive

charges are flowing in the wire).

66

Sample Problem

• A wire is in a magnetic field that is directed out of the

page. What is the magnetic field strength if the

current in the wire is 15 A and the force is downward

and has a magnitude of 40 N/m? What is the direction

of the current?

ANS: 2.67 T, right

67

Which way will this loop of wire rotate

B

a) if the current is clockwise?

b) if the current is counterclockwise?

a) The right side of the loop will rotate out of the page.

b) The right side of the loop will rotate into the page.

70

QuickCheck 24.23

The horizontal wire can be levitated—held up against the

force of gravity—if the current in the wire is

A. Right to left.

B. Left to right.

C. It can’t be done with

this magnetic field.


71

QuickCheck 24.23

The horizontal wire can be levitated—held up against the

force of gravity—if the current in the wire is

A. Right to left.

B. Left to right.

C. It can’t be done with

this magnetic field.


72

Example 24.11 Magnetic force on a power line

A DC power line near the equator runs east-west. At this

location, the earth’s magnetic field is parallel to the ground,

points north, and has magnitude 50 µT. A 400 m length of

the heavy cable that spans the distance between two towers

has a mass of 1000 kg. What direction and magnitude of

current would be necessary to offset the force of gravity and

“levitate” the wire? (The power line will actually carry a

current that is much less than this; 850 A is a typical value.)


73

Example 24.11 Magnetic force on a power line (cont.)

PREPARE First, we sketch a top

view of the situation, as in

FIGURE 24.38. The magnetic

force on the wire must be

opposite that of gravity. An

application of the right-hand

rule for forces shows that a

current to the east will result in

an upward force—out of the

page.


74

SOLVE The magnetic field is perpendicular to the current, so

the magnitude of the magnetic force is given by Equation

24.10. To levitate the wire, this force must be opposite to the

weight force but equal in magnitude, so we can write

mg = ILB

where m and L are the mass and length of the wire and B is

the magnitude of the earth’s field. Solving for the current,

we find

directed to the east.



75


ASSESS The current is much larger than a typical current, as

we expected.


76

Forces Between Currents

• Because a current produces

a magnetic field, and a

magnetic field exerts a

force on a current, it

follows that two current-

carrying wires will exert

forces on each other.

• A wire carrying a current I1

will create a magnetic field

1.


77


• A second wire with current

I2 will experience the

magnetic force due to the

wire with current I1.

• Using the right-hand rule for

forces, we can see that when

I2 is in the same direction as

I1, the second wire is

attracted to the first wire.

• If they were in opposite

directions, the second wire

would be repelled.


78


• The magnetic field created

by the wire with current I2

will also exert an attractive

force on the wire with

current I1.

• The forces on the two wires

form a Newton’s third law

action/reaction pair.

• The forces due to the

magnetic fields of the wires

are directed in opposite

directions and must have the

same magnitude.


79



80

Example 24.12 Finding the force between wires in jumper cables

You may have used a set of jumper cables connected to a

running vehicle to start a car with a dead battery. Jumper

cables are a matched pair of wires, red and black, joined

together along their length. Suppose we have a set of jumper

cables in which the two wires are separated by 1.2 cm along

their 3.7 m (12 ft) length. While starting a car, the wires

each carry a current of 150 A, in opposite directions. What

is the force between the two wires?


81

Example 24.12 Finding the force between wires in jumper cables (cont.)

PREPARE Our first step is to sketch the situation, noting

distances and currents, as shown in FIGURE 24.41. Let’s

find the force on the red wire; from the discussion above,

the force on the black wire has the same magnitude but is in

the opposite direction.


82


The force on the red wire is found using a two-step process.

First, we find the magnetic field due to the current in the

black wire at the position of the red wire. Then, we find the

force on the current in the red wire due to this magnetic

field.


83


SOLVE The magnetic field at the position of the red wire,

due to the current in the black wire, is

According to the right-hand rule for fields, this magnetic

field is directed into the page. The magnitude of the force on

the red wire is then


84


The direction of the force can be found using the right-hand

rule for forces. The magnetic field at the position of the red

wire is into the page, while the current is to the right. This

means that the force on the red wire is in the plane of the

page, directed away from the black wire. Thus the force

between the two wires is repulsive, as we expect when their

currents are directed oppositely.


85


ASSESS These wires are long, close together, and carry very

large currents. But the force between them is quite small—

much less than the weight of the wires. In practice, the

forces between currents are not an important consideration

unless there are many coils of wire, leading to a large total

force. This is the case in an MRI solenoid.


86

Forces Between Current Loops

• Just as there is an attractive force

between parallel wires that have

currents in the same direction, there is

an attractive force between parallel

loops with currents in the same

direction.

• There is a repulsive force between

parallel loops with currents in opposite

directions.


87


• The field of a current loop is very similar to that of a bar

magnet.

• A current loop, like a bar magnet, is a magnetic dipole

with a north and a south pole.


88



89

QuickCheck 24.24

The diagram below shows slices through two adjacent

current loops. Think about the force exerted on the loop on

the right due to the loop on the left. The force on the right

loop is directed

A. To the left.

B. Up.

C. To the right.

D. Down.


90

QuickCheck 24.24

The diagram below shows slices through two adjacent

current loops. Think about the force exerted on the loop on

the right due to the loop on the left. The force on the right

loop is directed

A. To the left.

B. Up.

C. To the right.

D. Down.


91

Example Problem

A 10 cm length of wire carries a current of 3.0 A. The wire

is in a uniform field with a strength of 5E-3 Tesla as in the

following diagram. What are the magnitude and direction of

the force on this segment of wire?


92

Summary: General Principles


Text: p. 794

93

Summary: Important Concepts


Text: p. 794

94

Summary: Applications


Text: p. 794

Section 24.5 Magnetic Fields Exert Forces on Moving...

Documents

Transcript of Section 24.5 Magnetic Fields Exert Forces on Moving...