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CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
79
Chapter 4
MAGNETIC FIELDS AND MAGNETIC FORCES
Properties of Magnet
1- A magnet attracts iron pieces or iron fillings. The filings cling
near the ends, at the (poles}) of the magnet.
2- The magnetization is zero at the middle of the magnet and
maximum at the poles.
3- When a magnet is freely suspended so that it can swing in a
horizontal plane, it comes to rest in N-S direction. N-pole points
towards the north, and S-pole points towards the south.
4- First law of magnetism: '' like poles repel, unlike poles attract”.
5- When a magnet is broken into pieces, each piece is found to be a
magnet with two poles, i.e. no monopole magnet.
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
80
What causes magnetism? It is the spinning of the electrons in the atoms
of materials, and the behavior of these electrons, which give a material
its magnetic properties.
Magnetic field: ''It is the region around a magnet in which a magnetic
force is exerted''. We can describe magnetic interactions in two steps:
1- A moving charge or a current creates a magnetic field in the
surrounding space (in addition to its electric field)
2- The magnetic field exerts a force F on any other moving charge
or current that is present in the field.
Magnetic field is a vector quantity and denoted by B . At any position
the direction of B is defined as the direction in which the north pole
of a compass needle tends to point.
The Magnetic Force on a Moving Charge:
The magnetic force, BF , on a test charge, oq , moving with velocity v in
a magnetic field, is
qv B F
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
81
The magnitude of the magnetic force is v sinF q B , where is the
angle measured from the direction of v to the direction of B . The force
is perpendicular both to the velocity and to the magnetic field. Since the
force is proportional to the velocity, charges at rest do not experience
magnetic forces. [B] = T (Tesla) = N
A.m. Also, 1T =10
4 Gauss.
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
82
EXTRA NOTICES:
1- The charge should be inmotion
(i.e. current).
2- Positive and negative charges
experience forces in opposite
directions.
3- The force is greatest when the
charge moves perpendicular to the
magnetic field and zero when the
charges move parallel to the field.
4- The size of the force also depends
on the magnitudes of the magnetic
field and the electric charge
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
83
(current) and on the speed of the moving charge.
5- The magnetic force does not accelerate the charge, but deflect it.
Magnetic Field Lines and Magnetic Flux
The magnetic flux lines or lines of force represent the magnitude and
direction of the magnetic field, i.e. it is vector quantities.
``Note that:
1- `The density of lines is
proportional to the
strength of the field (i.e.
the lines are close together
where the field is strong
and vice versa).
2- The lines go from North
Pole to the South Pole.
3- The lines never cross each
other.
Magnetic Flux and Gauss's Law for Magnetism
We can divide any surface into elements of area dA. For each element we
determine B , the component of B normal to the surface at the position
of that element. We define the magnetic flux Bd through this area as
cos .Bd B dA B dA B d A
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
84
Where is the angle between the direction of B and the line
perpendicular to the surface.
The total magnetic flux through the surface is the sum of the
contributions from the individual area elements:
cos .B B dA B dA B d A
Magnetic flux is a scalar quantity. In the special case in which B is
uniform over a plane surface with total area A, B and are the same at
all points on the surface, and
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
85
cosB B A BA
The SI unit of magnetic flux is called weber (Wb).
21 1 . ,1 1Wb T m T N A
Gauss's law for magnetism
The total magnetic flux through a closed surface is always zero.
Symbolically
. 0B d A
If the element of area dA is at right angles to the field lines, then B B ;
calling the area dA , we have
BdB
dA
That is, the magnitude of magnetic field is equal to flux per unit area
across an area at right angles to the magnetic field. For this reason,
magnetic field B is sometimes called magnetic flux density.
A Beam of Charged Particles in a Magnetic Field:
Consider a beam of positively charged particles with velocity v .If this
beam enters a magnetic field at right angle to its direction of motion, it
will experience a force perpendicular to both velocity and magnetic
field, i.e. it will be deflected.
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
86
Circular motion: The beam will move in circular path if the velocity is
small and the magnetic field is strong. In that case:
Magnetic force = Centripetal force
2v vv
mq B m B
R qR
where R is the radius of the circle. Also, the angular frequency ( ) and
the periodic time (T ) of a rotating charged particle are:
v,
2 2
v
qB
R m
mT
q B
The beam will move in helical (spiral) path if the velocity vector has
an angel with the magnetic field.
The beam of negative charges will be deflected in the reverse
direction.
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
87
Application of motion of charged particles
Velocity selector: In a region of
crossed or perpendicular
magnetic field B and electric
field E perpendicular to v, the
forces cancel when
v vE
q B qEB
Only particles with speeds equal
to E/B can pass through without
being deflected by the fields. By
adjusting E and B appropriately,
we can select particles having a
particular speed for use in other
experiments. Because v does not
depend on the charge, a velocity
selector works also for electrons or other negatively charged particles.
