# CHAPTER 4: MAGNETISM

Electronics - Devices

Oct 18, 2013 (4 years and 6 months ago)

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CHAPTER
4: MAGNETISM

4.2

Magnets and magnetic fields

4.2 Magnetic forces on moving charges particles

4.3
Magnetic forces and magnetic field due to current carrying wire

4.4
Magnetic Force between Wires

4.5

Torque on a Current Loop

4.6

Ampere's Law

4.7

Magnetic

materials

4.1

Magnets

and magnetic fields

A

phenomenon apparently unrelated to electricity is magnetism. We are familiar with
magnetism through the interaction of compasses with the earth's magnetic field, or
through fridge magnets or magnets on childr
en's toys. Magnetic forces are explained in
terms very similar to those used for electric forces:

There are two types of
magnetic poles
, conventionally called North and South

Like poles repel, and opposite poles attract

However, magnetism differs from e
lectricity in one important aspect:

Unlike electric charges, magnetic poles always occur in North
-
South pairs; there
are no
magnetic monopoles
.

As in the case of electric charges, it is convenient to introduce the concept of a
magnetic
field

(vector quan
tity)
in describing the action of magnetic forces. Magnetic field lines
for a bar magnet are pictured below.

Figure 4.1:

Magnetic field lines of a bar magnet

One c
an interpret these lines as indicating the direction that a compass needle will point if
placed at that position.

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The symbol for magnetic field is
B
.
The strength of magnetic fields is measured in units
of
Teslas

(T). One tesla is actually a relatively st
rong field
-

the earth's magnetic field is
of the order of 0.0001 T.

Figure 4.2

The Earth’s magnetic field resembles that achieved by burying a huge bar

magnet deep in
the Earth’s

interior

as in Figure 4.2
.The Earth’s
geographic

North Pole
corresponds

to a
magnetic south pole.

The Earth’s
geographic

South Pole
corresponds to a
magnetic
north pole.

4.2

M
agnetic forces on moving charged particles

a
moving

charge will experience a force

when move in an external magnetic field
.

the force on the charged p
article is always perpendicular to the
field(B) and the
direction

(V)

it is moving.

Thus magnetic forces cause charged particles to change their direction of motion,
but they do not change the
speed

of the particle.

-

The magnitude of the force is

where
is the angle between

the
and
vectors.

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I
f
the force is perpendicular to the motion, then no work is done. Hence magnetic
forces do no work on charged particles and cannot increase their kinetic energy.

If a charged particle moves through a constant magnetic field, its speed stays the
same, but it
s direction is constantly changing. A device in which this property is
used is the
mass spectrometer
, which is used to identify elements. A basic mass
spectrometer is pictu
red below in Figure
.

Figure 4.3:

Mass spectrometer

In this device a beam of charged particles (
ions
) enter a region of a magnetic field, where
they experience a force and are bent in a circular path. The amount of bending
depends
on the mass (and charge) of the particle, and by measurin
g this amount one can infer the

type of particle that is present by comparing to the bending of known elements.

Consider a particle moving in an external magnetic field so that

its velocity is
perpendicular to the field

The force is always directed toward the center of the circular path

The magnetic force causes a centripetal acceleration, changing the direction of the
velocity of the particle

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Equating the magnetic and centrip
etal forces:

Solving for r:

o

r is proportional to the momentum of the particle and inversely
proportional to the magnetic field

Negatively charged particles circulate in the opposite direction as positively
charged particles. The direction can be found usin
g the
right
-
hand
-
rule

applied to
the perpendicular component of the velocity.

Example

4.
1
:

A particle with a charge of

moves at a speed of

through a
magne
tic field in the direction at which the magnetic force on the particle is maximum. If
the force on the particle is

what is the magnitude of the magnetic field?

Solution:

For maximum force,
F

=
qvB

sin

=
qvB
,

so
B

=

F

qv

=

1.8

10

6

N

(
4.0

10

8

C
)
(
3.0

10
2

m/s
)

=
0.15 T
.

Example

4.
2
:

Find the direction of the magnetic field acting on the positively charged particle moving
in the various situations shown in figure below, if the direction of the magn
etic force
acting on it is as indicated.

Solution:

Since the particle is positively charged, use the right hand rule. In this case, start with the
thumb of the right hand in the direction of
v
and the palm facing the direction of
F
. The
fingers will po
int in the direction of
B
. The results are

(a)

(b)

(c)

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4.3

Magnetic forces and magnetic field due to current
carrying wire

It follows from the fact that moving charges experience a force in a magnetic field
that a curren
t
-
carrying wire will also experience such a force, since a current
consists of moving charges. An illustration of this effect appears below.

The magnitude of the force is given by

F = B I

sin θ

w
here θ is the direction of current to the direction of
magnetic field

a
nd

is the length of wire in magnetic field

The direction of the force is according to the Fleming left hand rule

or right hand
rule#1
.

