CHAPTER - 2

Electromagnetism

Learning objectives

 Magnetic flux lines

 Force on current carrying conductor
 Biot Savart's law
 Magnetomotive force
 Permeability and relative permeability
 Reluctance
 Comparison of electric circuits
and magnetic circuits
 Composite series magnetic circuits
 Leakage coefficient
 Electromagnetic induction
 Faraday's laws
 Lenz law
 Dynamically and statically induced emf
 Self and mutual inductance
 Coefficient of coupling
 Energy in a magnetic field
45
 Electromagnetism 47

Coloumb first determined experimentally the quantitative expression for the magnetic
force between two isolated poles. In reality magnetic poles cannot exist in isolation. Thus,
the concept is purely theoretical. However, poles of a long thin magnet may be assumed
to be isolated poles. The force between two magnetic poles placed in a medium is

(i) directly proportional to their pole strengths m,

(ii) inversely proportional to the square of the distance d between them
(iii) inversely proportional to the absolute permeability of the medium.

m1 m2 Km1 m2
F� ∝ or F =
µd 2 µd 2
1
In SI system of units the value of K is 4π

m1 m2 m1 m2
F� = 2
= N (2.1)
4πµd 4πµ0 µr d 2

where m1 , m2 are the pole strengths, d is the distance in m, µ0 is permeability of free

space = 4π × 10−7 H/m, µr is relative permeability of the medium.
Thus, theoretically a unit magnetic pole may be defined as that pole which when placed
in vaccum at a distance of one meter from a similar and equal pole repels it with a force
1
of 4πµ 0
N . Oersted discovered in 1820 that a magnetic field is produced around a current
carrying conductor.

2.2.1. Biot-Savart Law

The expression for the magnetic field d B� produced at a point P by an elemental length d l�
of a conductor carrying a current of I amperes is given by Biot-Savart’s law. Referring to
Fig. 2.2.

µI dl sin θ
d B� = Wb/m2
4πr 2
or

µI d l� × a�r
d B� = Wb/m2 (2.2)
4πr 2
48 Basic Electrical Engineering

dl ar

θ
P

Figure 2.2 Biot-Savart’s law.

where a�r is the unit vector along lines joining dl to P . The direction of d B� is perpendicular
to the plane containing both d l� and a�r . The field at a distance r due to an infinitely long
straight conductor carrying a current I amperes is given by

µI
B� = Wb/m2 .
2πr

The flux lines are in the form of concentric circles around the conductor. If the conductor
is held with the thumb pointing in the direction of the current, the encircling fingers give
the direction of the magnetic field.

2.2.2. Force on a current carrying conductor

It was further observed that another current carrying conductor experiences a force when
placed in the field. Now we can recollect that current is nothing but flow of electrons
(charges!). Thus magnetic fields are produced by moving charges (current carrying con-
ductor) and exert a force on moving charges. The characteristics of this magnetic force on
a moving charge are as follows:

• Its magnitude is proportional to the magnitude of the charge.

• The magnitude of the force is proportional to the magnitude or strength of the field.
• The magnetic force depends on the particle’s (charge’s) velocity v�. This is different
from the electric field force which is the same whether the charge is moving or not. A
charged particle at rest experiences no magnetic force.
• By experiment it is found that the force is always perpendicular to both the magnetic
field B� and the velocity v�.
 Electromagnetism 49

The above characteristics can be put compactly as,

F� = q v� × B� (2.3)

Similarly the force experienced by a current carrying conductor in a magnetic field

is found to be proportional to the magnetic field B,� the current I and the length of the
conductor and is perpendicular to the field and the length of the conductor. Thus,

F� = I l� × B� (2.4)

Since, the direction of the conductor, fixes the direction of the current (in space) (2.4) is
more commonly written as

F� = l I� × B� (2.5)

Let F� be the force in Newtons, I� the current in amperes and l the length of the conductor
in meters, at right amperes to the magnetic field. Then the magnetic field B� or flux density
is the density of a magnetic field such that a conductor carrying a current of 1 ampere at
right angles to the field has a force of 1 newton per meter acting upon it. The unit is Tesla
(T), after the scientist Nikola Tesla. The force on a current carrying conductor is given by,

F� = lI B sin θ (2.6)

where θ is the angle between the magnetic field and the current carrying conductor. Thus
a current carrying conductor experiences a force in the presence of a magnetic field. This
principle is used in all electric motors.
The direction of the force may be found from Fleming’s left-hand rule as shown in Fig. 2.3.

force F

magnetic field B

current I
left hand

Figure 2.3
50 Basic Electrical Engineering

Hold out your left hand with the fore finger, middle finger and thumb at right angles
to each other. If the fore finger represents the direction of the field and the middle finger
the direction of the current, the thumb gives the direction of the force on the conductor.
From (2.5) it is obvious that no force is exerted on the conductor when it is parallel to the
magnetic field (θ = 0◦ ).

