Module 1
Magnetism and
Electromagnetism
Engr. Gerard Ang
School of EECE
Magnets
1. Magnet - it is a substance that attracts pieces of iron
and its alloys and which possesses the property of
orientation.
2. Magnetism - it is the phenomenon by which a
magnet attracts iron and its alloys.
3. Magnetic Poles – these are regions on the body of
the magnet in which its magnetic power is said to
exist.
Types of Magnets
1. Natural Magnets - these are iron ores (magnetite
Fe3O4) which are obtained from mines and which have
the property of attracting iron pieces naturally.
2. Artificial Magnets - these magnets are those which are
created by artificial means.
Two Types:
a. Temporary Magnets. Magnets which retain their
power of magnetism by the flow of current through
it.
b. Permanent Magnets. Man-made magnets which are
created from steel and its alloys like cobalt steel,
tungsten steel, etc.
Properties of Magnets
1. A magnet always attracts iron and its alloys.
2. A magnet has two poles, the North and the South
pole
3. Like poles repel and unlike poles attract
4. If a magnet is broken into pieces, each piece
becomes an independent magnet.
5. A magnet can impart its properties to any magnetic
material.
6. A magnet losses its properties when hammered,
heated up to Curie temperature or drop from a height.
Coulomb’s Law of
Electromagnetism
•
•
First Law. The force of attraction or repulsion between two magnetic poles is
directly proportional to their magnetic pole strength.
Second Law. The force of attraction or repulsion between two magnetic poles is
inversely proportional to the square of the distance between them.
Where:
F
= force of attraction or repulsion between the poles
d
= distance between the poles
m1 = magnetic pole strength
m2 = magnetic pole strength
k
= constant of proportionality
μ0
μ0
μr
𝒌𝒎𝟏 𝒎𝟐
𝑭=
𝒅𝟐
= permeability of free space or vacuum
= 4π x 10-7 H/m
= relative permeability of the medium surrounding
the poles (μr = 1 for air)
Value of k:
𝟏
𝟒𝝅𝝁𝟎 𝝁𝒓
𝒌=𝟏
𝒌=
 MKS System
 CGS System
Magnetic Units
Symbol
MKS System
CGS System
m
Weber (Wb)
cgs unit pole
Magnetic Flux
𝜙
Weber (Wb)
Maxwell (Mx)
Magnetic Flux Density
B
Tesla (T)
Gauss (G)
Magnetizing Force
H
Newton per Weber
or ampere per meter
Oersted (Oe)
mmf
Ampere-turn (AT)
Gilbert (Gb)
ℜ
Ampere-turn per
Weber
Gilberts per
Maxwell
Quantity
Magnetic Pole
Strength
Magnetomotive Force
Reluctance
1 Weber (Wb) = 108 Maxwells (Mx)
1 Tesla (T) = 104 Gauss (G)
Sample Problems
1. Two magnetic north poles of strength 0.005 and 0.01 Wb
are placed at a distance of 5 cm in air. Determine the
force between them and its nature.
Solution:
𝑘𝑚1 𝑚2
𝐹=
𝑑2
1
𝑘=
4𝜋𝜇0 𝜇𝑟
𝑑 = 5 𝑐𝑚 = 0.05 𝑚
0.005 0.01
𝑚1 𝑚2
=
𝐹=
4𝜋 4𝜋 × 10−7 1 0.05
4𝜋𝜇0 𝜇𝑟 𝑑 2
𝑭 = 𝟏, 𝟐𝟔𝟔, 𝟓𝟏 𝒅𝒚𝒏𝒆𝒔 (𝒓𝒆𝒑𝒖𝒍𝒔𝒊𝒐𝒏)
2
Sample Problems
2. Two North poles, one having a strength of 500 units and
the other a strength of 150 units, are placed a distance of
4 inches in air. Determine the force acting between these
poles and the direction in which the force is acting.
