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Chapter IV
Magnetic Force
Magnetic Field:
A magnetic field is generated when electric charge carriers such as electrons move through
space or within an electrical conductor.
The symbol H refers to the strength of magnetic
field.
Figure (4.1):
Magnetic field of an ideal cylindrical magnet
with its axis of symmetry inside the image plane.
The magnetic field is represented by magnetic
field lines, which show the direction of the field
at different points.
Magnetic Induction:
The Magnetic induction (B) is defined as follows:
B  0 H
Where:
0 is the permittivity and equals to  0  4 xx10 7 wb / A.m
H is the magnetic field strength.
The magnetic induction (B) is measured in Tesla (T) or in Newton/ Ampere. Meter
(N/A. m).
Magnetic Flux:
The Magnetic flux (φ) is defined as follows:
B A
Where:
B is the magnetic induction and A is the area
ℓ
H
N
HN
Hs
S
b
a
Figure 4.1: Calculation of magnetic field at the point b.
The strength of magnetic field (H) at the point b can be expressed as follows:
H
1 2 Pm
4 a3
Where:
Pm is called the magnetic moment and equals to Pm = m ℓ
m is the strength of the magnet
ℓ is the length of the magnet
Note that:
No.
Symbol
Title
Units
1
H
Magnetic field strength
Ampere/m
2
B
Magnetic induction
Weber/m2 = Tesla
3
Φ
Magnetic flux
Weber
4

Permeability
Weber/Ampere.m
Biot-Savart Law:
Biot and Savart arrived at a mathematical
I
expression that gives the magnetic field at
dB
some point in space in terms of the current
that produces the field. That expression is
P
based on the following experimental

θ
observations for the magnetic field d B at a
r
point P associated with a length element
dℓ
ℓℓ

d  of a wire carrying a steady current I.


 The vector d B is perpendicular both to d  (which points in the direction of the

current) and to the unit vector r^ directed from d S toward P.

 The magnitude of d B is inversely proportional to r2, where r is the distance

from d 
to P.

 The magnitude of d S is proportional to the current I and to the magnitude dℓ of

the length element d  .

 The magnitude of d B is proportional to sin θ, where θ is the angle between the

vectors d  and r^.
These observations are summarized in the mathematical expression known today as
the Biot–Savart law:
dB 
 0 I d sin 
4
r2
Applications on Biot-Savart Law:
 Magnetic Induction of a Straight Conductor:
Consider a thin, straight wire of finite length
carrying a constant current I and placed
y

d  =dx
P
along the x axis as shown in the Figure.
Determine the magnitude and direction of
r
θ
the magnetic field at point P due to this
a

current.
x
d 
In this case, the magnetic field (H) can be
x
O
I
expressed as follows:
H
1
2
1
a
and the magnetic induction (B) is expressed
as follows:
B  0 H 
0 1
2 a
Example (4.--)
A current of 5 A passes through in a long conductor.
Calculate the magnetic field
strength and magnetic induction at a point 10 cm away from the middle of the wire.
Solution:
1- H 
1
2
1
1
1

 7.96 A / m
a 2 x3.14 10 x10 2
2- B   0 H  4 x3.14 x10 7 x 7.96  10  5 Wb / m 2
 Magnetic Induction of a Circular Conductor:
In this case, the magnetic field (H) in the center of the coil can be expressed as
follows:
H
NI
2r
Where,
N: number of turns,
I is the current
r is the radius.
and the magnetic induction (B) is expressed as follows:
B  0 H  0
NI
2r
and the magnetic moment (Pm) is expressed as follows:
P m  NAI  N  a 2 I
Example (4.--)
A current of 3.5 A passes through in a circular coil with number of turns 200 turn and
a radius of 20 cm. Calculate the magnetic field strength (H) in the center of the coil,
magnetic induction (B) and the magnetic moment (Pm).
Solution:
 The magnetic field (H)
H
NI
200 x 3.5