2.2.3. Force between two current carrying conductors

Consider two conductors carrying currents I1 and I2 respectively, separated by a distance
of dm. The force between the conductors is attractive if the currents flow in the same
direction and repulsive if the currents flow in opposite directions. Let us consider the
force on the second current due to the first. The field produced by conductor 1 is given by

µI1
B= T
2πd
The force experienced by conductor 2, from (2.5) is given by

µlI2 I1
F = N
2πd
or the force per unit length is given by

µI1 I2
F = N/m.
2πd

2.2.4. Magnetic flux

For a magnetic field having a cross-sectional area A m2 and a uniform flux density of B
Teslas, the total flux in Webers (Wb) passing through a plane at right angles to the flow is
given by

φ = BA
(Webers) = (Tesla) × (m2 )

or
φ
B= (2.7)
A
 Electromagnetism 51

Hence the unit of B is also Wb/m2

1Tesla = 1Wb/m2

Example 2.1 A conductor carries a current of 500A at right angles to a magnetic field
having a density of 0.4T. Calculate the force per unit length on the conductor. What
would be the force if the conductor makes an angle of 45◦ to the magnetic field?

Solution:
F = l I� × B� = lI B sin θ

When conductor is at right angles to the magnetic field, θ = 90◦ .

F = (1m)(500A)(0.4T) = 200N/m.

When θ = 45◦ ,

F = (1m)(500A)(0.4T) × sin 45◦

= 141.42N/m.

Example 2.2 A rectangular coil 100mm by 150mm is mounted so that it rotates about
the mid points of the 150mm sides. The axis of rotation is at right angles to a magnetic
field with a flux density of 0.02T. Calculate the flux in the coil when
(i) Maximum flux links with the coil. What is the position at which this occurs?
(ii) The flux through the coil when the 150mm sides make an angle of 30◦ to the
direction of flux.

Solution:

(i) This is shown in Fig. 2.4(a). The maximum flux passes through the coil when the
plane of the coil is at right angles to the direction of the flux.

φ = BA = 0.02 × (100 × 10−3 ) × (150 × 10−3 )

= 0.3mWb.
52 Basic Electrical Engineering

axis of rotation

30°
0.02T 0.02T
(a) (b)

Figure 2.4 Example 2.1.

(ii) This is shown in Fig. 2.4(b).

φ = BA sin θ = (0.3 × 10−3 ) × sin 30◦

= 0.15mWb.

2.3. Magnetomotive force and magnetic field strength

The magnetic flux is present in a magnetic circuit due to the existence of a magnetomotive
force (mmf), caused by a current flowing through one or more turns. It is analogous to emf
in an electric circuit which is responsible for the electric current.

mmf = N I (2.8)

where N is the number of turns. N is a dimensionless quantity. Hence, the unit of mmf is
actually Ampere, though more commonly the unit is said to be ampere-turns (AT).
Consider a coil as shown in Fig. 2.5. If the magnetic circuit is homogeneous and has a
uniform cross sectional area, the mmf per metre length of the magnetic circuit is called
the magnetic field strength H .

NI
H = AT/m (2.9)
l

The unit of H in SI units is A/m.

The ratio B/H is the permeability, µ0 , in free space

B
µ0 = (2.10)
H
 Electromagnetism 53

N turns

Figure 2.5 Coil with N turns on a toroid.

This value is almost same when the conductor is placed in free space, air or in any other
non-magnetic material like water, wood, oil etc.
µ0 = 4π × 10−7 H/m (2.11)

For magnetic materials,

B
H =
µ
µ = µ0 µr (2.12)

where µr is the relative permeability.

The relative permeability is defined as the ratio of the flux density produced in a material
to the flux density produced in vacuum by the same magnetic field strength.
The relative permeability of non-magnetic materials is close to 1. The relative
permeability of magnetic materials is very high, as shown in Table 2.1.

Table 2.1 Relative permeability of magnetic materials.