Thomson's e/m experiment
When a particle is accelerated through a potential difference, V ,
energy conservation requires that the kinetic energy equals the loss
of electric potential energy
221 v
v2 2
qqV m
m V
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
88
In this experiment, a beam of electron is used, so
221 v 2
v2 2
e eVeV m v
m V m
The electron passes between the plates and strike the screen at the end of
the tube, which is coated with a
material that fluoresces at the
point of impact. The electrons
pass straight through the plates
when
2
2
2 e so
m 2
E eV E
B m VB
All the quantities on the right
side can be measured, so the
ratio e/m of charge to mass can
be determined.
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
89
The Magnetic Force on a Current Carrying Wire:
The magnetic force BF on a segment of wire, L , carrying current, I, in a
magnetic field, B , is given by
vt
oB o o
qq q I
t
LF B B L B L B
The total force on the wire is the vector sum of the forces on the
segments.
Examples
Example1: An electron is projected into a uniform magnetic field given
by ˆ ˆ1.4 i + 2.1 j T.B . Find the vector expression for the force on the
electron when its velocity is 5 ˆ3.7 10 j m/s v .
Answer:
5 5 14
x y z
ˆ ˆ ˆ ˆ ˆ ˆ i j k i j k
ˆ ˆv v v v 0 3.7 10 0 (1.4)(3.7 10 )( k) (8.3 10 k) N
1.4 2.1 0
B e e
x y z
q q
B B B
F B
Example 2: A proton is moving with a velocity of 6 ˆ6.0 10 i m/s v at a
point where the magnetic field is given by ˆ ˆ ˆ3.0 i 1.5 j 2.0 k T B .
What is the magnitude of the acceleration of the proton at this point?
Answer:
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
90
2 2 2 15 21.44 10 m/sx y za a a a
Example 3: An electric field and a magnetic field normal to each other.
The electric field is 4.0 kV/m and the magnetic field strength is 2.0 mT.
They are act on a moving electron to produce no force,
a- Calculate the electron speed.
b- Calculate the kinetic energy of the electron.
Answer:
a- 3
6
3
4.0 10v 2.0 10 m/s
2.0 10
E
B
b-
2 31 6 2 181 1v (9.11 10 )(2.0 10 ) 1.82 10 J 11.4 eV
2 2k m
H.W. A proton with velocity 6 ˆ2.0 10 i v (m/s) moves horizontally
into a region of space in which there is an electric field
3 ˆ5.0 10 j N/C E and a magnetic field B. Find the smallest magnetic
field such that the proton will continue to move horizontally, i.e. un-
deflected. [Ans: 3 ˆ2.5 10 k T B ]
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
91
Example 4: A proton with a velocity of 6.0×106 m/s Travels at right
angles to magnetic field of 0.5 Tesla. What is the frequency of the
proton's orbit?
Answer:
2 196
27
1.6 10 0.57.62 10 Hz
2 2 2 2 1.67 10
v v qB v qBqvB m f
r r m r m
Example 5: Alpha particles (m = 273.3 10 kg , 2q e ) are accelerated
from rest through a potential difference of 1.0 kV. They then enter a
region of magnetic field B = 0.2 T perpendicular to their direction of
motion. What is the radius of the bath?
Answer:
21 2v v
2
q Vq V m
m
27
2
19
2 3.3 10 kg 1000 Vv 1 2 V 12.2 10 m
0.2 T 2 1.6 10 C
m mR
qB B q
Example 6:
A wire 72 cm in length has a mass of 15 g. It is suspended by a pair of
flexible leads in a magnetic field B = 0.54 T pointing out of the page, as
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
92
shown in the following figure. What current must exist in the wire for the
tension in the supporting leads to be zero?
Answer: At the balance, tension Magnetic force = Gravitational force
(i.e. B GF F ), then we have:
0.015 9.80.38 A from Q to P
0.72 0.54
B
in
L B mgI
L L L
mgI
LB
F
Example 7: A wire bent into a semicircle of radius R = 2.0 m forms a
closed circuit and carries a current of 1.5
A. The circuit lies in the xy-plane, and a
uniform magnetic field B = 3.0 T is
present along the y axis, as shown in the
figure. Find the magnitude of the
magnetic force on the curved portion of
the wire.
Answer:
Magnetic Dipoles: A current loop, of area A, behaves like a magnet
and has a magnetic dipole moment
ˆIAM n
0
(2 )3 18 N
circle straight circle straight
straight
F F F F
F i l B i R
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
93
Where n̂ is a unit vector normal to the plane of the loop. Its direction is
found by placing the fingers of the right hand along the current; the
right thumb then indicates the direction of n̂ and M . For N -current
loops
ˆNIAM n .