Solution:
𝑘𝑚1 𝑚2
𝐹=
𝑑2
𝑚1 𝑚2
𝐹=
𝑑2
𝑘=1
𝑑 = 4 𝑖𝑛 = 10.16 𝑐𝑚
500 150
=
10.16 2
𝑭 = 𝟕𝟐𝟔. 𝟓𝟔 𝒅𝒚𝒏𝒆𝒔 (𝒓𝒆𝒑𝒖𝒍𝒔𝒊𝒐𝒏)
Magnetic Terminologies
1. Unit Pole - it is defined as that pole which when placed in air from
a similar and equal pole repel it with a force of 1/4πμ0 Newtons.
2. Magnetic Induction - it is the phenomenon by which a magnetic
substance becomes a magnet when placed near a magnet.
3. Magnetic Field - it is the region outside the magnet in which its
poles experiences a force of attraction or repulsion.
4. Magnetic Lines of Force - these are imaginary lines drawn
through a region of space where a tangent at any point gives the
direction of that field at that point. It is the path of a free running
pole.
Properties of Magnetic
Lines of Forces
• A line of force is said to originate from
a unit North Pole and terminate at a
unit South Pole.
• The direction of a line of force is the
direction in which a unit North Pole
would be urged along it.
• No two lines of forces can ever cross
each other.
• Magnetic lines of forces behave like stretched rubber bands and tend
to contract lengthwise but expand laterally.
• Magnetic lines of force tend to take and easy path.
Magnetic Terminologies
5.
Magnetic Flux (ϕ) - it is the totality of
magnetic lines of force in a magnetic
field.
6.
Magnetic Flux Density (B) - it is the
number of magnetic lines of force per
unit area through any substance in a
plane at right angles to the direction of
the field.
7.
Magnetizing Force (H) - it is defined as
the force acting on a unit pole placed at
any point in a magnetic field. It is also
called as magnetic field intensity,
magnetic intensity and magnetic force.
𝑭
𝑯=
𝒎
𝝓
𝑩=
𝑨
Where:
𝜙 = magnetic flux
A = cross-sectional area of the material
B = magnetic flux density
𝒎
𝑯=
𝟒𝝅𝝁𝟎 𝝁𝒓 𝒅𝟐
𝒎
𝑯= 𝟐
𝒅
 MKS System
 CGS System
Magnetic Terminologies
8.
Relationship between Magnetic Flux Density (B) and Magnetizing Force
(H)
The relationship between the magnetic flux density and magnetizing force
can be derived as:
𝑩 = 𝝁𝟎 𝝁𝒓 𝑯
𝑩 = 𝝁𝟎 𝑯 (in air)
𝑩 = 𝑯 (in air)
9.
 MKS System
 CGS System
Flux Emanating from a Pole
The flux originating from a pole is:
𝝓=𝒎
 MKS System
𝝓 = 𝟒𝝅𝒎
 CGS System
10. Intensity of Magnetization (I or J) - it is the number of unit poles per unit
area at each end of a bar magnet.
𝒎
𝑰=
𝑨
Sample Problems
3. When a small pole having a strength of 25 unit poles is placed in a
magnetic field, it is acted on with a force of 200 dynes. Determine
the field intensity at this point.
Solution:
𝐹
𝐻=
𝑚
200
=
25
𝑯 = 𝟖 𝑶𝒆
Sample Problems
4. A total flux of 200,000 maxwells exists in air between two
parallel pole faces, each 8 cm square. The field is uniform.
Determine the force in grams acting on an N pole having a
strength of 100 cgs units, when placed on this field.
Solution:
𝜙
𝐵=
𝐴
𝐵=𝐻
200,000
=
8 8
𝐵 = 3,125 𝐺
𝐻 = 3,125 𝑂𝑒
𝐹
𝐻=
𝑚
𝐹 = 𝑚𝐻
= 100 3,125
𝐹 = 312,500 𝑑𝑦𝑛𝑒𝑠
𝑭 = 𝟑𝟏. 𝟖𝟔 𝒈𝒓𝒂𝒎𝒔
Sample Problems
5. A cylindrical bar magnet is 20 cm long and has a diameter of 0.5
cm. The N and S poles at the ends have a strength of 40 cgs units.
