UNIT 19 : MAGNETIC FIELD - PowerPoint PPT Presentation

1 / 92
About This Presentation
Title:

UNIT 19 : MAGNETIC FIELD

Description:

UNIT 19 : MAGNETIC FIELD 19.1 Magnetic field 19.2 Magnetic field produced by current- carrying conductor 19.3 Force on a moving charged particle in a – PowerPoint PPT presentation

Number of Views:637
Avg rating:3.0/5.0
Slides: 93
Provided by: matrikNe1
Category:

less

Transcript and Presenter's Notes

Title: UNIT 19 : MAGNETIC FIELD


1
UNIT 19 MAGNETIC FIELD
19.1 Magnetic field 19.2 Magnetic field
produced by current- carrying
conductor 19.3 Force on a moving charged
particle in a uniform magnetic
field 19.4 Force on a current-carrying conductor
in a uniform magnetic field 19.5 Forces
between two parallel current- carrying
conductors 19.6 Torque on a coil 19.7 Motion of
charged particle in magnetic field and
electric field
2
19.1 Magnetic Field (1 hour)
  • Learning Outcomes
  • At the end of this lesson, the students should be
    able to
  • Define magnetic field.
  • Identify magnetic field sources.
  • Sketch the magnetic field lines.

3
19.1 Magnetic field
  • is defined as a region surrounding a magnet
  • or a conductor carrying current where a
  • magnetic force is experienced.
  • Magnets always have two poles
  • a) North and south poles.
  • b) Like poles repel and unlike poles attract.

Magnetic field lines
  • A magnetic field can be represented by
  • magnetic field lines (straight lines or curves).
  • Arrows on the lines show the direction of the
  • field the arrows point away from north poles
  • and towards south poles.

4
Magnetic field lines
19.1 Magnetic field
  • A uniform field is represented by parallel
  • lines. This means that the number of lines
  • passing perpendicularly through unit area
  • at all cross-sections in a magnetic field
  • are the same as shown below.

unit cross-sectional area
5
Magnetic field lines
19.1 Magnetic field
  • A non-uniform field is represented by non-
  • parallel lines. The number of magnetic field
  • lines varies at different unit cross-sections
  • as shown below.

A1
A2
6
19.1 Magnetic field
Magnetic field lines
  • Magnetic field lines do not intersect
  • one another.
  • The tangent to a curved field line at a
  • point indicates the direction of the
  • magnetic field at that point.

7
19.1 Magnetic field
Magnetic field lines
  • The number of lines per unit cross section
  • area is an indication of the strength of
  • the field. The number of lines per unit
  • cross-sectional area is proportional to
  • the magnitude of the magnetic field.

A1
A2
stronger field in A1
8
Magnetic field lines
19.1 Magnetic field
  • Magnetic field can also be represented by
  • crosses or by dotted circles as shown
  • below.

Magnetic field lines enter the page
perpendicularly
Magnetic field lines leave the page
perpendicularly
B into the page
B out of page
9
Field Patterns
19.1 Magnetic field
  • The magnetic field lines pattern can be
  • obtained by using iron filings or a plotting
  • compass.

the arrowhead of a compass needle is a north pole.
10
19.1 Magnetic field
Field Patterns
  • The direction of the magnetic field at a point
  • is defined as the direction of a compass
  • needle points when placed at that point.

11
19.1 Magnetic field
Field Patterns
a. A bar magnet
b. Horseshoe or U magnet
c. Two bar magnets (unlike pole) - attractive
12
19.1 Magnetic field
Field Patterns
d. Two bar magnets (like poles) - repulsive
Neutral point (point where the resultant magnetic
force/field strength is zero)
13
19.1 Magnetic field
Field Patterns
e. A circular coil
f. A long straight wire
g. A solenoid
14
19.1 Magnetic field
Field Patterns
h. Earth Magnetic Field
15
19.1 Magnetic field
Magnetic Flux, ?
  • is a measure of the number of field lines
  • that cross a surface area.
  • is defined as the scalar product between
  • the magnetic flux density, B and the vector
  • of the surface area, A.

16
19.1 Magnetic field
Magnetic Flux, ?
  • scalar quantity.
  • unit weber(Wb)/ tesla-meter squared(T.m2)
  • 1 T.m2 1 Wb
  • Consider a uniform magnetic field B
  • passing through a surface area A as shown
  • in figure below.

