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Magnetism
Magnetic Flux
Symbol  (Phi) Units webers (Wb)
Starting in this section we will define the various important magnetic quantities, their basic
units, and their interrelationships. In addition we will draw analogies between the magnetic
quantities and their electrical counter-parts.
The first magnetic quantity to be considered is magnetic lines of force or magnetic flux. The
letter symbol for the quantity of magnetic flux is the Greek letter  (phi). In the SI system of
units, the basic unit for magnetic flux is the weber, abbreviated Wb. Thus, if we write
 = 2 Wb
this indicates that a flux of 2 webers is present.
A weber is actually a relatively large unit for flux. One weber equals 108 lines of flux. That
is,
1 Wb = 108 lines
The microweber (Wb) is a smaller unit that is often used.
1 Wb = 1 x 10-6 (Wb) = 1 x 10-6 x (108 lines)
= 102 lines
1 Wb = 100 lines
Flux is Analogous to Electric Current. We know that electric current must have a complete
closed path in order to flow. Similarly, flux lines always form dosed loops or paths. In
electricity the paths for current constitute an electric circuit. Similarly, we can define the
paths traced out by magnetic flux lines as magnetic circuits. Thus, we can state that
Magnetic flux follows a magnetic circuit in the same manner that current follows
an electric circuit.
The analogy between current and magnetic flux is not an exact one because magnetic flux
does not actually flow. However, the similarities are many and it will help us to a better
understanding of electromagnetism.
Flux Density Symbol B Units weber/m2 (Wbm-2) or Tesla T
A magnetic field is stronger if its flux lines are concentrated into a smaller area. To illustrate,
consider Fig. 1. In part (a) of the figure we have 10 flux lines passing through a crosssectional area of 1 cm x 1 cm = 1 cm2 = l0-4 m2. In (b) we have 10 flux lines passing through
a larger cross-sectional area of 2 cm x 2 cm = 4 x l0-4 m2. Thus, the flux is more concentrated
in (a) and therefore represents a stronger magnetic field than in (b).
1
2cm
1cm
(a)
1cm
(b)
2cm
 = 10 lines = 10-7 Wb
B = /A 10 -3 Wb/m2 =10 -3T
 = 10 lines = 10-7 Wb
B = /A = 1/4 x 10 -3 T
Dots indicate flux lines coming Out of the page toward observer
FIGURE 1 Difference between flux, , and flux density, B.
The flux concentration is more often called the flux density and represents the amount of flux
per unit of cross-sectional area. The letter symbol for flux density is B. Since B equals flux
per unit area, then it follows that
B = /A (1)
where  is the total flux in Wb passing through a cross-sectional area A in square metre (m2).
From this relationship it is apparent that flux density is given in units of webers/square meter
or Wb/m2. However, the SI chooses to use teslas (T) as the basic unit for flux density. That
is,
1 tesla = 1T = 1 Wb/m2 We will use teslas and Wb/m2 interchangeably.
EXAMPLE 1 Determine the flux density for each of the cases in Fig. 1.
Solution:
(a) The flux is equal to 10 lines. We saw earlier that 100 lines equal 1 Wb.
Thus, 10 lines represent 0.1 Wb. The cross-sectional area is 1 cm x 1 cm = 1 cm2 = 10-4 m2.
Thus, using Eq. (l) we have
B=
0.1 Wb
l0-4 m2
=
l0-7 Wb
l0-4 m2
B = l0-3T
(b)
 is the same, but A is now increased to 4 cm2 = 4 x l0-4 m2 so that
B=
0.lWb
4 x 10-4m2
= 1 x 10-3 T
4
The same flux spread out over a larger area produces a smaller flux density.
2
Permanent magnets and electromagnets are often described in terms of their flux density. A
bar magnet might have a flux density B equal to l0-3 Wb/m2 = l0-5 T at its poles. The earth's
magnetic field has a flux density roughly equal to 10-5 Wb/m2 = 10-5 T. A large laboratory
magnet could produce B = 5 Wb/m2 = 5 T.
