Magnetism

Magnetism
Prepared and Presented by:
Victor R. Oribe
Ph.D. Sci. Ed.-Student

Magnetism

• Is the study of magnetic fields and their
effect on materials.
• The effect is due to unbalanced spin of
electrons in atom.
• It is readily observed every day – from
the simple magnet that attracts nails
and other metals to cassette tapes to
magnet-driven trains.

Magnetism
• In terms of applications, magnetism is one of the
most important fields in physics.
• Large electromagnets are used to pick up heavy
loads.
• Magnets are used in such devices as meters,
motors, and loudspeakers.
• Magnetic tapes and disks are used routinely in
sound-and video-recording equipment and to
store computer data.
• Intense magnetic fields are used in magnetic
resonance imaging (MRI) devices to explore the
human body with better resolution and greater
safety than x-rays can provide.

Magnetism
• Giant superconducting magnet are used in the
cyclotrons that guide particles into targets at nearly
the speed of light.
• Magnetism is closely linked with electricity.
• Magnetic fields affect moving charges, and moving
charges produce magnetic fields.
• Changing magnetic field can even create electric
fields.
• These phenomena signify an underlying unity of
electricity and magnetism, which James Clerk
Maxwell first described in the 19th century.
• The ultimate source of any magnetic field is electric
current.

Nature of Magnetism

• In the ancient country of Lydia, in
western Asia Minor, now Turkey, was
a city called Magnesia.
• The Greeks discovered that certain
iron ores found in the place could
attract other pieces or iron, they called
it magnetites.
• Magnetites are classified as natural
magnet.

Nature of Magnetism
It is now believed that
magnetism is due to the
spin of electrons within
the atoms.
Since the electron is a
charged particle, the
concept implies that
magnetism is a property
of a charged particle in
motion.

Nature of Magnetism

• The power of attraction of a magnet
depends on the arrangement of the
atoms.
• All atoms are in themselves tiny
magnet formed into groups called
DOMAINS.
• The magnetic strength is increased if
the domains are induced to fall into
line by the action of another magnet.

General Properties of Magnet

• The properties of naturally occurring
magnets (magnetites) have been
known for over 2,000 years.
• Several studies on magnetism were
made, but the first thorough
investigation was done by William
Gilbert in 1600.
• Experimental results led to the
discovery of the many properties of
natural and artificial magnets.


1. Magnets usually have two poles.
• The end of the magnet which points north when
magnet is free to turn on a vertical axis is the
north-seeking pole, simply the N pole.


• The opposite end which points south is the
south-seeking pole or S pole.

• Magnets come in many shapes and sizes, but
each has at least two poles.

• If you cut a magnet into pieces, every piece will
still have at least two poles.

2. Like Magnetic poles repel and
unlike poles attract.

Charles Augustine de Coulomb,
a French physicist, was the first
recognized scientist to study
quantitatively the force exerted
by magnets.

The result of his experiments are
summarized in what is known as
Coulomb’s Law of Magnetism:

“The force of attraction/repulsion
between two magnetic poles is directly
proportional to the strength of the
poles and inversely proportional to the
square of the distance between them”.

The result of his experiments are
summarized in what is known as
Coulomb’s Law of Magnetism:

“The force of attraction/repulsion
between two magnetic poles is directly
proportional to the strength of the
poles and inversely proportional to the
square of the distance between them”.

Coulomb’s Law

• In the MKS system of units, the unit of
charge is the coulomb, the force is
expressed in newtons and the distance
in meter.

• A coulomb is a very large unit of charge.
A smaller unit is the statcoulomb.
1 coul = 3 x109 statcoul
1 coul = 106 microcoul

Coulomb’s Law

• Experiments have been undertaken to calculate
the approximate value of k. This was found to
be:

• The smallest quantity of charge is the charge
on the electron. This is called electronic
charge.
• Robert Millikan was able to obtain the value of
this charge to be e = 1.6 x 10-19 coul.
• This is also the charge of proton.