In Figure below, ? 0?
17
19.1 Magnetic field
Magnetic Flux, ?
If ? 90?
Magnetic Flux Density, B
  • is defined as the magnetic flux per unit
  • area at right angles to the magnetic field.

18
19.1 Magnetic field
Magnetic Flux Density, B
  • vector quantity and its direction follows the
    direction of the magnetic field.
  • unit weber per metre squared (Wb m-2)
  • or tesla (T).

19
19.2 Magnetic field produced by current-carrying
conductor (1 hour)
  • Learning outcomes
  • At the end of this lesson, the students should be
    able to
  • Apply magnetic field formula
  • (i) for a long straight wire
  • (ii) for a circular coil
  • (iii) for a solenoid

20
19.2 Magnetic field (B) produced by
current - carrying conductor
  • B is a vector quantity.
  • Magnitude

A long straight wire
(4?? x 10 -7 H m-1)
View from the top
21
19.2 Magnetic field (B) produced by current -
carrying conductor
B magnetic field strength / flux density
(T) I current in the wire (A) r
perpendicularly distance of P from the wire
(m) µo constant of proportionality
known as the permeability of free space
(vacuum) 4p x 10-7 Henry per metre (H
m-1)
r
P
Direction right-hand grip rule
out of the page
22
19.2 Magnetic field (B) produced by current -
carrying conductor
Example 19.2.1
Determine the magnetic field strength at point X
and Y from a long, straight wire carrying a
current of 5 A as shown below.
X
2 cm
I 5 A
6 cm
Y
5.0 x 10-5 T , into the page
BY 1.67 x 10-5 T , out of the page
23
19.2 Magnetic field (B) produced by current -
carrying conductor
Exercise (DIY)
1. Two straight parallel wires are 30 cm apart
and each carries a current of 20 A. Find the
magnitude and direction of the magnetic field at
a point in the plane of the wires that is 10 cm
from one wire and 20 cm from the other if the
currents are (i) in the same direction, (ii)
in the opposite direction.
24
19.2 Magnetic field (B) produced by current -
carrying conductor
A circular coil
N
S
25
19.2 Magnetic field (B) produced by current -
carrying conductor
A circular coil
r
Magnetic field strength at the center given as
r radius of the coil (m)
For N loops / number of turns on the coil
26
19.2 Magnetic field (B) produced by current -
carrying conductor
Example 19.2.3
A circular coil having 400 turns of wire in air
has a radius of 6 cm and is in the plane of the
paper. What is the value of current must exist in
the coil to produce a flux density of 2 mT at its
center ?
27
19.2 Magnetic field (B) produced by current -
carrying conductor
A solenoid
28
19.2 Magnetic field (B) produced by current -
carrying conductor
A solenoid
Magnetic field strength at the center
L
number of turns per length
29
19.2 Magnetic field (B) produced by current -
carrying conductor
Example 19.2.4
An air-core solenoid with 2000 loops is 60 cm
long and has a diameter of 2.0 cm. If a current
of 5.0 A is sent through it, what will be the
flux density within it ?
30
19.2 Magnetic field (B) produced by current -
carrying conductor
Exercise
1. A solenoid is constructed by winding 400 turns
of wire on a 20 cm iron core. The relative
permeability of the iron is 13000. What current
is required to produce a magnetic induction of
0.5 T in the center of the solenoid ?
31
  • 2. A student is provided with a 3.0 m long wire
    with a current of 0.15 A flowing through it. What
    is the strength of the magnetic field at the
    centre of the wire if the wire is bent into a
    circular coil of one turn ? ( B 1.97 x
    10-7 T )
  • 3. A circular coil has 15 turns and a diameter of
    45.0 cm. If the magnetic field strength at the
    centre of the coil is 8.0 x 10-4 T, find the
    current flowing in the coil. ( I 19.1 A
    )
  • ( µ0 4p x 10-7 Hm-1 )

32
19.3 Force on a moving charged particle
in a uniform magnetic field (1 hour)
  • Learning Outcomes
  • At the end of this lesson, the students should be
    able to
  • Use force,
  • Describe circular motion of a charge in a uniform
    magnetic field.
  • Use relationship