EXAMPLE 2 A certain bar magnet has a cross-sectional area of 9 cm2. The flux density at
the poles of this magnet is 0.0025 T. How much flux is emanating from this magnet's north
pole?
Solution:
Since B = /A, then it is also true that
 =B x A
In this example, B = 0.0025 T = 2.5 x 10-3 Wb/m2 and
A = 9 cm2 = 9 x l0-4m2. Thus,
= 2.5 x l0-3 Wb/m2 x 9 x l0-4 m2 = 2.25 x 10-6 Wb
= 2.25 Wb
EXAMPLE 3 A certain air-core solenoid consists of 100 turns of wire wound around a
cardboard cylinder. With 1 A of current through the coil, the flux produced in the centre of
the solenoid is only 0.4 Wb. Determine the flux density if the core has a diameter of 2 cm.
Solution. A = D2 =  x (0.02 m)2
4
4
=
0.000314 m2
=
3.14 x 10-4 m2
=
0.4 Wb (given)
B = 
A
0.4 x 10-6 Wb
3.14 X 10-4 m2
= 1.27 x 10-3 Wb/m2
= 1.27 x 10-3 T
EXAMPLE 4 If the cardboard core in Ex.3 is replaced by a soft-iron toroidal core, the flux
density increases by a factor of 1000. What will be the amount of flux in the toroidal core if it
has the same cross-sectional area as the cardboard core'?
Solution: The flux density has increased by 1000 times. This means that there is 1000 times
as much flux passing through the same area. Thus, the new value for  will be 1000 times
that which was present in the air-core solenoid.
=
=
1000 x 0.4 Wb
400 Wb (40,000 lines)
3
Magnetomotive Force, F
We know that an electric current will flow in a circuit when a potential difference or voltage
is applied to the circuit. This applied voltage is often called an electromotive force, emf. In a
like manner, we have seen that magnetic flux lines will be produced in a magnetic circuit
when a current is passed through a wire or coil. Furthermore, we have seen that an increase in
the coil current or in the number of turns in the coil causes the flux to increase. We can say
that the flux has been established by a magnetomotive force, mmf, that depends on the
current and the number of turns of wire in the coil. In other words, magneto-motive force
equals the product of amperes X turns. Stated formally
F=IxN
(2)
where F is the symbol for mmf, N is the number of turns, and I is the current in amperes. The
basic units for mmf are ampere. as defined. (The number of terms is unitless)
An mmf applied to a magnetic circuit produces flux in the same manner that an emf
produces current in an electric circuit.
Thus, we can say that mmf is analogous to emf. Neither of them is actually a force in the
strictest sense; instead they respectively represent the necessary applied energy needed to
establish flux in a magnetic circuit and current in an electric circuit.
FIGURE 2
EXAMPLE 5 Calculate the magnitude of the mmf for each case in Fig. 2
Solution:
(a) In Fig. 2(a) there are three turns of wire wound around the core so that
N = 3. Thus,
F=NxI
=3 turns x 2 A
=6A
(b)
Here there are six turns in the coil, which produces
F= 6 x 2
= 12 A
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Magnetic Field Intensity, H
Refer to Fig. 3 where two rectangular soft-iron cores of different lengths are shown. In each
case, a coil of five turns is wound around the core and a current of ten amperes is flowing
through the coil. Thus, for both cores the applied mmf is 50 A. Using the right`-hand rule, the
direction of flux will be cw for both cores.
FIGURE 3 The same mmf applied to cores of different magnetic path lengths produces
different magnetic field intensities (If) and therefore different amounts of flux.
The core in part (a) of the figure provides a magnetic circuit path length of 0.1 m while the
core in part (b) provides a path length of 0.2 m. Both cores have the same cross-sectional
area. This means that the flux produced in (a) travels a shorter path than the flux produced in
(b). For the same applied mmf, the flux will be greater for the core with the shorter length.
This is analogous to the electric circuit situation where the same voltage (emf) is applied to
two wires of different lengths. The shorter wire has less resistance and therefore will pass
more current than the longer wire, for the same voltage.
Thus, the same mmf applied to cores of the same material and cross-sectional area, but of
different lengths, will produce different amounts of flux  and flux density B = /A.