Material µr Application

Ferrite U60 8 UHF chokes

Ferrite M33 750 Resonant circuit RM cores
Nickel (99% pure) 600
Ferrite N41 3,000 Power circuits
Ferrite T38 10,000 Broadband transformers
Silicon steel 40,000 Dynamos, mains transformer
54 Basic Electrical Engineering

When working with non magnetic materials, the permeability is close to µ0 , making
it difficult to characterize them by permeability. We make use of magnetic susceptibility
defined as

ψm = µr − 1 (2.13)

Example 2.3 A coil of 100 turns is wound uniformly over a wooden ring having a
mean circumference of 500mm and a uniform cross sectional area of 500mm2 . If the
current through the coil is 2.0A calculate
(i) the magnetic field strength
(ii) the flux density
(iii) the flux
(iv) mmf

Solution:

(i) Mean circumference = 500mm = 0.5m.

NI 100 × 2
H = = = 400AT/m or A/m
l 0.5

(ii) B = µ0 H = 4π × 10−7 × 400 = 502.65 µT

(iii)

φ = BA = 502.65 × 10−6 × 500 × 10−6

= 0.2513 µWb

(iv) mmf = N I = 100 × 2 = 200AT.

Example 2.4 Calculate the mmf required to produce a flux of 0.01Wb across an airgap
2mm long, having an effective area of 100cm2 .
 Electromagnetism 55

Solution:

A = 100cm2 = 100 × 10−4 = 0.01m2

φ 0.01
B= = = 1T
A 0.01
B 1
H = = AT/m
µ0 4π × 10−7
l = 2mm = 2 × 10−3 m
2 × 10−3
mmf = H × l = = 1591.55AT
4π × 10−7

2.4. Reluctance
Consider the toroid shown in Fig. 2.5, with a cross-sectional area A m2 and a mean
circumference of l metres, with N turns carrying a current I amperes. We know

φ = BA
mmf = H l
φ BA A
∴ = = µr µ0
mmf Hl l
or
mmf mmf
φ= =
l/µr µ0 A S
where
l
S= (2.14)
µ0 µr A
S is the reluctance of the magnetic circuit and is indicative of the opposition of a magnetic
circuit to creation of magnetic flux through it. From (2.14) we can write

mmf = φS (2.15)

similar to Ohm’s law. Unit of S is AT/Wb. It is analogous to resistance in electric circuits.

The reciprocal of reluctance is called the permeance of the magnetic circuit. Its unit is
Wb/AT. It is analogous to conductance in electric circuits.
56 Basic Electrical Engineering

2.5. Comparison of magnetic and electric circuits

Table 2.2 gives the analogous quantities between electric circuits and magnetic circuits.

Table 2.2 Comparison of magnetic and electric circuits.

Magnetic circuit Electric circuit

1 φ-Flux (Webers) I -Current (Amperes)

2 B-Flux density (Wb/m2 ) J -Current density (A/m2 )
l ρl
3 S-Reluctance = (At/Wb) R-Resistance = (Ohms)
µ0 µr A A
1 1
4 P -Permeance = (Wb/AT) G-Conductance = R (mhos)
S
5 mmf = φS (AT) emf = I R (Volts)
6 Permeability (µ) Conductivity (σ )
 
1
7 Reluctivity Resistivity (ρ)
µ

There are however some differences between electric circuits and magnetic circuits:

• The flux does not flow through the magnetic circuit like the current does in an electric
circuit.
• In electric circuits if the temperature is maintained a constant, the resistance is a constant
and independent of the current. In a magnetic circuit, the reluctance depends on the
flux established through it. The reluctance is small for small values of B and larger for
larger values of B. This is because the B–H curve is not a straight line.
• Flow of electric current requires continuous expenditure of energy but in a magnetic
circuit energy is expanded only in creating the magnetic flux but not in maintaining it.
• A magnetic circuit stores energy in its field, while an electric circuit dissipates its energy
as heat.

2.6. Composite magnetic circuits

We will now discuss composite magnetic circuits; circuits connected in series and parallel.
 Electromagnetism 57

la

l1

l2

Figure 2.6 Composite series magnetic circuit.