33
19.3 Force on a moving charged particle
in a uniform magnetic field.
  • A charge q moving with speed v at angle ? with
    the direction of a uniform magnetic field of
    magnitude B experiences a magnetic force of
    magnitude,

Where ? angle between B and v
For electron , q e.
34
19.3 Force on a moving charged particle in a
uniform magnetic field.
Direction of F
  • Flemings right hand rule - negative charge
  • Flemings left hand rule - positive charge

positive charge
Thumb direction of Force (F) First finger
direction of Magnetic field (B) Second finger
direction of Velocity (v)
35
19.3 Force on a moving charged particle in a
uniform magnetic field.
Example 19.3.1
  • Determine the direction of the magnetic
    force, exerted on a charge in each problem below.

a.
b.
d.
c.
36
19.3 Force on a moving charged particle in a
uniform magnetic field.
Example 19.3.2
Determine the sign of a charge in each problem
below.
37
19.3 Force on a moving charged particle in a
uniform magnetic field.
Exercise
1. Calculate the magnitude of the force on
a proton travelling 3.1 x 107 m s-1 in the
uniform magnetic flux density of 1.6 Wb m-2,if
(i) the velocity of the proton is
perpendicular to the magnetic field. (ii)
the velocity of the proton makes an angle 60?
with the magnetic field. (charge of the proton
1.60 x 10-19 C)
38
19.3 Force on a moving charged particle in a
uniform magnetic field.
Example 19.3.4
A charge q1 25.0 µC moves with a speed of
4.5 x 103 m/s perpendicularly to a uniform
magnetic field. The charge experiences a magnetic
force of 7.31 x 10-3 N. A second charge q2
5.00 µC travels at an angle of 40.0 o with
respect to the same magnetic field and
experiences a 1.90 x 10 -3 N force. Determine
(i) The magnitude of the magnetic field
and (ii) The speed of q2.
39
19.3 Force on a moving charged particle in a
uniform magnetic field.
Solution 19.3.4
q1 25.0 µC , v1 4.5 x 103 m/s, ?1 90.0 o
F1 7.31 x 10-3 N, q2 5.00 µC, ?2 40.0 o,
F2 1.90 x 10 -3 N force.
(i)
(ii) v2 9.10 x 103 m/s
40
19.3 Force on a moving charged particle in a
uniform magnetic field.
Circular Motion of a Charged Particle in a
Uniform Magnetic Field
  • Consider a charged particle moving in a uniform
    magnetic field with its velocity (v)
    perpendicularly to the magnetic field (B).
  • As the particle enters the region, it will
    experience a magnetic force (F) which the force
    is perpendicular to the velocity of the particle.
    Hence the direction of its velocity changes but
    the magnetic force remains perpendicular to the
    velocity.
  • This magnetic force causes the particle to move
    in a circle.

41
19.3 Force on a moving charged particle in a
uniform magnetic field.
Circular Motion of a Charged Particle in a
Uniform Magnetic Field
  • The magnetic force provides the centripetal
    force for the particle to move in circular motion.

r? m?
42
19.3 Force on a moving charged particle in a
uniform magnetic field.
Circular Motion of a Charged Particle in a
Uniform Magnetic Field
  • The time for one rotation (period),

and
and
43
19.3 Force on a moving charged particle in a
uniform magnetic field.
Circular Motion of a Charged Particle in a
Uniform Magnetic Field
Exercise (DIY)
  • 1. A proton is moving with velocity 3 x 10 5 m/s
  • vertically across a magnetic field 0.02 T.
  • (mp 1.67 x 10 -27 kg)
  • Calculate
  • kinetic energy of the proton
  • the magnetic force exerted on the proton
  • the radius of the circular path of the
  • proton.