Figure 4.4
:

Force on a current
-
carrying wire in a magnetic field

The
magnetic field

lines around a
long wire

which carries an electric c
urrent form
conc
entric circles around the wire.
The direction of the magnetic field is
perpendicular to the wire and is in the direction the fingers of your right hand
would curl if you wrapp
ed them around the wire with your thumb in the direction
of the
current
.
(right hand rule#2)

Figure 4.5

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The
magnetic field

of an infinitely long straight wire can be obtained by applying
Ampere's law
. The expression for the magnetic field is

The
earth's magnetic field

permeability

of free space is

Example
4.
3:

A 2.0
-
m length of straight wire carries a current of 20 A in a uniform magnetic field of 50
mT whose direction is at an angle of 37° from the direction of the current
. Find the force
on the wire.

F

=
ILB

sin

= (20 A)(2.0 m)(0.050 T) sin 37

=
1.2 N perpendicular to the plane of
B

and
I

.

Example
4.
4:

A thin, horizontal copper rod is 1.00 m long and has a mass of 50.0 g. What is the

minimum current in the

rod that can cause it to float in a ho
rizontal magnetic field of
2.00
T?

Solution:

If the rod is to float, the magnetic force must be direc
ted upward and have a magnitude
equal to the weight of the rod. Thus,
, or

For minimum current,

giving

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4.4

Magnetic Force
between

Wires

Figure 4.6

Once the magnetic field has been calculated, the
magnetic force expression

can be
used to calculate the force.

The direction is o
btained from the
right hand rule
. Note that two wires carrying
current in the same direction attract each other, and they repel if the currents are
opposite in directio
n.

Once you have calculated the force on wire 2, of course the force on wire 1 must
be exactly the same magnitude and in the opposite direction according to
Newton's third law
.

Example
4.
5:

Two long, straight, parallel wires carry current in the same direction.

(a)

Use the right
-
hand source and force rules to determine whether the forces on the
wires are (1) attractive or (2) repulsive.

(b)

If the wires are 24 cm apart and carry cur
rents of 2.0 A and 4.0 A, respectively,
find the force per unit length on each wire.

Solution:

(a)

The forces on the wires are
attractive
. Assume the currents are upward in both
wires. The magnetic field by the left wire on the right wire is direc
ted into the
page, according to the right
-
hand source rule. Then the magnetic force on the
right wire is to the left according to the right
-
hand force rule. Vice versa, the
force on the left wire is to the right so they attract.

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(b)

F

L

=

o
I
1
I
2

2

d

=

(
4

10

7

T∙m/A
)
(
2.0 A
)
(
4.0 A
)

2

(
0.24 m
)

=
6.7

10

6

N/m

4.5

Torque on a Current Loop

The torque on a current
-
carrying coil, as in a
DC motor
, can be related to the
characteristics of the coil by the "
magnetic moment
" or "magnetic dipole moment". The
torque

exerted by the
magnetic force

(including both sides of the coil) is given by

Figure 4.
7

Application: Motor

Electric Motor

An electric motor converts electrical energy to mechanical energy

o

The mechanical energy is in the form of rotational kinetic energy

An electric motor consists of a rigid current
-
carrying loop that rotates when
placed i
n a magnetic field

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Figure 4.
8

The torque acting on the loop will tend to rotate the loop to smaller values of θ
until the torque becomes 0 at θ = 0°

(plane of area is perpendicular to B)

If the loop turns past this point and the current remains in the sa
me direction,
the torque reverses and turns the loop in the opposite direction

To provide continuous rotation in one direction, the current in the loop must
periodically reverse

o

In ac motors, this reversal naturally occurs

o

In dc motors, a
split
-
ring commu
tator

and brushes are used

Actual motors would contain many current loops and
commutators

Just as the loop becomes perpendicular to the magnetic field and the torque
becomes 0, inertia carries the loop forward and the brushes cross the gaps in
the ring, c
ausing the current loop to reverse its direction

o

This provides more torque to continue the rotation

o

The process repeats itself

Example

4.
6
:

A thin copper wire of length ‘L’ is bent to form a single turn plane circular loop and is
suspended in a uniform ma
gnetic field ‘B’ which is directed parallel to the plane of the
loop. The torque acting on the loop when a current ‘I’ passes in it is 0.1 N. If the same
wire were bent to form a plane circular loop of 10 turns, the torque would be

(a) 0.01 N

(b) 0.1 N

(
c) 1 N

(d) 0.001 N

(e) 10 N

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Solution:

The torque is maximum (equal to nIAB) since the plane of the coil is parallel to the
magnetic field. In the first case, since the number of turns (n) is one,

torque, τ = IAB = I
×
π(L/2π)
2
×B since the radius of the s
ingle turn coil is
L/2π.

Therefore, τ = IL
2
B/4π

When the coil is of ‘n’ turns (with the wire of the same length L),

torque, τ’ = nIA’B = nI
×
π(L/2πn)
2
×B =
IL
2
B/4πn.