2.6.1. Composite series circuits

Consider the composite circuit of Fig. 2.6, made up of two different sections in series and
an air gap. Each section is made of a different material and has its own reluctance. The
total reluctance is sum of the individual reluctances

ST = S1 + S2 + Sa
l1 l2 la
= + + (2.16)
µ0 µr1 A1 µ0 µr2 A2 µ0 Aa
mmf
= (2.17)
ST

To find the ampere-turns or the total mmf,

• Find the H of each section using

B
H = (if it is air gap)
µ0
B
= (for magnetic material)
µ0 µr
• Find the mmf (AT) for each section by

AT = H l

• Add these ampere-turns to get the total ampere turns. (Similar to adding emf’s in series!)
 Electromagnetism 59

Leakage flux
Iron ring
Fringing Flux lines
(useful flux)

Air gap

Figure 2.8 Leakage flux and fringing.

Solution: The circuit is shown in Fig. 2.9.

φ = 0.6 × 10−3 Wb
A = 10 × 10−4 m2
φ 0.6 × 10−3
B= =
A 10 × 10−4
= 0.6Wb/m2

Cast iron

0.15mm
20cm air gap

Cast steel

Figure 2.9 Example 2.5.

60 Basic Electrical Engineering

Air gap
B 0.6
H = = = 4.77 × 105 AT/m
µ0 4π × 10−7
Total air gap length = 0.15 + 0.15 = 0.3mm
= 0.3 × 10−3 m
AT required to set up flux in air gap = H l
= 4.77 × 105 × 0.3 × 10−3
= 143.1AT

Cast iron section

B 0.6
H = = = 596.83AT/m
µ0 µr 4π × 10−7 × 800
πD π × 20 × 10−2
length of the path = πr = = = 0.314m
2 2
AT required = H l = 596.83 × 0.314 = 187.41AT

Cast steel section

B 0.6
H = = = 2876.29AT/m
µ0 µr 4π × 10−7 × 166
πD
length of the path = = 0.314m
2
AT required = 2876.29 × 0.314 = 903.16AT
Total AT required = 143.1 + 187.41 + 903.16 = 1233.67AT.

Example 2.6 A mild steel ring having a cross sectional area of 600mm2 and a
mean circumference of 500mm has a coil of 300 turns wound uniformly around it.
Calculate
(i) the reluctance of the ring
(ii) the current required to produce a flux of 800µWb in the ring, if the relative
permeability is 400.
 Electromagnetism 61

Solution:

(i)
l 500 × 10−3
S= =
µ0 µr A 4π × 10−7 × 400 × 600 × 10−6
= 1.658 × 106 AT/Wb

(ii)
mmf = φ × S
= 800 × 10−6 × 1.658 × 106
= 1326.4AT
mmf 1326.4
I= = = 4.42A
N 300

Example 2.7 A magnetic circuit comprises three parts in series as follows:

(a) A length of 60mm with a cross section area of 50mm2
(b) A length of 30mm with a cross section area of 80mm2
(c) An air gap of length 0.3mm and cross section area of 150mm2 .
A coil of 2500 turns is wound on part (b) and the flux density in air gap is 0.3T.
Assuming that there is no leakage, and the relative permeability µr = 1500, estimate
the current required in the circuit to produce the flux density.

Solution:

φ = Bc Ac = 0.3 × 150 × 10−6 = 45 × 10−6 Wb

φla 45 × 10−6 × 60 × 10−3
Fa = mmfa = φSa = =
µ0 µr Aa 4π × 10−7 × 1500 × 50 × 10−6
= 28.6AT
φlb 45 × 10−6 × 30 × 10−3
Fb = mmfb = φSb = =
µ0 µr Ab 4π × 10−7 × 1500 × 80 × 10−6
= 8.95AT
62 Basic Electrical Engineering

φlc 45 × 10−6 × 0.3 × 10−3

Fc = mmfc = φSc = =
µ0 µr Ac 4π × 10−7 × 1 × 150 × 10−6

= 71.62AT

F = Fa + Fb + Fc = 28.6 + 8.95 + 71.62 = 109.17AT

F 109.17
I= = = 0.04367A = 43.67mA.
N 2500
(* Note the large mmf required to set up the flux through air gap as compared to a magnetic
material.)

Example 2.8 A wooden ring has a circular cross section of 200 sq mm and a mean
diameter of 200mm. It is uniformly wound with 600 turns. If the µr = 1, find (1)
the field strength produced by a current of 2A (ii) magnetic flux density (iii) current
required to produce a flux density of 0.015Wb/m2 .