Determine (a) intensity of magnetization (b) flux density at the
surface of a sphere having a radius of 2 cm and its center at either
pole, assuming point poles; (c) force on a 8 units strength at
surface of the sphere; (d) flux density at center zone of bar magnet.
Solution:
(a) For the intensity of magnetization
𝑚
𝐼=
𝐴
40
𝐼=
0.20
𝜋
𝜋 2
𝐴 = 𝑑 = 0.5
4
4
𝑨
𝑰 = 𝟐𝟎𝟑. 𝟕𝟐
𝒎
2
𝐴 = 0.20 𝑐𝑚2
(b) For the flux density at the surface of a sphere having a
radius of 2 cm and its center at either pole, assuming point
poles
𝐴 = 𝜋𝑟 2 = 𝜋 2
𝐴 = 4𝜋 𝑐𝑚2
2
𝜙 = 4𝜋𝑚 = 4𝜋 40
𝜙
𝐵=
𝐴
160𝜋
=
4𝜋
𝜙 = 160𝜋 Mx
𝑩 = 𝟒𝟎 𝑮
(c) For the force on a 8 units strength at surface of the sphere
𝐵=𝐻
𝐻 = 40 𝑂𝑒
𝐹 = 𝑚𝐻
= 8 40
𝑭 = 𝟑𝟐𝟎 𝒅𝒚𝒏𝒆𝒔
(d) For the flux density at center zone of bar magnet.
𝜋 2
𝐴= 𝑑
4
𝜋
= 0.5
4
𝜙 = 160𝜋 Mx
𝜙
𝐵=
𝐴
160𝜋
=
0.20
𝑩 = 𝟐, 𝟓𝟔𝟎 𝑮
2
𝐴 = 0.20 𝑐𝑚2
Magnetic Terminologies
11. Magnetic Potential (M) - it is the work done in moving a unit magnetic
pole from infinity to a point against the forces of a magnetic field.
𝒎
𝑴=
𝟒𝝅𝝁𝟎 𝝁𝒓 𝒓𝟐
𝒎
𝟏
𝟏
𝑴=
−
𝟒𝝅𝝁𝟎 𝝁𝒓 𝒓𝟐 𝒓𝟏 𝒓𝟐
12. Permeability (μ) of a material - it is the ability of a material to conduct
magnetic flux through it.
13. Relative Permeability - it is the ratio of the absolute permeability of a material
to the permeability of air or vacuum.
𝝁
𝝁𝒓 =
𝝁𝟎
Magnetic Terminologies
14. Magnetic Susceptibility (χm) - it is a dimensionless parameter which is a
measure of the ability of a material to be magnetized. It is the ratio of the
intensity of magnetization produced in a substance to the magnetizing
force producing it.
𝝌𝒎 =
𝑰
𝑯
For ferromagnetic and paramagnetic materials, χm is positive
For diamagnetic materials, χm is negative
For non-magnetic materials, χm is equal to zero
15. Relationship between Permeability and Susceptibility
The relationship between the relative permeability and magnetic
susceptibilty can be derive as:
𝝁𝒓 = 𝟏 + 𝝌𝒎
Sample Problems
6. Find the magnitude of flux density in a material for which (a)
the magnetization is 2.8 A/m and the magnetic susceptibility
is 0.0025, (b) the magnetic field intensity is 1,300 A/m and
the relative permeability is 1.006.