Problems to Illustrate
Coulomb’s Law

Problem # 1

Two point charges of 2.5 x 10-10 coul and -3.0 x
10-10 coul are 10 cm apart in air. Calculate the
magnitude and direction of the force on each
charge.
10 cm or .10 m

-3.0 x 10-10 coul 2.5 x 10-10 coul

N.m2 (-3.0 x 10-10 coul)(2.5 x 10-10 coul)
F = 9 X109 Coul2
(.10m)2
F = -6.75 x 10-8 N

• The result (F= -6.75 x 10-8 N) means that:
• F12 has a magnitude of 6.75 x10-8 N and is
directed toward q1.

• F21 has a magnitude of -6.75 x 10-8 N and is
directed toward q2.

Therefore: F12 = - F21

Problem # 2
Calculate the force between two point
charge of +4.5 x 10-6 coul and -5.0 x
10-6 coul which are 15 cm apart in air.

Answer:
- 9.0N
The negative sign in the answer
indicates that the force between the
point charges is an attractive force.

Problem # 3
Given the three point charges q1, q2, and q3 as
shown in the figure. If q1 = +1.5 x 10-6 coul, q2 =
+2.5 x 10-6 coul and q3 = -2.0 x 10-6 coul and 300 ,
d12 =10cm, d13 = 15cm.
Find: a) the force between q1 and q3
b)The resultant force on q1
q3 = -2.0x10-6 coul

F13 = ?

Ø d13 = 0.15m

d12 =0.10m
F12 =? q1 = +1.5x10-6 coul q2 = +2.5 x10-6 coul

Solution: a) the force between q1 and q3
k (q1 q3)
F13 =
r2
9 x 109 N.m2 / coul2 (+1.5x10-6 coul) (-2.0x10-6 coul)
F13 =
(0.15m)2
F13 = -1.2N The negative value of F13 indicates an attractive
force on q1 due to q3.

k (q1 q2)
F12 =
r2 q3 = -2.0x10-6 coul
9 x 109 N.m2 / coul2 (+1.5x10-6 coul) (-2.0x10-6 coul)
(0.01m)2
F12 = 3.375N F13 = ?
d13 = 0.15m
The positive value of F12
indicates a repulsive force q2 = +1.5x10-6 coul
F12 = ?
on q1 due to q2. d12 =0.10m

q1 = +1.5x10-6 coul

Solution: b)The resultant force on q1
To find the resultant force (Fnet ) on q1, add F12 and F13
vectorially.

The sum of the x-component is:
∑X = -3.375N + 1.2 cos 600
∑X = -2.775
The sum of the y-component is:
∑Y = -1.2 sin 600
∑X = 1.039 FNet = ?

300
Fnet =√(-2.775)2 + (1.039)2
F13 = 1.2 N
Fnet = 2.96 N
F12 = -3.375N

Solve the following problems:
1. Find the force between two eqaul charges of 15 x 10-8
coul if they are 20 cm apart in air. -3
5.06 X10 N
2. Find the magnitude and direction of the force in each of
two charges of 3.5 x 10-6 coul and -2.8 x 10-6 coul, 10 cm
apart. 8.82 X102 N, attractive
3. Find the magnitude and kind of force between two
charges of 20 x 10-6 coul which are 25 cm apart in air.
432 N, repulsive
4. A charge of 2.8 x 10-6 coul is at a distance of 12 cm from
another charge of -8.4 x10-6 coul. Find the magnitude
and direction of the force on the first charge that is due to
the second charge. +14.7 N
5. An unknown charge is attracted by a force of 25N when it
is at a distance of 10 cm from another charge of 25 x 10-6
coul. What is the magnitude of the un known charge?
q1 = 1.11 x 10-6 coul


3. A piece of
magnetite, when
made to hang and
swing freely,
would align itself
with the magnetic
field of the earth
following a north-
south direction.


4. Permanent magnets are
magnets made from alloys of
cobalt and nickel.

These magnets retain
their magnetism for a
long time.


5. Other metals like iron can be
magnetized by Induction.
When a piece of iron nails
touches a permanent magnet,
the nails becomes a magnet.