In order to take into account the magnetic path length, we can define a new magnetic quantity
called magnetic field intensity, which will equal mmf per unit length. The symbol for
magnetic field intensity is H. The defining relationship for H is
H = F = NI
l
l
(3)
where  is the applied mmf and 1 is the path length in metres. The basic units for H can be
determined from this expression as ampere-turns per meter (A/m).
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EXAMPLE 6 Determine the values of H for the two cases in Fig. 3.
Solution:
(a) For Fig. 3(a) we have
H = NI
l
= (5 x l0)A = 500
0.lm
(b)
A/ m
For Fig. 3(b) we have
H = NI - (5 x l0) A = 250A/m
l
0.2
The magnetic field intensity H is useful because it can be shown that for the same H applied
to cores of the same material, the flux density will be the same in each core. This point will
be discussed further in our investigation of magnetic permeability.
Magnetic Permeability, 
When a certain value of H is applied to a core, this specifies how much magnetic field
intensity is available to produce magnetic flux. The actual amount of flux which the applied
H can produce depends on the core material. Every material has a characteristic called its
magnetic permeability, which is a measure of how much flux density, B, can be established in
it for a specified value of field intensity, H. Stated mathematically,
B=xH
(4)
where  is the letter symbol for magnetic permeability. Equation (4) can also be written as
 = B (5)
H
which shows that  is the ratio of the flux density to the magnetic field intensity that
produces it. A larger value of permeability indicates a greater flux density for a given H. The
units for  can be determined from (5) as
units for  = B = Wb/m2 = T
H
A /m A /m
The permeability of air and most nonmagnetic materials has a value of 4 x l0-7 = 1.26 x 10-6
and is given the symbol o. Thus,
o = 1.26 x 10-6 A /m
EXAMPLE 7 A 10-cm long air-core coil consists of 200 turns of wire. What will be the flux
density in the coil if a current of 0.25 A flows in the coil?
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Relative Permeability, r
The permeability of magnetic materials can ran from 50 to 80,000 times that of air. The
relative permeability r of a magnet material is simply the ratio of its absolute magnetic
permeability,  to the permeability of air, o. That is,
r = 
o
(6)
For example, the relative permeability of cobalt is 60, which means it has permeability of 60
times that of air. r has no units but is a pure number
EXAMPLE 8 If the core in Fig. 4(a) is made of silicon iron (r = 7000), determine the flux
density in the core.
Solution.' The value of H in Fig. 4(a) was previously determined to be 500 A/m. To
determine B we have to know the value of  Since
r = 
o
then we have,
o r = 
= 7000 x 1.26x10-6 T
= 8.8 x 10-3 A/m Thus,
B = H
= 8.8 x 10-3x 500 = 4.4 T
The greater the permeability of a material, the better it is as a "conductor" of magnetic flux
lines, and vice versa. Thus,  is analogous to electrical conductivity,  (Greek letter "sigma").
 is the reciprocal of resistivity  and is a measure of a material's ability to conduct electrical
current. A low resistivity  means a high conductivity  and a better conductor of electricity.
B-H Curves
The value of permeability for air and other nonmagnetic materials is essentially constant at a
value equal to o. The permeability of ferromagnetic substances, however, is not constant but
can vary as the applied H varies. In other words, the flux density in such materials is not
directly proportional to the magnetic field intensity. This is much like the situation with a
non-linear resistor where the current is not directly proportional to the applied voltage.
For such materials it is often helpful to plot a curve showing how flux density B varies with
magnetic field intensity H. Such a curve is called a B-H curve (or magnetisation curve). An
example of a B-H curve for cast iron is shown in Fig. 5. The value of H is easily varied by
varying the current in the coil (by varying R). The resultant values of flux density B produce
the B-H curve as shown.
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FIGURE 6 B-H magnetisation curve for cast iron.
It is apparent that the curve is non-linear, indicating that B does not vary in proportion to H.
For example, the curve shows that B 0.3 T, when H = 1000 A /m (see point x on the curve).
If H is tripled in value to 3000 A4/m, the value of B does not triple but increases to only 0.75
T (point y).