2r
2 x 20 x10 2
=1.75x103 A/m
 The magnetic induction (B)
B   0 H  4 x 3.14 x10 7 x 1.75 x103 = 2.2x10-3 Wb/m2
 The magnetic moment (Pm)

P m  NAI  N  a 2 I  200 x 3.14 x 20 x10 2

2
x3.5 = 88 A. m2
 Magnetic Induction of a Solenoid:
In this case, the magnetic field (H) in the middle of the coil can be expressed as
follows:
H
NI

Also,
The magnetic field (H) at the end of the coil can be expressed as follows:
H
NI
2
 Where,
 N: number of turns,
 I is the current
 ℓ is the length of the solenoid.
and the magnetic induction (B) is expressed as follows:
Example (4.--)
A solenoid has 200 turns per centimeter and carrying a current of 1.5 A. Calculate
the magnetic field strength and magnetic induction in the middle and at the end of
solenoid.
Solution:
Firstly,
The magnetic field (H) in the middle of the solenoid can be expressed as follows:
H
NI
200 x 1.5

 3 x10 4 A.m
2

10
The magnetic induction (B) in the middle of the solenoid can be expressed as follows:
B   o H  4 x3.14 x10 7 x 3x10 4  3.768 x10 2 Wb / m 2
Secondly,
The magnetic field (H) at the end of the solenoid can be expressed as follows:
H
NI
200 x 1.5

 1.5 x10 4 A.m
2
2 x 10 2
The magnetic induction (B) at the end of the solenoid can be expressed as follows:
B   o H  4 x3.14 x10 7 x 1.5 x10 4  1.884 x10 2 Wb / m 2
Note That:
No. Status
Magnetic Field (H)
1
H
Straight Conductor
1
Magnetic Induction (B)
1
a
2
Circular Conductor
2
NI
H
2r
3
Solenoid
In the middle
At the end
B  0 H 
0 1
2 a
B  0 H  0
H
NI

In the middle
H
NI
2
At the end
NI
2r
B  0 H 
B  0 H 
 0 NI

 0 NI
2
The Magnetic Force Between Two Parallel Conductors
Consider two long, straight, parallel wires separated by a distance a and carrying
currents I 1 and I 2 in the same direction as in the Figure 4.3. Let’s determine the
force exerted on one wire due to the magnetic field set up by the other wire. Wire 2,
which carries a current I2 and is identified arbitrarily as the source wire, creates a

magnetic field B 2 at the location of wire 1, the test wire. The magnitude of this
magnetic field is the same at all points on wire 1.

The direction of B 2 is perpendicular to
1
ℓ
wire 1 as shown in the Figure.
I1
According to the following equation,


B2

FB I LxB
2
F1
a
the magnetic force on a length ℓ of wire 1


I2

is F 1 = I1  x B 2 .
Figure 4.2

Because ℓ is perpendicular to B 2 , in this


situation, the magnitude of F 1 is F 1 = I1  B2 .

Because the magnitude of B 2 is given by
0 I
2 a
Therefore,

B2 
 I 
F1  I1  B2  I1   0 2 
 2a 
I I 
F1  0  1 2  
 2a 
Example (4.--):
A conductor carrying a current of 10 A is placed in a magnetic field with induction of 1.2
Wb/m2. Calculate the force per unit length (F/ℓ) in the following cases:
 (a) If the direction of wire length is perpendicular to the direction of magnetic field.
 (b) If the direction of wire length is Inclined at an angle of 00, 300, 600 to the
direction of magnetic field.
Solution:
Note that:
F
 I B sin 

(a)
F
 10 x 1.2 x sin 90  12 N / m

(b)
F
 10 x 1.2 x sin 0  0 N / m

F
 10 x 1.2 x sin 30  6 N / m

F
 10 x 1.2 x sin 60  10.39 N / m

Example (4.--):
Two parallel conducting plates, a current of 10 A passes through one of them and a current
of 20 A passes through in the other one, calculate the following:
 The force per unit length (F/ℓ).
 Define the point at which B1= B2 , if the currents (I1 and I2) in the same directions.
 Define the point point at which B1= B2 , if the currents (I1 and I2) in opposite
directions.
I2=20 A
I1=10 A
Solution:
 The force per unit length (F/ℓ)
F 0 I 1 I 2