It retains in this condition for as
long as it is within the magnetic
field.

The nail is a temporary magnet
and its magnetism is described
as induced magnetism.

Magnetic Field of Force
• Experiment show that a stationary charged
particle doesn’t interact with a static magnetic
field.
• When a charged particle is moving through a
magnetic field, however, a magnetic force
acts on it.

• The force has its maximum value when the
charge moves in a direction perpendicular to the
magnetic field line, decreases in value at other
angles, and becomes zero when moves along
the field of lines.

• Magnetic force on a moving charge is directed
perpendicular to the magnetic field.
• It is found experimentally that the strength of the
magnetic force on the particle is proportional to
the magnitude of the charge q, the magnitude of
the velocity v, the strength of the external
magnetic field B, and the sine of the angle
between the direction of v and the direction of B.
F = qvB sin θ
• This expression is used to define the magnitude
of the magnetic field as:
F
B=
qv sin θ

Earth’s Magnetic Field

• A small bar magnet is said to have north
and south poles, but is it more accurate
to say it has a “north-seeking” pole and
“south-seeking” pole.
• By this expressions, we mean that if
such a magnet is used as a compass,
one end will “seek” or point to, the
geographic North Pole of Earth and the
other end will “seek” or point to, the
geographic South Pole of Earth.

• We therefore conclude that:
The Geographic North Pole of Earth corresponds
to a magnetic south pole, and the geographic
South Pole of Earth corresponds to a magnetic
north pole.


• The magnetic field pattern of Earth is
similar to the pattern that would be set
up by a bar magnet placed at its center.
• An interesting fact concerning Earth’s
magnetic field is that its direction
reverses every few million years.

Earth’s Magnetic Field and its Present
Application – Labeling Airport Runways
• The magnetic field of Earth is used to label
runways at airports according to their direction.
• A large number is painted on the end of the
runway so that it can be read by the pilot of an
incoming airplane.

Earth’s Magnetic
Field and its
Present Application
– Labeling Airport
Runways

• This number describes the direction in which
the airplane is traveling, expressed as the
magnetic heading, in degrees measured
clockwise from magnetic north divided by 10.
• A runway marked 9 would be directed toward
the east (900 divided by 10), whereas a runway
marked 18 would be directed toward magnetic
south.

• If F is in newton, q in coulombs, and v in meter
per second, the SI unit of magnetic field is the
tesla (T), also called the weber (Wb) per square
meter (1 T = 1 Wb/m2).

• The unit of B is:
B = T = Wb/m2 = N/C.m/s = N/A.m

• The cgs unit of magnetic field is the gauss (G).
1 T = 104 G

• From equation F=qvB sin θ, we see that the force
on a charged particle moving in a magnetic field
has its maximum value when the particle’s motion
is perpendicular to the magnetic field,
corresponding to θ=900 , so that sin θ = 1.
• The magnitude of this maximum force has the
value:
Fmax = qvB
• Experiment also show that the direction of the
magnetic force is always perpendicular to both v
and B. To determine the direction of force, we
employ the right-hand rule # 1.

Right-Hand Rule #1

The implications of this expression include:

1. The force is perpendicular to both the velocity v of
the charge q and the magnetic field B.

2. The magnitude of the force is F = qvB sinθ where θ
is the angle < 180 degrees between the velocity and
the magnetic field. This implies that the magnetic
force on a stationary charge or a charge moving
parallel to the magnetic field is zero.

3. The direction of the force is given by the right hand
rule.

Right-Hand Rule # 1
Right-Hand Rule #1 determines the directions
of magnetic force, conventional current and the
magnetic field. Given any two of these, the
third can be found.
Using your right-hand:
1. point your index finger in the
direction of the charge's
velocity, v, (recall conventional
current).

2. Point your middle finger in the
direction of the magnetic field,
B.
3. Your thumb now points in
the direction of the magnetic
force, Fmagnetic.

Right-Hand Rule # 1

When the magnetic force relationship is applied
to a current-carrying wire, the right-hand rule
may be used to determine the direction of force
on the wire.