Example 8: The current loop, in the
following figure, consists of one loop with
two semicircles of different radii. If the
current in the circuit is 2 A, a = 3.0 cm and
b = 5.0 cm, then the magnetic dipole
moment of the current loop is:
Answer:
In a uniform magnetic field, a magnetic dipole behaves much like an
electric dipole in uniform electric field, and the net force on the dipole
will be zero. The torque, , on the dipole and the potential energy, U, are
given by:
, cosU MB M B
The magnetic torque on a flat current-carrying loop of wire by a uniform
magnetic field B is maximum when the plane of the loop is parallel to B .
2 2
2
(1)(2) (1.05) (0.3)2
0.01 A.m , into the page.
NI
M A
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
94
Example 9: The plane of area 4.0 cm2 rectangular loop of wire is
parallel to a 2.0 T magnetic field. The loop carries a current of 6.0 A.
Calculate the magnitude of the torque acts on the loop.
Answer:
4 3
sin
(1)(6)(4 10 )(2)sin90 4.8 10 N.m
NI NIAB
M B A B
Example 10: A 100 turns coil, lies in
xz-plane, has an area of 2.0 m2 and
carries a current I = 0.3 A in the
direction indicated in the figure. The
coil lies in a magnetic field directed
along the x-axis and has a magnitude
of 1.5 T. What is magnitude and
direction of the torque on the coil?
Answer:
ˆ ˆ100 0.3 2j 1.5 i
ˆ ˆ ˆ90 j i 90 k, 90 N.m along the positive z axis
NI
M B A B
Example 11: A 100-turn circular coil of wire with radius 1.0 cm carries
a current of 0.5 A. What torque will be exerted on the coil when it is
placed in a magnetic field of 5.0 mT which makes an angle of 60o with
the plane of the coil?
Answer:
x
y
z
I
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
95
2 2 ˆ(100)(0.5) (1.0 10 ) k
ˆ0.0157 k.
NI
M A
3 50.0157 5.0 10 sin30 3.93 10 N.m M B
Example12: A square loop (L=1.00 m)
consists of 100 closely wrapped turns of
0.20 A. The loop is oriented as shown in
figure (5) in a uniform magnetic field of
0.10 T directed in the positive x- direction.
What is the torque (in N.m) on the loop? (j
is a unit vector in the +y-direction.)
Answer:
ˆ ˆk i sin150
ˆ(100)(0.2)(1.0)(0.1)(0.5) 1.0 j
NI NI A B
M B A B
Example 13: A current of 17 mA is maintained in a circular loop of 2 m
circumference which is parallel to the y-z
plane (see Figure 4). A magnetic field B =
(- 0.8 k) T is applied. Calculate the torque
exerted on the loop by the magnetic field.
(i, j and k are the unit vectors in x, y and z
directions, respectively).
Answer:
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
96
2 2
3
3
1 1 1 ˆ2 2m m m i
1 ˆ ˆ1 17 10 ( 0.8) i k
ˆ4.33 10 j
r r A r
NI A B
A
M B A B
Example 14: A current of 16 mA is maintained in a single circular loop
of radius 0.32 m. An external magnetic field of 0.8 T is directed parallel
to the plane of the loop.
a- Calculate the magnetic moment of the current loop.
Answer:
23
3 2
16 10 0.32
5.1 10 C m
IA
M
b- What is the magnitude of the torque exerted on the loop by the
magnetic field?
3
3 2
sin 90 5.1 10 0.8
4.1 10 C m T
M B M B
Sources Of Magnetic Field
1. Magnetic field of a moving charge
To find the magnetic field B of a point charge q moving with constant
velocity v at the point P, let's call the location of the moving charge at a
given instant the source point and the point P where we want to find the
field the field point. Experiments show that the magnitude of B is
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
97
proportional to q , the particle
velocity v , 21 r and to the sin .
But the direction of B is
perpendicular to the plane
containing the line from source
point to field point and the
particle velocity v .
0
2
sin
4
q vB
r
where 0
4
is proportionality
constant? B has the greatest
magnitude when sin 1 or
90 . If the charge q is
negative, the direction of B is
opposite.
The unit of B is tesla (T).
2. Magnetic Field of a Current
Element
There is a principle of
superposition of magnetic fields:
The total magnetic field caused
0
2
ˆ
4
qv rB
r
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
98
by several moving charges is the vector sum of the fields caused by the
individual charges.
We begin by calculating
the magnetic field caused
by a short segment dl of a
current carrying a
conductor, the total
moving charge dQ in the
segment is
dQ nqAdl
Where n is the number of
the moving charged
particles per unit volume,
A is the cross sectional
area, and Adl is the
volume.