2 a
30 cm
F 4 x x107 x10 x 20
= 1.33x10-4 N/m


2 x  x 30 x10 2
 The point at which B1= B2 , if the currents (I1 and I2) in the same directions
I1
I2


x  0.3  x 
10
20

x  0.3  x 
20x  10 x  0.3  x 
20x  10x  3
30x  3 
x 
3
 0.1 m
30
 The point at which B1= B2 , if the currents (I1 and I2) in opposite directions
I1
I2


x  0.3  x 
10
20

x  0.3  x 
20x  10 x  0.3  x 
20x 10x  3
10x  3 
x 
3
 0.3 m
10
Ampere’s Law
Ampere’s law states that “ The linear integral of magnetic induction (B) around closed
loop equals to the total currents (I) inside this loop multiplying by the permeability
(0)”.
 B cos 
d   o
I
Applications on Ampere’s Law
Note That:
No. Status
Magnetic Field (H)
Magnetic Induction (B)
1
H
NI

NI
H
2 r
B  0 H 
2
Solenoid
Toroid
0 N I

 NI
B  0 H  0
2 r
Magnetic Force:

The existence of a magnetic field B at some point in space can be determined by

measuring the magnetic force FB exerted on an appropriate test particle placed at that
point. This process is the same in defining the electric field. If we perform such an
experiment by placing a particle with charge q in the magnetic field, it is found the
following results that are similar to those for experiments on electric forces:
 The magnetic force is proportional to the charge q of the particle.
 The magnetic force on a negative charge is directed opposite to the force on
a positive charge moving in the same direction.
 The magnetic force is proportional to the magnitude of the magnetic field

vector B .
Also, the following results, which are totally different from those for experiments:
 on electric forces should taken in our consideration:
 The magnetic force is proportional to the speed v of the particle.
 If the velocity vector makes an angle θ with the magnetic field, the magnitude
of the magnetic force is proportional to sin θ.
 When a charged particle moves parallel to the magnetic field vector, the
magnetic force on the charge is zero.
 When a charged particle moves in a direction not parallel to the magnetic field


vector, the magnetic force acts in a direction perpendicular to both v and B


that is, the magnetic force is perpendicular to the plane formed by v and B .
F
This relation can be written as follows:



F q v x B
B
The magnitude of the magnetic force on a charged particle is
F  qvB sin 
θ
v
So:


FB is zero when v is parallel or antiparallel to B (θ = 00 and 1800) and maximum when


v is perpendicular to B (θ = 900).
Let’s compare the important differences between the electric and magnetic versions
of the particle in a field model:
 The electric force vector is along the direction of the electric field, whereas the
magnetic force vector is perpendicular to the magnetic field.
 The electric force acts on a charged particle regardless of whether the particle
is moving, whereas the magnetic force acts on a charged particle only when the
particle is in motion.
 The electric force does work in displacing a charged particle, whereas the
magnetic force associated with a steady magnetic field does no work when a
particle is displaced because the force is perpendicular to the displacement of
its point of application.
The SI unit of magnetic field is the newton per coulomb-meter per second, which is
called the Tesla (T):
1T  1
N
C
.m
sec
or
1T  1
N
A. m
Example (4.--):
An electron in an old-style television
picture tube moves toward the front of the
tube with a speed of 8x106 m/s along the
x axis. Surrounding the neck of the tube
z
are coils of wire that create a magnetic
field of magnitude 0.025 T, directed at an
-e
angle of 600 to the x axis and lying in the
the electron.
Solution:
FB = |q|vB sin θ
= (1.6x1019 C) (8x106 m/s) (0.025 T)(sin 600)
= 2.8x 10-14 N
y
v
xy plane. Calculate the magnetic force on
x
F
600
B
Motion of a Charged Particle in a Uniform Magnetic Field
The particle moves in a circle because the