7.52 x 10-17 J , 9.6 x 10 -16 N, 0.16 m
44
19.3 Force on a moving charged particle in a
uniform magnetic field.
Circular Motion of a Charged Particle in a
Uniform Magnetic Field
Exercise
1. An electron is projected from left to right
into a magnetic field directed into the page. The
velocity of the electron is 2 x 10 6 ms-1 and
the magnetic flux density of the field is 3.0 T.
Find the magnitude and direction of the magnetic
force on the electron. (charge of electron
1.6 x 10-19 C) (9.6 x 10-13 N,
downwards)
45
  • 2. A proton with a mass of 1.67 x 10-27 kg is
    moving in a circular orbit perpendicular to a
    magnetic field. The angular velocity of the
    proton is 1.96 x 104 rad s-1 . Determine
  • (i) the period of revolution,
  • (ii) the magnetic field strength of the field.
  • (charge of proton 1.6 x 10-19 C)
  • (T 3.2 x 10-4 s , B 2.05 x 10-4 T)

46
19.4 Force on a current- carrying
conductor in a uniform magnetic field
(1 hour)
  • Learning Outcomes
  • At the end of this lesson, the students should be
    able to
  • (i) Use force,

47
19.4 Force on a current- carrying
conductor in a uniform magnetic field
  • When a current-carrying conductor is placed in a
    magnetic field B, thus a magnetic force will act
    on that conductor.
  • The magnitude of the magnetic force exerts on the
    current-carrying conductor is given by
  • In vector form,

48
19.4 Force on a current-carrying conductor in a
uniform magnetic field.
  • Direction of F Flemings left hand rule.

Thumb direction of Force (F) First finger
direction of Magnetic field (B) Second finger
direction of Current (I)
49
19.4 Force on a current-carrying conductor in a
uniform magnetic field.
  • F 0 when ?0
  • F is maximum when ?90o

50
19.4 Force on a current-carrying conductor in a
uniform magnetic field.
Example 19.4.1
  • Determine the direction of the magnetic
    force, exerted on a conductor carrying current, I
    in each problem below.
  • a. b.

b.
a.
51
19.4 Force on a current-carrying conductor in a
uniform magnetic field.
Example 19.4.2
  • A wire of length 0.655 m carries a current of
    21.0 A. In the presence of a 0.470 T magnetic
    field, the wire experiences a force of 5.46 N .
    What is the angle (less than 90o) between the
    wire and the magnetic field?

52
19.4 Force on a current-carrying conductor in a
uniform magnetic field.
Exercise
  • 1. A square coil of wire containing a single
    turn is placed in a uniform 0.25 T magnetic
    field. Each side has a length of 0.32 m, and the
    current in the coil is 12 A. Determine the
    magnitude of the magnetic force on each of the
    four sides.

B
90o
0.96 N (top and bottom sides) 0 N (left and right
sides)
I
53
  • 2. A straight wire with a length of 0.65 m and
    mass of 75 g is placed in a uniform magnetic
    field of 1.62 T. If the current flowing in the
    wire is perpendicular to the magnetic field,
    calculate the current required to balance the
    wire ?
  • (g 9.81 ms-2 ) ( I 0.70 A )

54
19.5 Forces between two parallel
current- carrying conductors (1 hour)
  • Learning Outcomes
  • At the end of this lesson, the students should be
    able to
  • Derive force per unit length of two parallel
    current-carrying conductors.
  • Use force per unit length,
  • Define one ampere.

55
19.5 Forces between two parallel current-
carrying conductors.
  • Consider two identical straight conductors X and
    Y carrying currents I1 and I2 with length L are
    placed parallel to each other as shown below.
  • The conductors are in vacuum and their separation
    is d.

56
19.5 Forces between two parallel current-
carrying conductors.
  • The magnitude of the magnetic flux density, B1 at
    point P on conductor Y due to the current in
    conductor X is given by

into the page
  • Conductor Y carries a current I2 and in the
    magnetic field B1 then conductor Y experiences
    a magnetic force, F12.

57
19.5 Forces between two parallel current-
carrying conductors.
  • The magnitude of F12

to the left (towards X)
  • The magnitude of F21

to the right (towards Y)
Attractive force
58
19.5 Forces between two parallel current-
carrying conductors.
  • If the direction of current in conductor Y is
    changed to upside down as shown below

Repulsive force
  • The currents are in the same direction
  • (2 conductors attract each other)
  • The currents are in the opposite direction
  • (2 conductors repel each other)