Solution:
(i)

mmf = NI = 600 × 2 = 1200AT

Mean length = πd = π × 200 × 10−3 = 0.628m

NI 1200
H = = = 1910.83AT/m
l 0.628
(ii)

B = µ0 µr H = 4π × 10−7 × 1 × 1910.83 = 2.4 × 10−3 Wb/m2

(iii) The flux density is proportional to the current. A current of 2A produces

0.0024Wb/m2 . Therefore current required to produce 0.015Wb/m2 is

2 × 0.015
= 12.5A.
0.0024
 Electromagnetism 63

Example 2.9 A magnetic core in the form of a closed circular ring has a mean length
of 20cm and a cross sectional area of 1cm2 . The relative permeability of the material
is 2200. What current is needed in the coil of 2000 turns wound uniformly around
the ring to create a flux of 0.15mWb in the iron? If an air gap of 1mm is cut through
the core in a direction perpendicular to the direction of this flux, what current is now
needed to maintain the same flux in the air gap?

Solution: Reluctance of core

l 20 × 10−2
S= =
µ0 µr A 4π × 10−7 × 2200 × 1 × 10−4

= 723431.5 AT/Wb

φ = 0.15 × 10−3 Wb

mmf = φS = 0.15 × 10−3 × 723431.5 = 108.5AT

mmf = NI
mmf 108.5
I= = = 0.05425A
N 2000
1 × 10−3
Reluctance of 1mm air gap = = 7957747.1 AT/Wb
4π × 10−7 × 1 × 1 × 10−4
mmf required to set up flux in air gap = 0.15 × 10−3 × 7957747.1 = 1193.66AT
Total mmf = 108.5 + 1193.66
= 1302.16AT
1302.16
Current required = = 0.65108A
2000

Example 2.10 An iron ring of mean diameter 20cm, having a cross section area of
3 sq cm is required to produce a flux of 0.45mWb. If µr = 1800 find the mmf required.
If an air gap of 1mm is made in the ring, how many extra ampere turns are required
to maintain the same flux?
64 Basic Electrical Engineering

Solution:

Length of the mean path = πd = π × 0.2

= 0.6283m
φ = 0.45 × 10−3 Wb
B = φ/A = 0.45 × 10−3 /3 × 10−4 = 1.5Wb/m2
B 1.5
H = = = 663.154AT/m
µ0 µr 4π × 10−7 × 1800
mmf = H l = 663.154 × 0.6283 = 416.65AT

An air gap of 1mm will need extra mmf.

Bg 1.5
Hg = =
µ0 4π × 10−7
1.5 × 1 × 10−3
mmf = Hg × lg = = 1, 193.66AT
4π × 10−7

So additional mmf required when a 1mm air gap is cut is 1,193.66AT.

2.7. Electromagnetic induction—Faraday’s law

In 1831, Michael Faraday discovered the principle of electromagnetic induction. Two of
his experiments are well known. The first experimental setup is shown in Fig. 2.10(a).

S A
B G
G S N

(a) (b) A

Figure 2.10 Electromagnetic induction.

66 Basic Electrical Engineering

Magnetic Motion of the conductor

field Thumb
Forefinger

Induced current

Middle finger

Figure 2.11 Fleming’s right hand rule.

Consider a coil of N turns. Let the flux linking the coil change from φ1 Wb to φ2 Wb in t
seconds. Now,

Initial flux linkage ψ1 = Nφ1 Wb turns

Final flux linkage ψ2 = Nφ2 Wb turns
Nφ2 − Nφ1
emf induced, e = volts
t
N(φ2 − φ1 )
=
t

The above relationship can be written as

dφ dψ
e=N = volts (2.19)
dt dt

To incorporate Lenz’s law we often write e = −N dφ dt , meaning the induced emf is set up
in a direction such that it opposes the rate of change of flux.
The emf can be induced in two ways

(i) Dynamically induced emf (or motional emf)

(ii) Statically induced emf (or transformer emf)
68 Basic Electrical Engineering

Average induced emf in the coil is LI

t volts. This current also produces a change in flux,
from 0 to φWb whose average rate is given by φt . From Faraday’s law the induced emf
= Nφt . Thus, we have

LI Nφ
=
t t

or

Nφ ψ
L= = (2.21)
I I

Thus inductance is nothing but the flux linkage per ampere. An alternative expression is

dφ
L=N
dI

We can also define 1H as the inductance of a coil when a current of 1 ampere through the
coil produces a flux linkage of 1Wb turn. Now from (2.15)

mmf NI NI
φ= = =
reluctance S l/µ0 µr A
Nφ NI N 2 µ0 µr A
∴ L= =N = H (2.22)
I I (l/µ0 µr A) l

Equation (2.22) gives the expression for L from the geometric parameters of the coil.