Solution:
(a) For the flux density if the magnetization is 2.8 A/m and the
magnetic susceptibility is 0.0025
𝐴
2.8
𝐼
𝐻 = 1,120
0.0025 =
𝜒𝑚 =
𝑚
𝐻
𝐻
𝜇𝑟 = 1 + 𝜒𝑚
𝐵 = 𝜇 0 𝜇𝑟 𝐻
= 1 + 0.0025
𝜇𝑟 = 1.0025
𝐵 = 4𝜋 × 10−7 1.0025 1,120
𝑩 = 𝟏. 𝟒𝟏 × 𝟏𝟎−𝟑 𝑻
(b) For the flux density if the magnetic field intensity is
1,300 A/m and the relative permeability is 1.006
𝐵 = 𝜇0 𝜇𝑟 𝐻
𝐵 = 4𝜋 × 10−7 1.006 1,300
𝑩 = 𝟏. 𝟔𝟒 × 𝟏𝟎−𝟑 𝑻
Electromagnetism
1. Electromagnetism - it is the branch of engineering which deals
with the magnetic effects of electric current.
2. Magnetic Effects of Electric Current through a Current
Carrying Conductor
•
A magnetic field is set up when an electric current flows
though a conductor.
• The magnetic lines of force are circular in a plane
perpendicular to the current.
• The magnetic field near a conductor is stronger and becomes
weaker as we go farther away from the conductor.
• The direction of the field is reversed if direction of current is
reversed.
•
Direction of Magnetic Field
There are two methods which are used in
determining the direction of magnetic field.
These are:
• Ampere’s Right Hand Rule (Right Hand
Thumb Rule). Grasp the conductor in your
right hand with the thumb pointing in the
direction of current and the encircling finger
will point in the direction that the magnetic
field encircles the conductor.
• Right Hand Corkscrew Rule. Held the
corkscrew in your right hand and move it in
such a way that it moves forward in the
direction of current and the direction of
which the fingers move in the direction of
magnetic field round the conductor.
Fleming’s Left Hand Rule
Fleming’s left hand rule is used to
determine the direction of the force
acting on a conductor. With your left
hand, stretch out the thumb, forefinger
and middle finger so that these are at
right angles with each other. This was
invented by the English electrical
engineer and physicist Sir John
Ambrose Fleming (1849 – 1945).
Thumb
 direction of force
Forefinger
 direction of field
Middle finger  direction of current
Direction of Current
Symbolic Representation of a Current Carrying Conductor
•
A cross represents the direction of current flow in the conductor going into the
plane of paper. A cross represents the tail of an arrow.
•
A dot represents the direction of current flow in the conductor going outwards
the plane of paper. A dot represents the head of an arrow.
Force on a Conductor in a Magnetic Field
When a current carrying conductor is placed in a magnetic field, the conductor
experiences a mechanical force which acts in a direction perpendicular to both the
direction of current and the field.
The force on the conductor is given by,
𝑭 = 𝑩𝑳𝑰
Where:
B = magnetic flux density of field produced by the poles in Tesla
L = length of the conductor in meter
I = current flowing through the conductor in ampere
F = mechanical force on the conductor in Newton
If the conductor makes an angle  with the field then,
𝑭 = 𝑩𝑳𝑰 𝐬𝐢𝐧 𝜽
Sample Problems
7. A straight long conductor of 1 meter length carrying a
current of 60 amperes is placed at right angle to a uniform
magnetic field strength of 2.5 T. Determine the mechanical
force acting on the conductor.
Solution:
𝐹 = 𝐵𝐿𝐼
𝑭 = 𝟏𝟓𝟎 𝑵
= 2.5 1 60
Magnetic Field Strength around a
Long Straight Conductor
Given a straight long conductor carrying a current of I amperes in
downward direction
The field strength at any point at a
distance of r meters from the center of
the conductor due to its field is:
𝑯=
𝑰
𝟐𝝅𝒓
If there are N conductors, then
𝑯=
𝑵𝑰
𝟐𝝅𝒓
Sample Problems
8. Calculate the magnetizing force and flux density at a distance of 5
cm from a long straight conductor carrying a current of 250 A and
placed in air.
Solution:
𝑟 = 5 𝑐𝑚 = 0.05 𝑚
𝐼
𝐻=
2𝜋𝑟
250
=
2𝜋 0.05
𝑨
𝑯 = 𝟕𝟗𝟓. 𝟕𝟕
𝒎
Magnetic Field Strength of a
Closely Wound Toroidal Core
The path in this case is a circle with
a radius r. The path links N turns
carrying currents of the same
direction.