59
19.5 Forces between two parallel current-
carrying conductors.
Rearrange,
If I1 I2 1 A and d 1 m , then
Definition of 1 Ampere
60
19.5 Forces between two parallel current-
carrying conductors.
Definition of 1 Ampere
One ampere is defined as the constant current
that, when it is flowing in each of two
infinitely long, straight, parallel conductors
which have negligible of cross sectional areas
and are 1.0 metre apart in vacuum, would produce
a force per unit length between the
conductors of 2.0 x 10-7 N m-1.
61
19.5 Forces between two parallel current-
carrying conductors.
Example 19.5.1
Two very long parallel wires are placed 2.0 cm
apart in air. Both wires carry a current of 8.0 A
and 10 A respectively. Find the magnitude of the
magnetic force in newton, on each metre length of
wire.
62
19.5 Forces between two parallel current-
carrying conductors.
Exercise
1. Two long parallel wires are 5.0 cm apart. They
each exerts a force of attraction per unit length
on the other of 6 x 10 -7 Nm-1 . The current in
one wire is 400 mA. (i) Calculate the current in
the second wire. (ii) In which direction is the
current in the second wire, relative to the first
?
I2 0.375 A (same direction)
63
19.6 Torque on a coil (1 hour)
  • Learning Outcomes
  • At the end of this lesson, the students should be
    able to
  • Use torque, where N number
    of turns.
  • Explain the working principles of a moving coil
    galvanometer.

64
19.6 Torque on a coil.
  • Consider a rectangular loop with length a and
    width b is pivoted so that it can rotate about a
    vertical axis (shown in figure) which is at right
    angle to a uniform magnetic field of flux density
    B.

Axis of rotation
Top view
65
19.6 Torque on a coil.
  • When a steady current I passes round the coil, a
    magnetic force acts on each side of the coil.
  • No magnetic forces act on sides 1 and 3 because
    these wires are parallel to the field.

Axis of rotation
Top view
66
19.6 Torque on a coil.
  • The two magnetic forces on sides 2 and 4 each of
    length a are equal and opposite and have the
    value F where,
  • F2 F4 BIL BIa

Side view
  • The forces exerted a torque that tends to rotate
    the coil clockwise.

67
19.6 Torque on a coil.
  • The magnitude of this torque for each side is

abA (area of the coil) ? angle between B
and the normal to plane of the coil
68
19.6 Torque on a coil.
Example 19.6.1
A 20 turns rectangular coil with sides 6.0 cm x
4.0 cm is placed vertically in a uniform
horizontal magnetic field of magnitude 1.0 T. If
the current flows in the coil is 5.0 A, determine
the torque acting on the coil when the plane of
the coil is (a) perpendicular to the field, (b)
parallel to the field, (c) at 60? to the field.
69
19.6 Torque on a coil.
Solution 19.6.1
N 20 turns, A 24 x 10-4 m2 , B 1.0 T, I
5.0 A
(a)
(b)
(c)
70
19.6 Torque on a coil.
Exercise
1. A rectangular loop of wire has an area of 0.30
m2 . The plane of the loop makes an angle of 30o
with a 0.75 T magnetic field. What is the torque
on the loop if the current is 7.0 A ?
Solution
71
19.6 Torque on a coil.
Exercise
2. Calculate the magnetic flux density required
to give a coil of 100 turns a torque of 0.5 Nm
when its plane is parallel to the field. The
dimension of each turn is 84 cm2 , and the
current is 9.0 A.
Solution
72
19.6 Torque on a coil.
A moving coil galvanometer
Structure of a moving-coil galvanometer
Structure of a moving-coil galvanometer
73
19.6 Torque on a coil.
A moving coil galvanometer
  • The galvanometer is the main component in analog
    meters for measuring current and voltage.
  • It consists of a magnet, a coil of wire, a
    spring, a pointer and a calibrated scale.

74
19.6 Torque on a coil.
A moving coil galvanometer
  • The coil of wire contains many turns and is
    wrapped around a soft iron cylinder.
  • The coil is pivoted in a radial magnetic field,
    so that no matter what position it turns, the
    plane of the coil is always parallel to the
    magnetic field.

75
19.6 Torque on a coil.
A moving coil galvanometer
  • The basic operation of the galvanometer uses the
    fact that a torque acts on a current loop in the
    presence of a magnetic field.
  • When there is a current in the coil, the coil
    rotates in response to the torque ( t NABI )
    applied by the magnet.
  • This causes the pointer ( attached to the coil)
    to move in relation to the scale.