The moving charges in
this segment are
equivalent to a single
charge dQ , travelling with
a velocity equal to a drift
velocity v . The magnitude
of the resulting field dB at any point P is
0 0
2 2
sin sin
4 4
d ddQ v dq nv AdldB
r r
,
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
99
but
dn dq v A I ,
Then 0
2
sin
4
IdldB
r
In the vector form, using the unit vector r̂ , we have
This is the Biot and Savart law.
We can use this law to find the total magnetic field at any point in space
due to the current in a complete circuit.
0 0
2 3
ˆ OR
4 4
Idl r Idl rB B
r r
3. Magnetic Field of a Straight Current-Carrying Conductor
An important application of the Biot and
Savart law is finding the magnetic field
produced by a straight current carrying
conductor. From the figure
2 2 2 2 and sin sin( )r x y x x y
0
2 2 3 24 ( )
a
a
xdyB
x y
The final result is
0
2 2
2
4
I aB
x x a
When the length 2a of the conductor is
very great in comparison to its distance x
0
2
ˆ
4
Idl rd B
r
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
100
from the point P, we can consider it to be infinitely long. When a is
much larger than x , 2 2x a is approximately equal to a , hence in the
limit a
0
2
IB
x
Thus, at all points on a circle of radius r around the conductor, the
magnitude of B is
0 (near a long, straight, current-carrying conductor)2
IB
r
Note:
The total magnetic flux
through any closed surface is
always zero. This implies that
there are no isolated magnetic
charges or magnetic
monopoles. Any magnetic
field lines that enters a closed
surface must also emerge from
that surface.
. 0B d A
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
101
4. Force Between Parallel Conductor
The lower conductor produces a B field that, at the position of the upper
conductor, has magnitude
0 2
IB
r
The force that this field exerts on a length L of the upper conductor is
'F I L B
Where the vector L is in the direction of the current 'I and has the
magnitude L. Since B is perpendicular to the length of the conductor and
hence to L , the magnitude of this force is
'' 0
2
II LF I LB
r
And the force per unit length is
'
0
2
IIF
L r
Applying the right-hand rule to 'F I L B shows that the force on the
upper conductor is directed downward.
Note:
1- The two conductors carrying current in the same direction attract
each other. If the direction of either current is reversed, the forces
also reverse.
2- Parallel conductors carrying currents in opposite directions repel
each other.
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
102
Defining the Ampere:
The attraction or repulsion between two straight, parallel, current-
carrying conductors is the basis of the official SI definition of the
ampere:
One ampere is that unvarying current that, if present in each of two
parallel conductors of infinite length and one meter apart in empty space,
causes each conductor to experience a force of exactly 72 10 Newton's
per meter of length.
5. Magnetic Field of a Circular Current Loop
We can use the law of Biot and Savart
to find the magnetic field at a point P
on the axis of the loop, at a distance x
from the center. As the figure shows,
d l and r are perpendicular, and the
direction of the field dB lies in the x-y
plane. Since
``2 2 2r x a
`0
2 24 ( )
I dldB
x a
0
2 2 2 2 1 2cos
4 ( ) ( )x
I dl adB dB
x a x a
0
2 2 2 2 1 2sin
4 ( ) ( )y
I dl xdB dB
x a x a
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
103
The situation has rotational symmetry about the x-axis, so there can’t be
a component of the total field B perpendicular to this axis. For every
element d l there is a corresponding element on the opposite side of the
loop, with opposite direction. These two elements give equal
contributions to the x-component of d B , but opposite components
perpendicular to the x-axis. Thus all the perpendicular components
cancel and only the x-component survive.
0 0
2 2 3 2 2 2 3 24 ( ) 4 ( )x
I IaadlB dl
x a x a
The integral of dl is just the circumference of the circle 2dl a , then
2
0
2 2 3 22( )x
IaB
x a
The direction of the magnetic field is given by a right-hand rule.
Magnetic field on the axis of a coil
2
0
2 2 3 2 (on the axis of circular loops with radius )
2( )x
NIrB N r
x r
0 (at the center of circular loops with radius )2
x
NIB N r
r
Magnetic field due to a current in a circular arc wire
, is the radius and is the angle of the arc4
o
NIB R
R
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
104
Ampere's Law
It states that: the line integral of B around any closed path equal to 0
times the net current through the area enclosed by the path.
0. enclBdl I
The positive sense of current is determined by a right-hand rule
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
105
Applications of Ampere’s Law
Example 1: Magnetic field of a long, straight, current-carrying
conductor:
.
2
2
B d s
o enc
enco
B ds B r i
iB
r
Example 2: Magnetic field inside a long straight wire with current (a
long cylindrical conductor): a cylindrical conductor with radius R
carries a current. The current is uniformly distributed over the cross
sectional area of the conductor.
.