𝑯=
𝑵𝑰
𝑳
Where:
L = is the length of the toroidal
core or solenoid
Sample Problems
9. An air-cored solenoid of length 50 cm has 1,000 turns and carries a
current of 5 A. Find (a) field strength and (b) flux density inside the
coil.
Solution:
(a) For the field strength
𝐿 = 50 𝑐𝑚 = 0.5 𝑚
1,000 5
𝑁𝐼
=
𝐻=
0.5
𝐿
𝑨
𝑯 = 𝟏𝟎, 𝟎𝟎𝟎
𝒎
(b) For the flux density inside the coil
𝐵 = 𝜇 0 𝜇𝑟 𝐻
= 4𝜋 × 10−7 1 10,000
𝑩 = 𝟎. 𝟎𝟏𝟐𝟔 𝑻
Magnetic Field Strength Along the
axis of a Square Coil
Consider a square coil with length equal to 2a meters carrying a
current of I amperes
The magnetic field strength at the
center of the square coil is:
𝑰 𝟐
𝑯=
𝝅𝒂
Magnetic Field Strength Along the
axis of a Circular Coil
Consider a single turn circular coil of radius r meters and carrying a
current of I amperes
The magnetizing force at the
axial point P is given by:
𝑰
𝑯=
𝒔𝒊𝒏𝟑 𝜽
𝟐𝒓
𝑵𝑰
𝑯=
𝒔𝒊𝒏𝟑 𝜽
𝟐𝒓
 For single turn coil
 For N turn coil
The magnetizing force at the center point of the coil is:
𝑰
𝑯=
𝟐𝒓
 For single turn coil
𝑵𝑰
𝑯=
𝟐𝒓
 For N turn coil
Sample Problems
10. A wire 2.5 m long is bent (a) into a square (b) into a circle. If the
current flowing through the wire is 100 A, find the magnetizing force
at the center of the square and the center of the circle.
Solution:
(a) When bent into a square
2.5
𝑎 = 0.3125 𝑚
2𝑎 =
4
𝑨
100 2
𝐼 2
𝑯 = 𝟏𝟒𝟒. 𝟎𝟓
=
𝐻=
𝒎
𝜋 0.3125
𝜋𝑎
(b) When bent into a circle
𝐶 = 2𝜋𝑟
𝐼
𝐻=
2𝑟
2.5 = 2𝜋𝑟
100
=
2 0.40
𝑨
𝑯 = 𝟏𝟐𝟓
𝒎
𝑟 = 0.40 𝑚
Sample Problems
11. A current of 15 A is passing along a straight wire. (a) Calculate the
force on a unit magnetic pole placed 0.15 m from the wire. (b) If the
wire is bent to form a loop, calculate the diameter of the loop so as
to produce the same force at the center of the coil upon a unit
magnetic pole when carrying a current of 15 A.
Solution:
(a) For the force on a unit magnetic pole placed 0.15 m from
the wire
𝐴
15
𝐼
𝐻 = 15.92
=
𝐻=
𝑚
2𝜋 0.15
2𝜋𝑟
𝐹 = 𝑚𝐻
= 1 15.92
𝑭 = 𝟏𝟓. 𝟗𝟐 𝑵
(b) If the wire is bent to form a loop, calculate the diameter
of the loop so as to produce the same magnetizing force at
the center of the coil upon a unit magnetic pole when
carrying a current of 15 A.
𝐼
𝐻=
2𝑟
15
15.92 =
2𝑟
𝒓 = 𝟎. 𝟒𝟕 𝒎
Force between Two
Parallel Conductors
The force between two parallel conductors is given by:
𝑭=
𝟐𝑰𝒂 𝑰𝒃 𝑳
× 𝟏𝟎−𝟕
𝒅
Where:
Ia, Ib = current carried by the two conductors a and b
respectively
d
= distance between the two conductors
L
= length of the conductor
F
= force between the two conductors
Ampere’s Work Law or Ampere’s Circuital Law
This law states that the magnetomotive force (mmf) around a closed
path is equal to the current enclosed by that path.