76
19.6 Torque on a coil.
A moving coil galvanometer
  • The torque experienced by the coil is
    proportional to the current in it the larger the
    current, the greater the torque and the more the
    coil rotates before the spring tightens enough to
    stop the rotation.
  • Hence, the deflection of the pointer attached to
    the coil is proportional to the current.
  • The coil stops rotating when this torque is
    balanced by the restoring torque of the spring.

77
A moving coil galvanometer
19.6 Torque on a coil.
  • The coil stops rotating when this torque is
    balanced by the restoring torque of the spring.
  • From this equation, the current I can be
    calculated by measuring the angle ?.

78
19.6 Torque on a coil.
  • This working principles of a moving coil
  • galvanometer also used in voltmeter
  • (multiplier), ammeter (shunt) , ohmmeter
  • and multimeter.

79
  • Exercise
  • The moving coil of a galvanometer has 100 turns
    and an area of 1.5 x 10-4 m2 . It is suspended
    by a wire with a torsional constant of 2.6 x
    10-8 Nm rad-1 . The coil is placed in a radial
    magnetic field of 0.1 T. Calculate the current
    flowing in the coil if a deflection of 1.2 rad is
    observed.
  • ( I 2.08 x 10-5 A )

80
19.7 Motion of charged particle in magnetic
field and electric field (1hour)
  • Learning Outcomes
  • At the end of this lesson, the students should be
    able to
  • Explain the motion of a charged particle in both
    magnetic field and electric field.
  • Derive and use velocity, in a
    velocity selector.

81
19.7 Motion of charged particle in
magnetic field and electric field
  • Consider a charged particle q moves with a
    velocity v in combined electric and magnetic
    fields (the electric and magnetic fields are
    perpendicular), the particle experiences no
    resultant force ( a 0).
  • The particle will continue to move in the same
    direction with the same velocity.
  • For this to happen, the electric force downward
    must balance the magnetic force upwards (refer
    diagram),

82
19.7 Motion of charged particle in magnetic
field and electric field
83
Velocity Selector
19.7 Motion of charged particle in magnetic
field and electric field
  • A velocity selector uses this property of crossed
    electric and magnetic fields to select a single
    velocity of particle only particles traveling at
    this velocity will be undeflected.

84
19.7 Motion of charged particle in magnetic
field and electric field
Velocity selector
85
19.7 Motion of charged particle in magnetic
field and electric field
Example 19.7.1
What is the velocity of protons (1 e) injected
through a velocity selector if E 3 x 105 V/m
and B 0.25 T ?
Solution
1.20 x 106 m/s
86
19.7 Motion of charged particle in magnetic
field and electric field
Exercise
1. A velocity selector is to be constructed to
select ions (positive) moving to the right at 6.0
km/s. The electric field is 300 Vm-1 upwards.
What should be the magnitude and direction of the
magnetic field?
Solution
87
19.7 Motion of charged particle in magnetic
field and electric field
  • When the magnetic field only is applied, the
    particle moves in an arc of a circle of radius r
    under the action of the centrally-directed
    magnetic,

88
19.7 Motion of charged particle in magnetic
field and electric field
Only B is applied
Both E and B are applied
89
19.7 Motion of charged particle in magnetic
field and electric field
90
19.7 Motion of charged particle in magnetic
field and electric field
A mass spectrometer
  • A mass spectrometer is a device used for
    separating atoms or molecules according to their
    mass.
  • The atoms or molecules are ionized and then
    accelerated through an electric field, giving
    them a speed which depends on their mass (their
    kinetic energies are all the same).
  • Then they enter a region of uniform magnetic
    field, which bends them in a circular path.

91
19.7 Motion of charged particle in magnetic
field and electric field
A mass spectrometer
  • The radius of the path depends on the momentum of
    the particle if the kinetic energies are the
    same and the masses are different, the momentum
    will be different as well.
  • A detector can be placed to detect particles
    whose path has a particular radius, thereby
    selecting a particular mass.

92
19.7 Motion of charged particle in magnetic
field and electric field
A mass spectrometer
Write a Comment
User Comments (0)
About PowerShow.com