2.7.7. Mutually induced emf

Consider two coils A and B placed as shown in Fig. 2.12. When the switch S is closed,
some flux produced by A, also links with B. This produces an induced emf in coil B and
a current flows through circuit of B, as indicated by the galvanometer deflection. Since
a change of current in one coil is accompanied by a change of flux linked with the other
coil inducing an emf in it, it is called mutually induced emf. The two coils are said to have
a mutual inductance.
 Electromagnetism 69

G
S

A B

Figure 2.12 Mutually induced emf.

2.7.8. Mutual inductance

The unit of mutual inductance is also Henry. Two coils have a mutual inductance of 1H
if an emf of 1volt is induced in one coil when the current through the other coil varies
uniformly at the rate of 1A/s.
Again consider the two coils of Fig. 2.12.
Let I1 be the current flowing through coil A, which produces a flux φ1 . All this flux
does not link with coil B. The flux φ11 which links only with coil A is called the leakage
flux. The flux φ12 which also links with coil B is called the mutual flux.

φ1 = φ11 + φ12 (2.23)

Coefficient of coupling, ‘K’ is defined as the ratio of mutual flux to total flux.

φ12
K1 = (2.24)
φ1

K ≤ 1. The induced emf in coil B is given by

dI1
e2 = M .
dt
Also
dφ12
e2 = N2
dt
dφ12
∴ M = N2
dI1
 Electromagnetism 71

where L1 , L2 are the self inductances of coil A and coil B respectively. If K1 �= K2 , then
in (2.27) we use the geometric mean of K1 and K2 ;

K = K1 K2

0 ≤ K ≤ 1. Larger values of coefficient of coupling are obtained with coils which are
physically closer, which are wound or oriented to provide a larger common magnetic flux
or which are provided with a common path through a material which serves to concentrate
and localize the magnetic flux. Coils with K close to unity are said to be tightly coupled.

Example 2.11 A coil consists of 750 turns. A current of 10A in the coil gives rise to
a magnetic flux of 1200µWb. Determine the inductance of the coil and the average
induced emf in the coil when the current is reversed in 0.01sec.

Solution:

N = 750; I = 10A; φ = 1200 × 10−6 Wb

Nφ 750 × 1200 × 10−6
L= = = 0.09H
I 10
Current reverses from 10A to −10A.

∴ dI = 10 − (−10) = 20A
dI 20
e=L = 0.09 × = 180V.
dt 0.01

Example 2.12 An air cored solenoid has a length of 50cm and diameter of 2cm.
Calculate its inductance if it has 1000 turns.

Solution:
N 2 µ0 µr A
L=
l
πd 2 π × (2 × 10−2 )2
A= = = 3.14 × 10−4 m2
4 4
72 Basic Electrical Engineering

l = 50 × 10−2 m
(1000)2 × 4π × 10−7 × 1 × 3.14 × 10−4
L=
50 × 10−2
= 78.9 × 10−5 H = 0.7892mH

2.8. Energy stored in magnetic field

In an electric field energy is continuously dissipated. The energy cannot be stored. In a
magnetic field on the other hand, we need energy only to set up the initial flux and no
energy is required to maintain it. Magnetic field stores the energy which has been used to
create the flux.
Let the current flowing through a coil of constant inductance L Henrys grow at an
uniform rate from zero to I amperes in t seconds. The average value of current is I2 and
 
the emf induced in the coil is L × It volts. The average power absorbed by the magnetic
field is
1 LI
I× Watts
2 t
and the total energy absorbed is
1 LI
average power × time = I × ×t
2 t
or
1
W = LI 2 Joules (2.28)
2
Now lets consider a more general case where the instantaneous current i increases in a coil
having a constant inductance LH. The rate of increase can be uniform or non-uniform. If
the current increases by di amperes in dt seconds, the induced emf is given by
di
e=L Volts
dt
The energy absorbed is
 
di
W =i L dt = Li · di Joules.
dt
 Electromagnetism 73

The total energy absorbed by the magnetic field when the current increases from 0 to I
amperes is given by
 I
1 1
W = Li · di = L × [i 2 ]I0 = LI 2 Joules
0 2 2

From (2.22) L = N 2 µ Al H . The energy per cubic meter Wf is

1 2 2µ 1 2 1 1 B2
Wf = I N 2 = µH = BH = Joules (2.29)
2 l 2 2 2 µ
Equation (2.29) can be used only if µr is a constant.
Now when the inductive circuit is opened, the current has to reduce to zero and the stored
energy released. If there is no resistor in the circuit the energy will be mostly dissipated
in the arc across the switch. If there is a resistor, the energy is dissipated as heat in the
resistor.