Sample Problems
12. Two parallel wires each of 3 meter length have a separation of 4 mm.
Calculate the force exerted on each of these wires when they carry a
current of 5 A in the same direction.
Solution:
2𝐼𝑎 𝐼𝑏 𝐿
𝐹=
× 10−7
𝑑
2 5 5 3
=
× 10−7
0.004
𝑭 = 𝟑. 𝟕𝟓 × 𝟏𝟎−𝟑 𝑵
𝑑 = 4 𝑚𝑚 = 0.004 𝑚
Sample Problems
13. Two infinite parallel conductors carry parallel currents 10 A each. Find
the magnitude and direction of the force between the conductors per
meter length if the distance between them is 20 cm.
Solution:
2𝐼𝑎 𝐼𝑏 𝐿
𝐹=
× 10−7
𝑑
𝐹 2 10 10
=
× 10−7
𝐿
0.2
𝑭=𝟏×
𝟏𝟎−𝟒
𝑵
𝒎
𝑑 = 20 𝑐𝑚 = 0.2 𝑚
Sample Problems
14. Two long straight parallel wires, standing in air 2 m apart, carry
currents I1 and I2, in the same direction. The magnetic intensity at
point midway between the wires is 7.95 AT/m. If the force on each wire
per unit length is 2.4 x 10-4 N, evaluate I1 and I2.
Solution:
2𝐼1 𝐼2 𝐿
𝐹=
× 10−7
𝑑
𝐹 2𝐼1 𝐼2
=
× 10−7
𝐿
𝑑
2.4 ×
10−4
2,400
𝐼1 =
𝐼2
2𝐼1 𝐼2
=
× 10−7
2
𝐼
𝐻=
2𝜋𝑟
𝐼1
𝐻1 =
2𝜋𝑟1
𝐼2
𝐻2 =
2𝜋𝑟2
𝐻𝑚𝑖𝑑𝑤𝑎𝑦 = 𝐻1 − 𝐻2
𝐻𝑚𝑖𝑑𝑤𝑎𝑦
𝐼1
𝐼2
=
−
2𝜋𝑟1 2𝜋𝑟2
𝐼1
𝐼2
7.95 =
−
2𝜋(1) 2𝜋(1)
𝐼1 − 𝐼2 = 49.95
2,400
𝐼1 =
𝐼2
Solving for I1 and I2:
𝑰𝟏 =
𝑨
𝑰𝟐 =
𝑨
Sample Problems
15. Two long parallel wires A and B, 12 cm apart carry currents of 750 A
and 500 A respectively in opposite directions. Determine the flux
density at the midpoint of the perpendicular line between the wires.
Solution:
𝐼
𝐻=
2𝜋𝑟
𝐼1
𝐻1 =
2𝜋𝑟1
𝐻𝑚𝑖𝑑𝑤𝑎𝑦 = 𝐻1 + 𝐻2
𝐼2
𝐻2 =
2𝜋𝑟2
𝑑 = 12 𝑐𝑚 = 0.12 𝑚
𝐻𝑚𝑖𝑑𝑤𝑎𝑦
𝐼1
𝐼2
=
+
2𝜋𝑟1 2𝜋𝑟2
𝐻𝑚𝑖𝑑𝑤𝑎𝑦
750
500
=
+
2𝜋(0.06) 2𝜋(0.06)
𝐻𝑚𝑖𝑑𝑤𝑎𝑦
A
= 3,315.73
m
𝐵𝑚𝑖𝑑𝑤𝑎𝑦 = 𝜇0 𝜇𝑟 𝐻𝑚𝑖𝑑𝑤𝑎𝑦
= 4𝜋 × 10−7 1 3,315.73
𝑩𝒎𝒊𝒅𝒘𝒂𝒚 = 𝟒. 𝟏𝟕 × 𝟏𝟎−𝟑 𝑻