magnetic force F B is perpendicular to v

and B and has a constant magnitude qvB.
r
q
FB
As Figure 4.2 illustrates, the rotation is
v
+
counterclockwise for a positive charge in
+
FB
a magnetic field directed into the page. If
q
FB
v
q were negative, the rotation would be
+
clockwise. We use the particle under a net
v
q
force model to write Newton’s second
law for the particle:
F F
B
 ma
Because the particle moves in a circle, we also model it as a particle in uniform
circular motion and we replace the acceleration with centripetal acceleration:
FB  qvB 
mv 2
r
This expression leads to the following equation for the radius of the circular path:
r
mv
qB
That is, the radius of the path is proportional to the linear momentum mv of the
particle and inversely proportional to the magnitude of the charge on the particle and
to the magnitude of the magnetic field. The angular speed of the particle is expressed
as follow:

v qB

r m
The period of the motion (the time interval the particle requires to complete one
revolution) is equal to the circumference of the circle divided by the speed of the
particle:
T
2


2r 2m

v
qB
These results show that the angular speed of the particle and the period of the circular
motion do not depend on the speed of the particle or on the radius of the orbit. The
angular speed v is often referred to as the cyclotron frequency because charged
particles circulate at this angular frequency in the type of accelerator called a
cyclotron.
Example (4.--):
A proton is moving in a circular orbit of radius 14 cm in a uniform 0.35-T magnetic
field perpendicular to the velocity of the proton. Find the speed of the proton.
Solution:
v
qBr
mP
Substitute numerical values:
q B r 1.6 x10 19 x 0.35 x 0.14
v

mP
1.67 x10 27
1.67 3 10227 kg
V= 4.7x106 m/s
Example (4.--):
An electron moves in circular path with radius of 2x10-5 cm under the effect of magnetic
field with induction of 1.5x10-3 Wb/m2. Calculate the following:
Qe= 1.6x10-19 C
and me= 9.11x10-31 kg
Solution:
 The magnetic force (F)
F  Q v B 1.6 x1019 x 52.68 x1.5x103 1.26 x1020 N
 The electron velocity (v)
eBr 1.6 x1019 x1.5 x103 x 2 x105 x102

m
9.11x10 31
v  52.68 m / sec
v
 The angular velocity of electron (ω)
eB 1.6 x1019 x1.5 x103
 
m
9.11x10 31
  263x109
rad / sec .
Electromagnetic Induction
Introduction
What is the physical basis for the static or noise produced in a radio when certain
appliances are used in its vicinity? What is the principle supporting the operation of
the ferrite rod antenna in a radio? How does the power company generate AC voltage?
The answers to all of these questions involve electromagnetic induction. In this part
we will study the physical phenomena associated with electromagnetic induction.
Induced EMF: Faraday's Law
Let us consider a simple experiment. The equipment needed consists of a coil of wire,
a galvanometer and a bar magnet. The coil of wire is connected to the galvanometer.
We observe the impulsive deflection of the galvanometer under a series of conditions.
First,
If we move the magnet towards the coil of wire, we see a deflection of the
galvanometer. The deflection means that there is a current in the galvanometer coil
circuit, hence an induced emf.
Second,
If we move the magnet away from the coil, we see another deflection of the
galvanometer. We observe in these cases that the deflections are in opposite
directions.
Third,
If we reverse the ends of the magnet and repeat the experiment described above, we
observe that the galvanometer deflection for approach is opposite to that in the first
case and likewise on taking the magnet away.
What conclusions can we reach from this series of experiments? Now let us try a set
of experiments in which the magnet remains fixed and the coil is moved toward and
away from the magnet - first toward one end of the magnet and then toward the other
end. We will find that when the coil and a given end of the magnet are approaching
each other the direction of deflection is the same regardless of which is moving. Also
the direction of deflection is the same when the coil and a given end are receding from
each other, regardless of which is moving.
We can get additional information relative to the time involved in the process.
If we move either the magnet or the coil faster, the deflection is increased. The results
of these experiments indicate that an emf is induced in a coil of wire whenever there is
a change in the magnetic flux interlinking it, and that there is a definite relationship
between the direction of the field and the direction of motion and between the
magnitude of the induced emf and how rapidly we change the magnetic flux. On the
basis of experiments such as these, Michael Faraday, one of the greatest experimental
investigators in the history of science, formulated his now famous law for
electromagnetic induction. Faraday envisioned the magnetic field associated with a
magnet to be ethereal tentacles that grasp the iron filings and pull them toward the
magnet. His induction law expressed the following idea: When the number of
tentacles cutting through the area capping a circuit changes in time, there will be an
induced electric field in the circuit and thus an induced voltage and current. We can
put this induction law in a useful quantitative form.
We define the magnetic flux φ
B A
B
as the magnetic induction field magnitude B,
times the area of the circuit that is perpendicular to B.
The magnetic flux φ can be written as
r
B into Paper
  r 2 B
  B A cos 
Where,
θ is the angle between the magnetic induction B and the line perpendicular to the area
A.
Faraday's law states that whenever the magnetic flux through an area changes, there
will be an emf induced in the perimeter of the area (or equivalently there will be an
electric field induced at the perimeter of the area). We can express this in the
following equation:
 