2.9. Dot convention

In an inductor, which is a two terminal element, if the current enters the terminal used
as positive reference, the induced voltage L didt is positive, as shown in Fig. 2.13(a). If it
enters the coil at the terminal used as negative reference, it is negative, as in Fig. 2.13(b).
The dot convention is used to fix the polarity of the voltage induced in a coil due to
mutual inductance. A dot is placed on each of the coils. The sign of the voltage due to
mutual inductance is as follows:

A current entering the dotted terminal of one coil produces a voltage with a positive
reference at the dotted terminal of the second coil.

i + i −

Ldi v Ldi v
v= v=−
dt dt
− +
(a) (b)

Figure 2.13 Induced voltage.

 Electromagnetism 75

M = 2H
i1 i2
+ +
•

v1 L1 L2 v2

•
− −

Figure 2.16 Example 2.13

Solution: (i) i2 enters undotted terminal of L2 . Hence, mutually induced emf is positive
at the undotted terminal of L1 . However, v1 is referenced positive at the dotted terminal.
Therefore,
di2
v1 = −M = −(2)(314)(10 cos 314t)
dt
= −6280 cos 314tV

Since i1 = 0, there is no self induced emf.

(ii) i1 enters the dotted terminal of L1 . Hence, v2 is positive at the dotted terminal of L2 .
However, it is referenced positive at the undotted terminal.
di1
∴ v2 = −M = −(2)(−1)(−8e−t )
dt
= −16e−t V.

2.10. Inductance in series

Consider two inductances connected in series as shown in Fig. 2.17(a).
Each of the coils has self induced emf and mutually induced emf.
di1 di2
v1 = L1 +M
dt dt
i1 = i2 = i
di
∴ v1 = (L1 + M)
dt
76 Basic Electrical Engineering

M M
L1 L2 L2
i i i • L1 i •
• • + v1 − + v2 −
+ v1 − + v2 −
(a) v (b) v

Figure 2.17 Inductances in series.

Similarly
di
v2 = (L2 + M)
dt
di
v = v1 + v2 = (L1 + L2 + 2M)
dt
di
= Leq
dt
Leq = L1 + L2 + 2M (2.30)

This is called series-aiding connection, where the mutual flux and leakage flux aid each
other.
In Fig. 2.17(b) M is negative
di
v1 = (L1 − M)
dt
di
v2 = (L2 − M)
dt
di
v = v1 + v2 = (L1 + L2 − 2M)
dt
Leq = L1 + L2 − 2M (2.31)

This is called series-opposing connection.

The instantaneous energy stored in a coupled circuit is given by
1 1
W = L1 i12 + L2 i22 ± Mi1 i2 (2.32)
2 2
√
As derived earlier M = K L1 L2 .
78 Basic Electrical Engineering

Example 2.15 The equivalent of two inductances connected in series is 0.6H or 0.1H,
depending on the connection. If L1 = 0.2H find (i) M (ii) K.

Solution:
L1 + L2 + 2M = 0.6 (i)
L1 + L2 − 2M = 0.1 (ii)
4M = 0.5 or M = 0.125H
L1 = 0.2H

From (i) L2 = 0.6 − 0.2 − 0.125 × 2 = 0.15H

 0.125
K = M/ L1 L2 = √ = 0.722.
0.2 × 0.15

Example 2.16 Two identical air cored solenoids have 200 turns, length of 25cm and
cross section area of 3cm2 each. The mutual inductance between them is 0.5µH. Find
the self inductance of each coil and the coefficient of coupling.

Solution:
N 2 µ0 µr A (200)2 (4π × 10−7 )(1)3 × 10−4
L= = = 60.318µH
l 0.25
L1 = L2 = L
M 0.5 × 10−6
K=√ = = 8.289 × 10−3 .
L1 L2 −6
(60.318 × 10 ) 2

Example 2.17 A closed iron ring of mean diameter 12cm is made from round iron
bar of 2cm diameter. It has a winding of 1000 turns. Calculate the current required
to produce a flux density of 1.5Wb/m2 given the relative permeability is 1250. Hence
find the self inductance.
80 Basic Electrical Engineering

(i)