2
2
2
2
2
B d s
o enc o
o
rB ds B r i i
R
iB r r R
R
Current density (current per unit area) 2
iJ
R , so
22
2 ( )enc
i ri J r
R
Magnetic field of a solenoid:
a coil of wire that acts as a magnet when an electric current flows
through it
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
106
inside solenoid, near centres o o
NB nI I
L
Where B is the magnetic field anywhere within the solenoid, its unit is
Tesla (T), o is the permeability of free space, 74 10 T.m/Ao , n is
the number of coils per meter in the solenoid, L is the length of the
solenoid, and I is the current passing through the solenoid.
0 outside solenoid B
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
107
Electromagnetic Induction
1. Electromagnetic Experiments
Here is what we observe: When there is
no current in the electromagnet, so that
0B , the galvanometer shows no current.
1- When the electromagnet is turned
on, there is a momentary current
through the meter as B increases.
2- When B levels off at a steady
value, the current drops to zero, no
matter how large B is.
3- With a coil in a horizontal plane,
we squeeze it so as to decrease the
cross-sectional area of the coil. The meter detects current only
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
108
during the deformation, not before or after. When we increase the
area to return the coil to its original shape, there is current in the
opposite direction, but only while the area of the coil is changing.
4- If we rotate the coil a few degrees about the horizontal axis, the
meter detects current during the rotation, in the same direction as
when we decrease the area. When we rotate the coil back, there is
a current in the opposite direction during this rotation.
5- If we jerk the coil out of the magnetic field, there is a current
during motion, in the same direction as when we decreased the
area.
6- If we decrease the number of turns of the coil by unwinding one
or more turns, there is a current during unwinding, in the same
direction as when we decrease the area. If we wind more turns
onto the coil, there is a current in the opposite direction during the
winding.
7- When the magnet is turned off, there is a momentary current in
the direction opposite to the current when it was turned on.
8- The faster we carry out any of these changes, the greater the
current.
9- If all these experiments are repeated with a coil that has the same
shape but different material and different resistance, the current in
each case is inversely proportional to the total circuit resistance.
This shows that the induced emfs that are causing the current do
not depend on the material of the coil but only on its shape and
the magnetic field.
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
109
Note: The common element in all these experiments is changing
magnetic flux through the coil connected to the galvanometer.
2. Faraday's Law
The common element in all
induction effects is changing
magnetic flux through the circuit.
Remember that
. cosB B A BA
If B is uniform over a flat area A .
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
110
Faraday's law of induction states:
The induced emf in a closed loop equals the negative of the time rate
of change of magnetic flux through the loop.
In symbols, Faraday's law is
Bd
dt
If we have a coil with N identical turns, and if the flux varies at the same
rate through each turn, the total rate of change through each turn is
BdN
dt
3. Lenz's Law
Lenz's Law is a convenient method for determining the direction of an
induced current or emf. Lenz's law is not an independent principle: it can
be derived from Faraday's law.
Lenz's law states: The direction of any magnetic induction effect is such
as to oppose the cause of the effect.
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
111
4. Motional Electromotive Force
Figure shows the rod moves to the right at a constant velocity v in a
uniform magnetic field B directed
into the page. A charged particle q
in the rod experiences a magnetic
force with magnitude F q vB .
This magnetic force causes the free
charges in the rod to move,
creating a positive charge at the
upper end and a negative charge at
the lower end. This in turn creates
an electric field E within the rod.
There is a downward electric force
with magnitude qE to cancel
exactly the upward magnetic force
with magnitude qvB . Then
qE qvB and the charges are in
equilibrium. The magnitude of
potential difference
ab a bV V V EL vBL
Now suppose the moving rod slides along a stationary U-shaped
conductor, forming a complete circuit. No magnetic force acts in the
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
112
charges, but the charge redistributes itself, creating an electric field
within it. This field establishes a current in the direction shown. The
moving rod has become a source of electromotive force. We call this emf
a motional electromotive force, denoted by .
vBL
Note: The motional emf; length and velocity perpendicular to uniform B .
Examples
Faraday’s Law of Electromagnetic Induction and Lenz’s Law
1. For the following scenarios, determine whether the magnetic flux
changes or stays the same. If the flux changes: indicate whether it is
increasing or decreasing (and in which direction). Explain your answer.
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
113
2. Find the direction of the induced current for the solenoid in the figure
below, when the magnet is _____.
3. A circular loop (radius of 10 cm or 0.10 m) is placed in a uniform
magnetic field of magnitude, B = 2.0 T, where the face of the loop is
perpendicular to the direction of the magnetic field.
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
114
a. Determine the magnetic flux through the loop.
2
B = B dA = BAcos = 0.628 T m
b. The loop is then rotated 90o in 3.0 seconds. What is the magnetic flux
through the loop at the end of the 3.0 seconds?
2
B = B dA = BAcos = 0 T m
c. What is the induced emf in the loop during the rotation?