d
dt
(Weber / sec)
Where is the electromotive force in volts and dφ is the change of magnetic flux in
webers in a change of time dt in seconds.
The negative sign in the equation is due to Lenz's law which states that the induced
voltage is in such a direction as to oppose its cause; that is, the magnetic field due to
the induced current opposes the changing magnetic flux. The induced current has an
associated magnetic flux which tries to maintain the status quo.
Applications of Faraday's Law
There are many different ways to produce induced currents. Faraday's law indicates
that a changing field B, a changing area, or a changing angle between B and the circuit
can each produce an induced emf. We now consider some specific quantitative
examples.
Consider a U-shaped conductor in a plane perpendicular to a magnetic field with
induction B. A rod of length ℓ is moving to the right with a velocity v as shown in the
Figure.
+
According to Faraday's law:
 
d
Bd

dt
t
I
  B v
where
d
v
-
B is the magnitude of the magnetic
induction in tesla, ℓ is the length of
the rod, d is the distance in meters
through which the rod moves in t seconds,
and v is the magnitude of the velocity of the rod
A Simple generator. When the rod is
moved to the right as shown an emf will
be induced across the length Blv. The
top end of the rod will be positive
Example (4.--)
Find the induced emf in a 0.50 m rod moving with velocity of 2 m/sec perpendicular
to a field of 2 x 10-4 Wb/m2, or tesla.
Solution:
  B v
= 2 x 10-4x 2 x 0.50 = 2 x 10-4 Volt
Suppose a coil of wire of 1cm radius is removed from a field of 0.1 Wb/min 0.01 sec.
The induced emf in the coil will be:

BA
,
t
where, A   d , then
0.1 x 3.14 x 0.01
= 
10 2
2
  3.14 x10 3 Volt   3.14 mVolt
If the coil has 100 turns of wire, the flux is increased by a factor of 100:
φ= NBA
= 100 BA
Such a coil can be used as a search coil to measure an unknown magnet
Self Inductance
A single coil of many turns has inductive properties due to the magnetic interaction of
adjacent loops in the coil. If the loops are wound in the same direction the total effect
of this interaction is an induced back emf that opposes the initial current change. This
self-inductance is a form of inertia resisting current change in the coil. The selfinduced emf can be expressed in equation form as follows:
 L
where
di
dt
L is the self-inductance
(di/dt) is the current change per unit time in the coil.
If  is measured in volts and (di/dt) in amperes per second, the inductance is given in
Henries, 1 H = 1 V sec/A. The negative sign is consistent with Lenz's law and
conservation of energy.
Example (4.--)
If the current changes in an air core inductor having a self-inductance of 25 x 10-5 H
from zero to 1.0 A in 0.10 sec, what is the magnitude and direction of the self induced emf?
Solution:
 L
di
1
 2.5 X 10 3 Volt
=  25 x10 5 x
dt
0.1
As the current is increasing, the direction of the induced emf opposes the growing
current; hence, the negative sign. The effect of an inductor in a circuit is to oppose any
change in current. Thus, inductance is analogous to inertia in a mechanical system.
To minimize self-inductance in wire-wound resistors, the loops are wound back on
themselves. This non-inductive winding tends to cancel out the inductive interactions
of the loops.
Inductors
An inductor is a device that stores energy in the magnetic field set up by the current
through a coil. By analogy the inductor plays the same role for magnetic field energy
that the capacitor does for the electric field energy. We can derive an equation for the
energy stored in an inductor as follows: the work done in moving charge q against the
back emf is
W   q  L
di
i dt ,
dt
where
q   i dt
If we start with the current at zero and the current increases at a constant rate to the
final value I, we can use the average value of I/2 to get the work done. We then get
the following equation for the work done in building up the magnetic field:
W
1
LI2
2
= energy stored in the field
This represents the work done in building up a current of I in the self-inductor and
establishing the magnetic field. It also represents the energy that is stored in the
magnetic field and that becomes available if the current is reduced to zero. This is
shown by the flash on opening a switch in a DC circuit in which there is an inductor.
Example (4.--)
The back emf induced in a coil when the current changes from 100 mA to zero in 1 m.
sec is 1 Volt. Find the self-inductance of the coil and the energy dissipated when the
current goes to zero.
Solution:
The self inductance of the coil is
L

di dt  100x10

1
3
1x10 3
  0.01 Henry
The energy dissipated when the current goes to zero is
W


2
1
1
L I 2  x 0.01 x 100 x10 3  0.5 x10 4 J
2
2
Combinations of Inductors
(a)
Series connection
(b)
Parallel connection
I
I
L1
V
L2
V
L1
L2
L3
I1
I2
I3
L3
When the inductors are connected in series (Figure a)
The current is the same through all of the inductors and the potential is the sum of the
individual emfs,
V  1   2   3
V L
di
di1
di
L 2L 3
dt
dt
dt
V  Leff
di
dt
Where
L1, L2 and L3 are the inductances and where dI is the change in the current in the
system in a time dt. For a series configuration, all of the currents are the same:
di1  di2  di3  di
 V   L1  L2  L3 
Thus,
di
dt
 Leff
di
dt
Leff   L1  L2  L3 
When inductors are connected in series, their effective inductance is the sum of the
individual inductances.
When the inductors are connected in parallel (Figure b),
The potential across all of the inductors is the same, but the current through the system
is divided into components so that the total current is the sum of the individual
components,
I  di1  di2  di3
and
V   L1
di
di
di1
di
  L2 2   L3 3   Leff
dt
dt
dt
dt
Then solving for, di1/dt, di2/dt and di3/dt, we have:
di
di di1
di

 2 3
dt
dt
dt
dt
V
V
V

 

Leff
L1
L2

V
L3
Thus
1
1
1
1



Leff
L1
L2
L3
When inductors are connected in parallel, their effective inductance is the reciprocal
of the sum of the reciprocals of the individual inductance.
Example (4.---)
Inductors of 3 H, 5 H, and 10 H are connected
a. in series
b. in parallel
c. with the 3H inductor in series with a parallel connection of the 5H and 10 H
inductors. What is the effective inductance of each configuration?
Solution:
(a) Series
Leff   L1  L2  L3 
= 3 H + 5 H + 10 H = 18 H
(b) Parallel
Parallel
1
1
1
1



Leff
L1
L2
L3
1
1
1
1 19




Leff
3 5
10 30
Leff  1.6 H
(c ) Combination
5 x10
1 1 
Leff  3      3 
5  10
 5 10 
Leff  6.3 H