Right-Hand Rule # 2
Right-Hand Rule #2 determines the direction
of the magnetic field around a current-
carrying wire and vice-versa
Using your right-hand:
Curl your fingers into
a half-circle around
the wire, they point in
the direction of the
magnetic field, B

Point your thumb in
the direction of the
conventional current.

Problem
• A proton moves with a speed of 1.00 x 105 m/s
through Earth’s magnetic field, which has a value of
55.0µT at a particular location. When the proton
moves eastward, the magnetic force acting on it is
directed straight upward, and when it moves
northward, no magnetic force act on it.
a) What is the direction of the magnetic field?
b) What is the strength of the magnetic force when the proton
moves eastward?
c) Calculate the gravitational force on the proton and compare it
with the magnetic force. If there were an electric field with a
magnitude equal to E=1.50x102 N/C at that location, a
common value at Earth’s surface. Note the mass of the
proton is 1.67 x10-27 kg.

Solution:
a) What is the direction of the magnetic field?

No magnetic force act on the proton when it’s going
North, so the angle such a proton makes with the
magnetic field direction must be either o0 or 1800.
Therefore, the magnetic field B must point either
north or south.
Now apply the right hand rule. When the particle
travels east, the magnetic force is directed upward.
Point your thumb in the direction of the force and
your finger in the direction of the velocity eastward.
When you curl your finger, they point north, which
must therefore be the direction of the magnetic
field.

Solution
a) What is the strength of the magnetic force when the
proton moves eastward?
Substitute the given values and the charge of a proton into
equation F=qvB sin θ. From part (a), the angle between the
velocity v of the proton and magnetic field B is 900

F= qvB sin θ
F= (1.60x10-19 C)(1.00X105 m/s) x (55.0x10-6 T) sin (900 )
F= 8.80 x 10-19 N

Solution:
Calculate the gravitational force on the proton and
compare it with the magnetic force and also with
the electric force if E=1.50 X102 N/C.
Fgrav. = mg
= (1.67x10-27 kg.) (9.8 m/s2 )
= 1.64 x10-26 N
Felec = qE
= (1.60x10-19 C) (1.50 X 102 N/C)
= 2.40 X 10-17 N

Problem # 2
A Proton Moving in a Magnetic Field
A proton moves at 8.00 x 106 m/s along the x-axis. It
enters a region in which there is a magnetic field of
magnitude 2.50 T, directed at an angle of 600 with
the x-axis and lying in the xy-plane.
a)Find the initial magnitude and direction of the
magnetic force on the proton.
b) Calculate the proton’s initial acceleration.

Hint: Finding the magnitude and direction of the magnetic
force requires substituting values into the equation for
magnetic force, (a) (F=qvB sin θ), and using the right-
hand rule. (b)Applying Newton’s second law.

Solution:
a) Find the magnitude and direction of the
magnetic force on the proton.
F=qvB sin θ
= (1.60x10-19C) (8.00X106 m/s) (2.50T) (sin 600)
= 2.77 x10-12 N

Point the fingers of the right hand in the x-direction
(direction of v) and then curl them toward B. The
thumb points (upward, in the positive z-direction.

Solution:

b) Calculate the proton’s initial acceleration.

am = F
(1.67 X10-27 kg) ( a ) = 2.77 x 10-12 N
a = 1.66 x 1015 m/s2

Magnetic Force on a Current-
Carrying Conductor
• If a straight conductor of length ℓ carries current,
the magnetic force on that conductor when it is
placed in a uniform external magnetic field B is:
F = BIℓ sin θ
• Where
 Ɵ is the angle between the direction of the current
and the direction of the magnetic field.
 B is the direction of the current.
 I is the current
 ℓ is the length of the wire

Magnetic Force on a Current-
Carrying Conductor
• Right-hand rule #1 also gives the direction of
the magnetic force on the conductor. In this
case, however, you must point your fingers in
the direction of the current rather than in the
direction of v.

Application:
• A magnetic force acting on a current-carrying
wire in a magnetic field is the operating
principle of most speakers in sound systems.

