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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 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
The Magnetic Field Created by a Long Current-Carrying Wire
----
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


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

B2 
0 I
2 a
I2

Therefore,
 I 
F1  I1  B2  I1   0 2 
 2a 
I I 
F1  0  1 2  
 2a 
Example (4.1):
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
F 4 x x107 x10 x 20
= 1.33x10-4 N/m

2

2 x  x 30 x10
30 cm
 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 
20x  10 x  0.3  x 
10
20

x  0.3  x 
20x 10x  3
10x  3 
x 
3
 0.3 m
10
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:


F
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
Example (4.2):
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
or
1T  1
N
A. m
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
y
v
xy plane. Calculate the magnetic force on
the electron.
x
F
600
B
Solution:
FB = |q|vB sin θ
= (1.6x1019 C) (8x106 m/s) (0.025 T)(sin 600)
= 2.8x 10-14 N
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 (4.3):
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
Substitute numerical values:
qBr
mP
v
q B r 1.6 x10 19 x 0.35 x 0.14

mP
1.67 x10 27
1.67 3 10227 kg
V= 4.7x106 m/s
Example (4.4):
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 .