2
2B T m
s
d 0.628 T m = = = 0.0209 or V
dt 3 s
4. A person moves a 2-m rod at a constant velocity of 3 m/s in a
magnetic field, B= 2.0 T. The rod is perpendicular to the direction of
the B field.
a. What is the direction of induced current in the rod?
`Ans. in the +z direction
b. Determine the induced emf in the
rod.
Ans. =vBL= 12 V
c. The resistance in the rod (and
connecting wires) is 2 . What is the current in the rod?
Ans.
12Vi = = = 6A
R 2
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
115
d. Determine the magnitude of the magnetic force acting on the rod.
BF =iL B =24N
5. Consider a 1-m conducting rod attached at each end by conducting
rails. The rails are connected at the top and the total loop has a
resistance of 5- . (see figure below). The rod falls to the ground at a
constant velocity, v. The apparatus is inside a constant magnetic field, B
= 3.0 T (directed out of the page). The mass of the rod is 0.5kg.
`
a) What is the magnetic force on the falling rod, due to the magnetic
field?
ˆ ˆB netF = F - mg = mg j = 4.9N j
b) What is the induced current in the rod?
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
116
BB
F 4.9NF = iLBsin = 4.9N i = = = 1.63A
LB (1m)(3T)
c) What is the induced electromotive force?
= iR = (1.63A)(5 ) = 8.15 V
d) What is the equation for the rate of change of magnetic flux for this
problem?
Bd dA d(Lh) dh
= B = B = BL = BLv = -dt dt dt dt
e) How fast is the rod falling?
ˆ ˆ ˆ
ms
8.15 Vv = - j = - j = -2.72 j
BL 3 T m
f) When the rail falls for 1 sec, verify that energy is conserved.
mmg ε s
εiP =P mgv=εi v = = 2.71 check with (e)
mg
Or
mg εP =P mgv=εi mgv=4.9 2.72 13.3, εi 13.3
Generator
6. A water powered generator, shown below, to convert mechanical
energy into electrical energy. A rotating wheel receives falling
water forcing a wire loop (N=500), located within a constant
magnetic field B=0.01 T (as shown), to rotate counter-clockwise at
a rate of 150 rpm. The length of the segment normal to the B field
(side a) are 0.20 m and the length of the segment parallel to the
field (side b) is 0.15 m.
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
117
a. What is the area of the
region of the coil within
the magnetic field?
Answer.
2
A = ab
A = (0.15m)(.20m)
A = 0.030 m
b. Determine the general
equation for the magnetic
flux through the coil in terms of area A, B, and angular velocity
.
. B= B dA=B A cos t
c. What is the angular velocity of the rotating coil?
Answer.
rot radmin s
1 min 2 rad = 150 = 15.7
60 s 1 rot
d. Calculate the induced electromotive force around the loop.
Answer.
T2
T2
B
0avg T 0
2
avg
d d =-N =-N B A cos t
dt dt
NBA sin t dt2NBA 4NBA
= = - cos t = -T T
= -1.5 V
The instantaneous emf is: (t)= NBA sin( t+ )
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
118
e. What direction does the induced current flow around the
coil? Explain.
Answer.
The current will flow clockwise (looking down on the armature),
accordance with RHR.
Self Inductance:
7. A solenoid, r=0.01 m, l=0.03 m (length) and N=100, is in series with
a 10 resistor, both of which are
in parallel with a 10 resistor, all
of these are in series with a 5 V
power supply.
a. Determine the inductance, L,
of the solenoid.
Answer.
2 -6oL = N A = 3.96x10 H
b. When the power supply is initially connected. What is the
current across the solenoid?
Answer.
io-solenoid = 0A
c. What is the initial current drawn from the power supply?
Answe.
o
Vi = = 0.50 A
R
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
119
d. After 1 minute, what is the current through the solenoid?
Ans. 60sRt
-73.96x10 sL--
solenoid max
5Vi = i 1-e = 1-e = 0.50 A
10
e. What is the total current drawn from the power supply?
Answer.
itot= 1.00A
f. How much energy is stored in the inductor after 1 minute?
Answer.
U = ½Li2 = 4.95x10
-7 J
Examples
Example 1: A cylindrical conductor of radius 2.5 cmR carries a
current 2.5 AI along its length; this current is uniformly distributed
through the cross section of the conductor. Calculate the magnetic field
midway along the radius of the wire (that is, at / 2r R ).
Answer:
2
7 5
2
, 2
2.5( / 2) 4 10 1.0 10 T
2 4 (0.05)
o
o
IB r r R
R
IB R
R
------------------------------------------------------------------------
Example 2: What current in a solenoid 15-cm long wound with 100
turns would produce a magnetic field equal to that of the earth, which is
5.1×10-5
T?