Application:
• Speaker is consists of a coil of wire called the voice
coil, a flexible paper cone that acts as the speaker,
and a permanent magnet.
• The coil of wire is surrounding the north pole of the
magnet is shaped so that the magnetic field lines are
directed radially outward from the coil’s axis.
• When an electrical signal is sent to the coil,
producing a current in the coil, a magnetic force to
the left acts on the coil. (can be seen by applying
right-hand rule #1 to each turn of wire).

Application:
• When the current reverses direction, as it would
for a current that varied sinusoidally, the
magnetic force on the coil also reverses
direction, and the cone, which is attached to the
coil, accelerates to the right.
• An alternating current through the coil causes
an alternating force on the coil, which results in
vibrations of the cone.
• The vibrating cone creates sound waves as it
pushes and pulls on the air in front of it.
• In this way, a 1-kHz electrical signal is
converted to a 1-kHz sound waves.

Additional Research Work

• Research on the other application of current-
carrying conductors, and explain how it works.
• What is electromagnetic pumps (artificial heart
and kidney). Explain how current-carrying
conductors is applied in the manufacture of
electromagnetic pump.

Problem:
• A wire carries a current of 22.0 A from
west to east. Assume the magnetic field
of Earth at this location is horizontal and
directed from south to north and it has a
magnitude of 0.500 x 10-4 T.
• a) Find the magnitude and direction of
the magnetic force on a 36.0-m long of
wire.
• b) Calculate the gravitational force on
the same length of wire if it’s made of
copper and has a cross-sectional area of
2.50 x 10-6 m2 .

Solution:
• a) Find the magnitude and direction of the
magnetic force on a 36.0-m long of wire.
F = BIℓ sin θ
= (0.500 x 10-4 T) (22.0 A) (36.0 m) sin 900
= 3.96 x 10-2 N
Applying the right-hand rule #1 to find the direction
of the magnetic force:
With fingers of your right hand pointing west
to east in the direction of the current, curl them
north in the direction of the magnetic field. Your
thumb points UPWARD.

Solution:
• b) Calculate the gravitational force on the
same length of wire if it’s made of copper
and has a cross-sectional area of 2.50 x
10-6 m2 . Density of the
metal
• m= ρv = ρ(Aℓ)
(2.50x10-6 m2 . 36.0m)
=8.92x103 kg.m3)
= 0.803 kg
• To get the gravitational force, multiple the mass by
the acceleration of gravity:
• Fgrav. = mg
= (0.803 kg) (9.8 m/s2)
= 7.87 N

Torque on a Current Loop and
Electric Motor
• The torque Ƭ on a current-carrying loop of wire in a
magnetic field B has magnitude:
Ƭ = BIA sin θ
• Where:
 I is the current in the loop
 A is the cross-sectional area
• The magnitude of the magnetic moment of the
current-carrying coil is defined by µ = IAN
 N is the number of loops
• The magnetic moment is considered vector, µ, that
is perpendicular to the plane of the loop.
• The angle between B and µ is θ.

Problem:
• A circular wire loop of radius 1.00 m is placed in
a magnetic field of magnitude 0.500 T. The
normal to the plane of the loop makes an angle
of 30.00 with the magnetic field.(see Illustration)
The current in the loop is 2.00 A in the direction
shown.
• a) Find the magnetic
moment of the loop and the
magnitude of the torque at
this instant.
• b) The same current is
carried by the rectangular
2.00-m by 3.00-m coil with
three loops. Find the
magnetic moment of the coil
and the magnitude of the
torque acting on the end of
the coil at that instant.