B 0.5
µ= = = 1.2 × 10−3
H 416.67
µ 1.2 × 10−3
µr = = = 954.9
µ0 4π × 10−7

(ii)

Nφ π(2 × 10−2 )2
L= ; φ = B × A = 0.5 ×
I 4
−4
= 1.57 × 10 Wb
250 × 1.57 × 10−4
L= = 0.0785H
0.5
(iii)

φ1 = BA = 1.57 × 10−4 Wb
φ2 = 10% of φ1 = 0.157 × 10−4 Wb
dφ = φ1 − φ2 = 1.413 × 10−4 Wb
dφ 1.413 × 10−4
e=N = 250 × = 35.325V
dt 0.001

Example 2.19 When a voltage of 220V is applied to a coil with a resistance of 50�,
the flux linking with the coil is 0.005Wb. If the coil has 1000 turns find the inductance
of the coil and the energy stored in the magnetic field.

Solution:
V 220
Current = = = 4.4A
R 50
Nφ 1000 × 0.005
L= = = 1.136H
I 4.4
1 1
Energy stored = LI 2 = × 1.136 × 4.42 = 11J
2 2
 Electromagnetism 81

Example 2.20 A mild steel ring has a mean diameter of 160mm and a cross section
area of 300mm2 . Calculate
(a) the mmf to produce a flux of 333µWb
(b) reluctance
(c) relative permeability.
The B-H data is given in table below.
B(T) 0.9 1.1 1.2 1.3
H(AT/m) 260 450 600 820

Solution:
φ 400 × 10−6
B= = = 1.2T
A 333 × 10−6

(a) From Table, H = 600AT/m

mmf = H l = 600 × π × (160 × 10−3 )

= 301.59AT

(b) mmf = φS
mmf 301.59
S= = = 9.057 × 105 AT/Wb
φ 333 × 10−6
(c)
B 1.2
µ= = = 2 × 10−3
H 600
µ 2 × 10−3
µr = = = 1591.5.
µ0 4π × 10−7

Example 2.21 A steel circuit has a uniform cross sectional area of 5cm2 and a length
of 25cm. A coil of 120 turns is wound uniformly over it. When the current in the coil
is 1.5A, the total flux is 0.3mWb. Find (i) H (ii) µr .
82 Basic Electrical Engineering

Solution:

mmf = NI = 120 × 1.5 = 180AT

mmf = H l
NI 180
∴ H = = = 720AT/m
l 2.5 × 10−2
φ 0.3 × 10−3
B= = = 0.6Wb/m2
A 5 × 10−4
B 0.6
B = µ0 µr H or µr = =
µ0 H 4π × 10−7 × 720
= 663.145

Example 2.22 A steel ring has a mean circumference of 750mm and a cross sectional
area of 500mm2 . It is wound with 120 turns (a) Using the table of example 2.20 find
the current required to set up a magnetic flux of 630µWb in the ring (b) If the air
gap in a magnetic circuit is 1.1mm long and 2000mm2 in cross section, calculate the
reluctance of the air gap and the mmf required to send a flux of 700µWb across the
air gap.

Solution:
(a)
φ 630 × 10−6
B= = = 1.26
A 500 × 10−6
From table, using interpolation, H = 732AT/m.

Hl 732 × 750 × 10−3

I= = = 4.575A
N 120
(b)
φ 700 × 10−6
B= = = 0.35Wb/m2
A 2000 × 10−6
B 0.35
H = = = 2.785 × 105 AT/m
µ0 4π × 10−7
84 Basic Electrical Engineering

(b)

N1 φ1 12, 000 × 0.05 × 10−3

L1 = = = 0.12H
I1 5
N2 φ2 15, 000 × 0.075 × 10−3
L2 = = = 0.225H
I2 5
(c)

M 0.0675
K=√ =√ = 0.411
L1 L2 0.12 × 0.225

Example 2.25 Two coupled coils of self inductances 0.6H and 0.16H have a
coefficient of coupling 0.8. Find mutual inductance and turns ratio.

Solution:
 √
M = K L1 L2 = 0.8 0.6 × 0.16
= 0.248H
N1 φ1
I1 =
L1
N2 Kφ1 N2 Kφ1 N2 L1 K
M= = =
I1 N1 φ1 /L1 N1
N2 M 0.248
∴ = = = 0.516
N1 KL1 0.8 × 0.6

Questions

(1) Define magnetic flux.

(2) What is Biot-Savart’s law?
(3) What is the force between two current carrying conductors?
(4) Explain mmf and its analogy with emf.