Because the B-H curve for cast iron is non-linear, the permeability M will different at
different points on the curve. For example,
(at point x) = B
H
= 0.3 T
1000 A/m
(at point y) = 0.75T
3 000A/m
=
2.5 x 10-4 A/m
The value of permeability decreases for higher values of H because B not increase
proportionally with H. In fact, at values of H above 5000 A/m, approximately, the value of B
increases at a very slow rate, as indicated by relatively flat slope of the curve. This effect
whereby a small change in density occurs for a large change in field intensity is called
saturation. As discussed earlier, saturation occurs when most of the magnetic domains of the
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Materials align themselves with the applied field so that very little additional flux can be
produced by increasing the applied mmf.
Many electronic applications of magnetic materials, such as transformers, inductors,
speakers, and magnetic amplifiers, require that the core material does not saturate or even
operate near saturation. This is because such applications require that the core's flux
variations should follow the current variations (H variations). In these applications the core is
operated on the linear portion of its B-H curve.
Hysteresis
The B-H curve in Fig. shows us only part of the behaviour of ferro-magnetic materials. We
are now going to take a closer look at B-H curves a the phenomenon of hysteresis.
Consider the B-H curve in Fig. 14-22(a). Note that it is a closed curve which is symmetrical
about the origin. Also, note that negative as well as position values of H and B are
represented on the graph, the negative values representing a reversal of current in the coil and
the accompanying reversal of the magnetic field.
The section of the curve 0 abc is similar to the B-H curve in Fig. 14-21, and represents the
curve which would be followed by an initially unmagnetised core. If the core is initially
unmagnetised and no current is supplied to the co then H = 0 and no net flux is produced in
the core because the magnetic domains in the core are randomly oriented. This is illustrated
in part (b) of the figure.
If current is supplied to the coil and is gradually increased, the flux density in the core will
follow the curve 0 abc until H reaches 3000 A/m or more as the core is saturated [see part (c)
of the figure.
At point c on the curve, all the domains in the core are aligned with the applied magnetic
field so that B is at its maximum value.
If the current is now gradually reduced, H will return to zero. However, the flux density B
does not return along the original path cba 0 but follows the
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FIGURE 1 22 Following a hysteresis curve.
path to point d instead. In other words, when H is returned to zero, there is still a significant
flux density in the core (B = 1.3 T at point d) representing a residual or remanent magnetism.
The value of B at point d is called the remanent flux density, BR. For this curve, BR = 1.3 T.
The remanent magnetism is due to the fact that, once aligned, the magnetic domains do not
return exactly to their original positions [Fig. 14-22(d)] when the magnetising force (H) is
removed. It's as if the domains were forced to move against an internal friction among the
atoms of the magnetic material.
This effect is called hysteresis. The word hysteresis means "to lag behind," which describes
how the change in B lags behind the change in H.
We can now reverse the current through the coil [see part (e) of the figure]. Since the current
is flowing through the coil in the opposite direction, it will attempt to magnetise the core in
the opposite direction. For these negative values of H, as we increase the current the flux
density follows the curve from d to e. At point e, the magnetic field intensity has become
large enough in the negative direction to nullify the remanent magnetism that was present at
point d. This value of H is called the coercive force, Hc. For this curve Hc = 1300 A/m.
If the reverse current is increased beyond this point, the domains become aligned in the
reverse direction (right to left) until the core is saturated in the reverse direction (point on the
curve). If the reverse current is then reduced to zero, the core will again retain some remanent
magnetism (point g on the curve), but in the opposite polarity.
Finally, we can reverse the current back to its original direction. Increasing the current will
then bring the core back to a demagnetized condition (point h), and eventually to saturation at
point c. Thus, the complete curve for this ferromagnetic material is a closed curve and is
called a hysteresis curve or loop.
Hysteresis Loss The hysteresis of a ferromagnetic substance is due to some of the domains
not wanting to return to their original orientation and having to be forced to by a reversed
magnetic field intensity (- H~). Thus, energy is required to rotate these domains and this
energy shows up as heat (sort of an internal friction). If an alternating current (ac) is applied
to the coil, then the energy is taken from the ac source which is responsible for the domains
having to rotate, The higher the frequency of the ac in the coil, the more rapidly the domains
have to rotate and, therefore, the greater the hysteresis energy loss.