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
120
Answer:
52
7
5.1 10 0.156.1 10 A
4 10 100
s o o
s
o
NB nI I
L
B NI
L
--------------------------------------------------------------------------
Example 3: A solenoid is formed by tightly winding a single layer of
wire. The wire is 1.0 mm in diameter. What is the magnitude of the
magnetic field inside the solenoid when there is a current of 0.081 A in
the windings?
Answer:
7 4
3
14 10 0.081 1.02 10 T
10
s o o
NB nI I
L
------------------------------------------------------
--------------------
Example 4: Three long wires parallel to the
x-axis carry currents as shown in the
following figure. If I = 20 A, what is the
magnitude of the magnetic field at the origin?
31 21 2 3
1 2 3
5
,2
34 2 31.2 10 T.
2 2 1 3 2
o
o o
II IB B B B
a a a
I I II
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
121
Example 5: A segment of wire is formed into
the shape shown in the following figure and
carries a current I. What is the resulting
magnetic field at the point P?
Answer:
1 2 3 2
2
0 0,
3 3 into the page
4 4 2 8o o o
B B B B B
I I IB B
R R R
Example 6: Figure 10 shows two concentric, circular wire loops, of radii
r1 = 15 cm and r2 = 30 cm, are located
in the xy plane. The inner loop
carries a current of 8.0 A in the
clockwise direction, and the outer
loop carries a current of 10.0 A in the
counter clockwise direction. Find the
net magnetic field at the center.
Answer:
.
1
2
7
1
7
2
1 2
4 10 833.5 T into the page.
2 2(0.15)
4 10 1020.9 T out of the page.
2 2(0.3)
33.5 20.9 12.6 T into the page.
or
or
IB
r
IB
r
B B B
-------------------------------------------------------------------
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
122
Example 7: A circular loop of radius 0.1 m has a resistance of 6 Ohms.
If it is attached to a 12 V battery, how large a magnetic field is produced
at the center of the loop?
Answer:
7 5
122 A,
6
24 10 2 1.3 10 T
4 4 0.1o
IV
R
IB
R
----------------------------------------------------------------------
Example 8: How many turns should be in a flat circular coil of radius
0.1 m in order for a current of 10 A to produce a magnetic field of
33.0 10 T at its center?
Answer:
3
7
2 2 3.0 10 0.148 Turns
2 4 10o
o
NI BRB N
R I
--------------------------------------------------------------------
Example 9: Two long parallel wires, a distance d apart, carry currents of
I and 5I in the same direction. Locate the point r, from I, at which their
magnetic fields cancel each other.
Answer:
1 2at equilibrium 2 2 ( )
Solve for r, one can find 6
o oI IF F
r d r
dr
---------------------------------------------------------------------------
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
123
Example 10: Two long parallel conductors, separated by a distance
10 cma , carry currents in the same direction. If 1 5.0 AI and
2 8.0 AI , what is the force per unit length exerted on each conductor by
the other?
Answer:
1 2
7 5
2
5.0 8.04 10 8.0 10 N/m attractive
2 0.1
o
I IF
l a
MAGNETIC PROPERTIES OF MATERIALS
All matter is composed of atoms and atoms are composed of protons,
neutrons and electrons. The protons and neutrons are located in the
atom's nucleus and the electrons are in “constant motion” around the
nucleus. Electrons carry a negative electrical charge and produce a
magnetic field as they move through space. A magnetic field is produced
whenever an electrical charge is in motion.
This may be hard to visualize on a subatomic scale but consider an
electric current flowing through a conductor. When electrons (electric
current) are flowing through the conductor, a magnetic field forms
around the conductor. The magnetic field can be detected using a
compass.
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
124
Since all matter is comprised of atoms, all materials are affected in some
way by a magnetic field. However, not all materials react the same way.
At the atomic level, the motion of an electron gives rise to current loop
magnetic dipole moment magnetic field
Magnetic dipole moment m [A.m2]
um m N i A n
Where N is the number of electrons, i is the current, and A is the area.
It is a vector quantity and its direction can be found by the right hand
screw rule.
The magnetic moments associated with atoms have three origins:
1 The electron orbital motion.
2 The change in orbital motion caused by an external magnetic
field.
3 The spin of the electrons.
Magnetization M [A.m-1]
It is equal to the magnetic dipole moment per unit volume
magnetic momentM
volume
Permeability
Permeability is a material property that describes the ease with which a
magnetic flux is established in the component. It is the ratio of the
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
125
magnetic flux density to the magnetic intensity and, therefore,
represented by the following equation: B
H
Magnetic susceptibility ( m )
It represents the response of a system to the external magnetic field.
When a material is placed within a magnetic field, the material's
electrons will be affected. However, materials can react quite
differently to the presence of an external magnetic field. This reaction is
dependent on a number of factors such as the atomic and molecular
structure of the material, and the net magnetic field associated with the
atoms. In most atoms, electrons occur in pairs. Each electron in a pair
spins in the opposite direction, so when electrons are paired together,
their opposite spins cause their magnetic fields to cancel each other.