Solution:
• a) Find the magnetic moment of the circular
loop and the magnetic torque exerted on it.
• Hint:
 First, calculate the enclosed area of the circular
loop.
 Calculate the magnetic moment of the loop
 Substitute values for the magnetic moment,
magnetic field, and θ into Ƭ = µB sin θ

A = ╥r2 = 1.00m)2 = 3.14m2
µ = IAN = (2.00A) (3.14m2) (1) = 6.28A.m2
Ƭ = µb sin θ = (6.28A.m2) (0.500T) (sin 30.00)
= 1.57 N.m

Solution:
• Find the magnetic moment of the rectangular
coil and the magnetic torque exerted on it.
• Hint:
 Calculate the area of the coil.
 Calculate the magnetic moment of the coil
 Substitute values into equation Ƭ = µB sin θ

A= L x H = (2.00m) (3.00m) = 6.00m2
µ= IAN = (2.00 A) (6.00 m2) (3) = 36.0 A.m2
Ƭ = µb sin θ = (.500T) (36.0 A.m2) (sin 300)
= 9.00 N. m

Application:
• Its hard to imagine life in the 21st century without motors.
• Some of the appliances that contain motors include
computer disk drives, CD players, DVD players, food
processors and blenders, car starters, furnaces, and air
conditioners.
• The motors convert electrical energy to kinetic energy of
rotation and consist of a rigid current-carrying loop that
rotates when placed in the field of a magnet.

Magnetic Field of a Long,
Straight Wire
During the lecture
demonstration in
1819, Danish Scientist
Hans Oersted found than
an electric current in a wire
deflected a nearby
compass needle.

This momentous discovery, linking a magnetic
field with an electric current for the first time,
was the beginning of our understanding of the
origin of magnet.

Straight Wire
• In the simple experiment carried by Oersted in
1820, several compass needles are placed in a
horizontal plane near a long vertical wire.

Straight Wire
• When there is no current in the wire, all needles
point in the same direction (that of Earth’s field),
as one would expect.

• When the wire carries a
strong, steady current,
however, the needles all
deflect in directions
tangent to the circle.

Straight Wire
• The observation of Oersted show that the
direction of B is consistent with the following
convenient rule, Right-Hand Rule # 2.

Straight Wire
• The magnetic field at distance ṙ from along,
straight wire carrying current I has the
magnitude:
µ0I
B=
2╥ṙ

• Where;
 µ0 = 4╥ X 10-7 T.m/A is the permeability of free
space
 The magnetic field lines around a long, straight
line wire are circles concentric with the wire.

Magnetic Field of a
Long, Straight Wire
• Ampere’s law can be used to find the magnetic
field around certain simple current-carrying
conductors. It can be written:
∑Bǁ ∆ℓ = µ0I
• Where:
 Bǁ is the component B tangent to a small
current element of length ∆ℓ that is part of a
closed path and I is the total current that
penetrates the closed path.

Sample Problem:
• A coaxial cable consists of an insulated wire
carrying current I1 = 3.00 A surrounded by a
cylindrical conductor carrying current I2 = 1.00 A
in the opposite direction.( see illustration)

a) calculate the magnetic
field inside the cylindrical
conductor at rint = 0.500cm.
Calculate the magnetic field
outside the cylindrical
conductor at rext = 1.50 cm.

Solution:
• a) Calculate the magnetic field Bint inside the
cylindrical conductor at rint = 0.500 cm.
• Hint:
 Write the Ampere’s law
 The magnetic field is constant on the given path
and the total path length is 2╥ṙint :
 Solve for Bint and substitute values:
µ0I1
∑Bǁ ∆ℓ = µ0I Bint =
2╥ṙint
Bint =(2╥ṙint )= µ0I1
(4╥ x 10-7 T. m/A) (3.00A)
=
2╥ (0.005m)
= 1.20 x 10-4 T

Solution:
• b) Calculate the magnetic field Bext outside the
cylindrical conductor at rext = 1.50 cm.
• Hints:
 Write the Ampere’s law
 The magnetic field is again constant on the given
path and the totla path length is 2╥rext
 Solve for Bext and sustitute values

∑Bǁ ∆ℓ = µ0I
Bext (2╥ṙext )= µ0(I1 – I2)
µ0(I1 – I2) (4╥ X 10-7 T.m/A) (3.00 A – 1.00 A)
Bext =
2╥ṙext 2╥ (0.015 m)

= 2.64 x 10-5 T

Magnetism

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