It turns out that the amount of energy loss in one complete magnetization cycle (once around
the hysteresis loop) is measured by the area inside the loop. In applications where current in
the coil will be continuously or often changing, such as in transformers or electromagnets, it
is desirable that the hysteresis loss be kept to a minimum. There is no ideal ferromagnetic
material that has no hysteresis loss. However, very soft and pure iron has a narrow hysteresis
loop with a low remanent flux density (BR) and a small coercive force (H~) for demagnetization. Thus, it is commonly used in such applications. Its B-H curve is like the one
in Fig. l4~2l.
On the other hand, if we wish to construct a permanent magnet, we look for a ferromagnetic
material with a high residual magnetism requiring a high coercive force to demagnetize it.
Permanent magnets are made from a wide variety of materials, but all contain some form of
10
ferromagnetic substance. Some of the popular permanent magnetic materials include Alnico
5 (an alloy of aluminum, nickel, cobalt, and copper) and hard cobalt steel.
In some applications a permanent magnet is desired which is not a metallic material, as in
cases where a high electrical resistance is required. In such applications a ceramic magnet
material can be used. This relatively new type of material is made from the oxides of certain
metals (iron, manganese, magnesium). These oxides are calledferrites. Ferrite materials find
useful application in memory banks of many modern digital computers.
14.16 Ohm's Law for Magnetic Circuits
As discussed earlier, magnetic flux ~ is analogous to electric current I, and magnetomotive
force ~ is analogous to electromotive force (voltage) E. In an electric circuit, for a given emf,
the current depends on the circuit's resistance R. Likewise in a magnetic circuit, for a given
mmf, the flux depends on the circuit's opposition. For a magnetic circuit, this opposition is
called reluctance and is given the symbol 6{ (script R).
For a resistive electric circuit, we have shown that I = E/)?, corresponding to Ohm's law. A
similar law may be expressed for magnetic circuits as follows:
(1~7)
This says that flux is directly proportional to mmf (the stimulus producing the flux) and
inversely proportional to reluctance, which is the opposition to flux lines.
The units for reluctance can be established by rearranging (14-7):
(l~8)
which indicates A-t/Wb as the units for reluctance.
EXAMPLE 14.9 A coil with 250 turns is wound around an iron core. With a coil current of
2 A, the flux in the core is 100 ~Wb. Calculate the reluctance of the core.
Solution:
~ = mmf _ NXI (250 x 2) A-t
100 X 10-6 Wb = 5 X 106 A-t/Wb
EXAMPLE 14.10 The same coil with an air core produces a flux of only 0.2 MWb. What is
its reluctance?
Solution: Since ~ is only 0.2 ~Wb, a decrease by a factor of 500, the reluctance must be 500
times greater. Thus,
=
500 X 5 X 106 A-t/Wb = 2.5 X l0~ A-t/Wb
These examples emphasize the fact that ferromagnetic materials have a much lower
reluctance than nonmagnetic materials.
Factors Determining Ot For a given magnetic circuit, the reluctance or opposition to flux
lines depends on the geometrical dimensions of the circuit and on the material in the circuit.
These are the same factors which determined the resistance of a resistor. The relationship
between reluctance and these factors is summarized in the formula below:
11
I
I
364
Chapter Fourteen
(l~9)
~ A
where M is the permeability of the material, l is the length of magnetic meters, and A is the
cross-sectional area of the path in square meters.
As with resistance, the reluctance is proportional to length and invers portional to crosssectional area. It is also interesting to note that ~ is ii proportional to permeability, M. Recall
that M is a characteristic of the and is greater for ferromagnetic materials. Thus, a larger M
would re~ lower reluctance for a given circuit. Reluctance is a characteristic of ti netic circuit
and depends on M as well as the circuit dimensions.