Therefore, no net magnetic field exists. Alternately, materials with some
unpaired electrons will have a net magnetic field and will react more to
an external field.
Most materials can be classified as ferromagnetic, diamagnetic or
paramagnetic.
Relation between the magnetic field (B), external magnetic field (H)
and magnetization (M):
m o m(1 )B H M H
where
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
126
m is the susceptibility, is the permeability, and o is the permeability
of free space (air).
Diamagnetic materials 0m
Small and negative susceptibility.
Slightly repelled by a magnetic field.
Do not retain the magnetic properties when the external field is
removed.
Magnetic moment – opposite direction to applied magnetic field.
Solids with all electrons in pairs - no permanent magnetic
moment per atom.
Properties arise from the alignment of the electron orbits under
the influence of an external magnetic field.
Most elements in the periodic table, including copper, silver, and
gold, are diamagnetic.
m(argon) ~ -1.010-8 m(copper) ~ -1.010-5
H
B
Diamagnetic material
m < 0 (small)
B = o (1+ m ) H
Permeability
= o (1+ m ) =slope of B-H line
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
127
Paramagnetic materials 0m small
Small and positive susceptibility.
Slightly attracted by a magnetic field.
Material does not retain the magnetic properties when the
external field is removed.
Properties are due to the presence of some unpaired electrons
and from the alignment of the electron orbits caused by the
external magnetic field.
Examples - magnesium, molybdenum, lithium, and tantalum.
m(oxygen) ~ 2.010-6 m(aluminium) ~ 2.110-5
Ferromagnetic materials
Large and positive susceptibility.
Strong attraction to magnetic fields.
Retain their magnetic properties after the external field has been
removed.
Ideal magnetic material
or paramagnetic material
m > 0 (small)
B = o r H = H
= constant = slope of B-H curve
B
H
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
128
Some unpaired electrons so their atoms have a net magnetic
moment.
Strong magnetic properties due to the presence of magnetic
domains. In these domains, large numbers of atomic moments
(1012 to 1015) are aligned parallel so that the magnetic force
within the domain is strong. When a ferromagnetic material is in
the demagnetized state, the domains are nearly randomly
organized and the net magnetic field for the part as a whole is
zero. When a magnetizing force is applied, the domains become
aligned to produce a strong magnetic field within the part.
Iron, nickel, and cobalt are examples of ferromagnetic materials.
o ( )B H M Magnetization is not proportional to the applied
field.
m(ferrite) ~ 100 m(iron) ~ 1000
Magnetic Domains
Ferromagnetic materials get their magnetic properties not only because
their atoms carry a magnetic moment but also because the material is
made up of small regions known as magnetic domains. In each domain,
all of the atomic dipoles are coupled together in a preferential direction.
This alignment develops as the material develops its crystalline
structure during solidification from the molten state. Magnetic domains
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
129
can be detected using Magnetic Force Microscopy (MFM) and images of
the domains like the one shown below can be constructed.
Magnetic Force
Microscopy (MFM) image
showing the magnetic
domains in a piece of heat
treated carbon steel.
During solidification a trillion or more atom moments are aligned
parallel so that the magnetic force within the domain is strong in one
direction. Ferromagnetic materials are said to be characterized by
"spontaneous magnetization" since they obtain saturation
magnetization in each of the domains without an external magnetic
field being applied. Even though the domains are magnetically
saturated, the bulk material may not show any signs of magnetism
because the domains develop themselves are randomly oriented
relative to each other. Ferromagnetic materials become magnetized
when the magnetic domains within the material are aligned. This can be
done by placing the material in a strong external magnetic field or by
passes electrical current through the material. Some or all of the
domains can become aligned. The more domains are aligned, the
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
130
stronger the magnetic field in the material. When all of the domains are
aligned, the material is said to be magnetically saturated. When a
material is magnetically saturated, no additional amount of external
magnetization force will cause an increase in its internal level of
magnetization.
demagnetized Material Magnetized Material
The Hysteresis Loop and Magnetic Properties
A great deal of information can be learned about the magnetic properties
of a material by studying its hysteresis loop. A hysteresis loop shows the
relationship between magnetization and external magnetic field. This
relationship, for many ferromagnetic materials, is different when external
field
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
131
is increasing from when it is decreasing. An example hysteresis loop is
shown below.
The materials of both (a) and (b) remain strongly magnetized when
external field is reduced to zero. Since (a) is also hard to demagnetized, it
would be good for permanent magnets. Since (b) magnetizes and
demagnetizes more easily, it could be used as a computer memory
material. The material of (c) would be useful for transformers and other
alternating devices when zero hysteresis would be optimal.
CHAPTER 4 MAGNETIC FIELDS AND MAGNETIC FORCES
132