The torque thus reduces to (1/n)
th

of the torque on the single turn coil. Since n = 10, the

correct option is 0.01 N.

Example

4.
7
:

A current
-
carrying loop is rectangular with dimensions of 20 cm by 30 cm. The loop
carries a current of 10 A and is in a uniform magnetic field of 50 mT directed parallel to
the plane of the loop. Find the torque on

the loop.

Solution:

=
NIAB

sin

= (1)(10 A)(0.20 m)(0.30 m)(0.050 T) sin 90

=
3.0

10

2

m∙N
,

where

is the angle between the normal to the loop and the field.

4.6

Ampere's Law

The
magnetic field

in space around an
electric current

is proportional to the electric
current which serves as its source, just as the
electric field

in space is proportional to the
charge

which serves as its source. A
mpere's Law states that for any closed loop path, the
sum of the length elements times the magnetic field in the direction of the length element
is equal to the
permeabi
lity

times the electric current enclosed in the loop.

In the electric case, the relation of field to source is quantified in
Gauss's Law

which is a
very powerful tool for calculating electric fields.

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Magnetic field of solenoid

Figure 4.
9

If a long straight wire is bent into a coil of several closely spaced
loops, the
resulting device is called a solenoid
.

It is also known as an electromagnet since it acts like a magnet only when it
carries a current
.

The field lines inside the solenoid are nearly parallel, uniformly spaced, and close
together

o

This indicates
that the field inside the solenoid is nearly uniform and
strong

The exterior field is nonuniform, much weaker, and in the opposite direction to
the field inside the solenoid

The field lines of the solenoid resemble those of a bar magnet

The magnitude of th
e field inside a solenoid is constant at all points far from its
ends

B = µ
o

n I

o

n is the number of turns per unit length

o

n = N / ℓ

Magnetic Field of a Current Loop

The strength of a magnetic field produced by a wire can be enhanced by
forming the wire into a loop

All the segments, Δx, contribute to the field, increasing its strength

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Figure 4.
10

Example

4.
8
:

A solenoid 10
cm long is wound with 1000 turns of wire. How much current must exist in
the windings to produce a magnetic field of

at the solenoid’s center?

Solution:

n

=

1000

0.10 m

= 10 000 /m.
B

=

o
nI
,

I

=

B

o
n

=

4.0

10

4

T

(
4

10

7

T∙m/A
)
(
10 000 /m
)

=
3.2

10

2

A
.

4.7

Magnetic materials

Materials may be classified according to

some of their basic magnetic pro
perties,
particularly whether or not they are magnetic and how they behave in the vicinity
of an
external magnetic field. We discuss here three particular type.

Non
-
magnetic materials

Most materials we encounter have no obvious magnetic properties
-

they are said to be
non
-
magnetic. In these materials, the magnetic fields of the individual atom
s are
randomly aligned and t
hus tend to cancel out, as shown in Figure 4.10
.

Figure 4.1
1
: A non
-
magnetic material

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Permanent magnets

In a
permanent magnet
, howeve
r, the magnetic fields of the individual atom are aligned
in one preferred direction, giving rise to a net magnetic field, as
in Figure 4.11
.

Figure 4.1
2
: A perman
ent magnet

We see in this picture why two magnets result when cutting a permanent magnet in half,
instead of separate North and South poles, since each individual atom is itself a magnet
with two poles.

Ferromagnets

A third type of magnetic material is

a
ferromagnet
. In this material, there are domains in
which the magnetic fields of the individual atoms align, but the orientation of the
magnetic fields of the domains is random, giving rise to no net magnetic field. This is
illustrated
in Figure 4.12
.

Figure 4.1
3
: A ferromagnet

A useful property of ferromagnets is that when an external magnetic field is applied to
them, the magnetic fields of the individual domains tend to line up in the direction of this
external field, due to the nature of the magnetic forces, which causes the
external
magnetic field to be enhanced. This is illustrated
in Figure.

Figure

4.14
:

A ferromagnet in an external magnetic field

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This enhancement of the external m
agnetic field is the reason one often finds the loops of
wire as used in an electromagnet
, are

wrapped around a ferromagnetic core.

Another area where ferromagnetic materials are employed is in magnetic recording
devices, such as for cassette tapes, flopp
y discs for computers, and the magnetic stripe on
the back of credit cards. These devices essentially take information in the form of
electrical signals and permanently encode it into a magnetic material. The way this is
done is illustrated
in Figure
.

Figure

4.15

As an (AC) electrical signal passes through the wire loop, a magnetic field is produced
which passes through the ferromagne
tic core, which in turn produces a magnetic field in
the vicinity of a moving magnetic tape. This magnetic field aligns the magnets of atoms
on the tape that happen to be passing by it at that instant.

A short time later the direction of the current reve
rses, which reverses the direction of
the magnetic field, which subsequently reverses the orientation of the next atom on the
tape which passes by. In this way information stored in the electrical signal is encoded in
a particular orientation of the magnet
ic fields of individual atom
.