EXAMPLE 14.11 Consider the magnetic circuit in Fig. 14-23. The c an average path length
of 20 cm and a cross-sectional area of 5 cm2. T material is soft iron with Mr = 500. Calculate
the core's reluctance an~ mine the coil current needed to produce ~ = 10 MWb.
20 cm A = 5 em2
N~40 ~=500
FIGURE 1~23
Solution: (a) First of all, the absolute permeability of the iron core r calculated.
M = Mr X Mo
=
T
=
500>< 1.26 x 10-6
6.3 x l0-~ A-t/m Now we can calculate reluctance as
I
MA
0.2 m
T
6.3
X lO~~
x 0 0005 m2
A-t/m
= 6.35 X 1O~ A-t/Wb
12
(b) Since ~ = mmf/G~, then we can write
mmf = ~ X ot
=
(10 x 106 x 6.35 X 105)A-t
= 6.35 A-t
Since mmf = N x I and N = 40 (given in Fig. 14-23), then 40 x 1 = 6.35 A-t
or
I = 0.159 A
Effect of an Air Gap in Magnetic Circuits In many applications of magnetic circuits, the
lines of force are not always traveling through a low-reluctance ferromagnetic material.
Magnetic circuits which are part of motors, generators, relays, loudspeakers, etc., usually
consist of low reluctance ferromagnetic material. However, the flux lines in these circuits
often have to cross an air gap to complete their loops. This gap is necessary in the case of
motors, meters, and loudspeakers in order to permit space in which moving parts can move.
Even in the case of stationary magnetic equipment, an air gap may be purposely inserted into
the magnetic circuit. The purpose here is to increase the total reluctance of the circuit and
thereby prevent the ferromagnetic portion of the circuit from saturating when a large current
flows through the coil.
Figure 14-24(a) is an example of a magnetic circuit which includes an air gap. The core is the
same one in Fig. 14-23 except that a small gap of 0.1 cm has been cut into it. Note that any
flux which is produced in the core by an applied mmf must also flow through the air gap. We
can say that the air gap is in series with the ferromagnetic core. As might be expected, we
can treat this series magnetic circuit in much the same way that we treat a series electrical
circuit.
Since ~ must travel through the core and the air gap, then the total reluctance in the circuit
path is equivalent to the reluctance of the core plus the reluctance of the gap. We can
represent this schematically as shown in Fig. 14-24(b) just like a series electric circuit. The
mmf (equal to N >< 1) is like the applied voltage. ~ is the reluctance of the iron core, ~gap is
the reluctance of the air gap; they are like series resistors. Carrying the analogy further, we
can say that
mmf
=
NXI
total reluctance
_ ~co~e + ~gap
(1~10)
Thus, in order to determine ~ for a given mmf we have to know the reluctance of both the
core and the air gap.
EXAMPLE 14.12 For the magnetic circuit in Fig. 14-24(a) determine: (a) the reluctance of
the core, (b) the reluctance of the air gap, (c) the amount of current needed to produce a flux
of 10 ~Wb.
13
Solution:
(a) The reluctance of the iron core was determined in Ex. 14.11 to be 6.35 ><
l0~ A-t/Wb when the core was 20-cm long. The 0.1-cm air gap removed from the core will
not change this value significantly. Thus,
ot=re ~ 6~35 X 10~ A-t/Wb
366
Chapter Fourteen
I = 20cm A= 5cm2 Mr 500
(a)
~core~ ~gap
NX l -~
(b)
FIGURE 1~24 Magnetic circuit with air gap can be treated like a series circuit.
(b)
To find ~ we use (14-9) with
Mo = 1.26 x 10-6, l = 0.1 cm = l0-~ m, and
A = S cm2 = 5 >< 10-~ m2
Thus,
10-~
~gap = 1.26 >< 10-6 x 5 >< l0-~
=
1.59 X 106 A-t/Wb
(c)
Rearranging (14-10) we have
14
mmf = ~ >< total reluctance
10 x 10-6 Wb >( (6.35 )< 10~ +1.59 x 106) A4/Wb
~ X (0.635 +1.59) X 106 A-t
22.3 A-t
Thus,
N>< I = 22.3 A-t
or
I=
= 0.577 A
40
15