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Supplementary Examples
Magnetism
Supplementary Example 6.
A loudspeaker coil is placed in a radial magnetic field of flux density 0.7T. The coil is circular in
shape, with a diameter of 30mm, and consists of 110 turns. Calculate the force acting upon the
coil when the current is 25mA.
m
d N
3
= × × ×
= ×
−
N
F B I
3
= × × ×
−
Supplementary Example 6.
A current of 10A in a 4000 turn coil produces a flux of 6mWb. Assuming the current and flux are
proportional calculate the emf induced in the coil if:
6.2.1 The current is reduced to 1A in 0.13 seconds.
6.2.2 The current is increased from 10A to 14A in 0.4 seconds.
6.2.
V
t t
N
E
3 3
1 0
1 0 1
− × × − ×
− −
10 A
1 A 0.6mWb
6mWb
0.13s
6.2.
V
t t
N
E
3 3
1 0
1 0 2
− × × − ×
− −
Supplementary Example 6.
A 600mm long conductor is moved at a uniform rate at right angle to its length and to the
magnetic field having a density of 0.5T. If the emf generated in the conductor is 3V and the
conductor forms part of a close circuit having a resistance of 0.6Ω calculate the following;
6.3.1 The velocity of the conductor.
6.3.2 The force acting on the conductor.
6.3.3 The work done when the conductor has moved 700mm.
( v = 10m.s
- ; F = 1.5N; W = 1.05J)
6.3.
1
3
−
−
× ×
m s
B
e v
e B v
6.3.
N
R
e F B I B
3
= × × ×
−
6.3.
J
W F s
3
= × ×
= ×
−
14A
10A 6mWb
8.4mWb
0.4s
Magnetic circuits
Supplementary Example 7.
A magnetic circuit consists of cast steel yoke which has a cross sectional area of 200mm 2 and a
mean length of 120mm. There are two air-gaps, each 0.2mm long. Calculate:
7.1.1 The mmf required to produce a flux of 50μWb in the air-gaps
7.1.2 The value of the relative permeability of cast steel at this flux density.
The magnetisation curve for the cast steel is given by the following:
B (T) 0.1 0.2 0.3 0.
H (A/m) 170 300 380 460
7.1.
Part A: Cast Steel Yoke
T
a
B
6
6
×
×
−
−
From the table, the corresponding straignt line velue of magnetic field strength of stell H will be
A m
H
The Air-gap
A m
B
H
3
7
0
= ×
×
−
A
H H
mmf mmf mmf
CS CS Ag Ag
T CS Ag
3 3
= × + × + ×
−
7.1.
MA Wb
S
S
A
mmf
CS
CS
CS
6
6
= ×
×
= ×
−
7 6 6
× × × × ×
− −
r
o r
CS a
S
Supplementary Example 7.
A steel ring has a mean diameter of 20cm, a cross section of 20cm 2 and an air-gap of 0.5mm
cut in it. The ring is uniformly wound with a coil of 1500 turns. Magnetizing current of 1.2A
produces a flux of 1000μWb in the air-gap. Neglecting the effect of magnetic leakage and
fringing, calculate:
7.1 The reluctance of the magnetic circuit (1.8MA/Wb)
7.2 The relative permeability of the steel. (156)
Determine the current necessary in a coil of 4000 turns wound on part B in order to produce
a flux density of 0.7T in the air-gap. Magnetic leakage and fringing may be neglected.
Air-gap Part A Part B
Φ
Wb
Ba
42 μ
6
6
= ×
= × ×
−
= 42 μ Wb = 42 μ Wb
B 0. 7 T
T
a
B
6
6
×
×
T
a
B
6
6
×
×
H
A m
B
H
3
7
0
= ×
×
−
A m
H
H = 340 A / m
mmf (^) T
A
H (^) Ag Ag
3 3
= × × ×
−
A
H A A
3
= × ×
−
A
H B B
3
= × ×
−
A
mmf (^) T mmfAg mmfA mmfB
mA
I
mmf NI
Supplementary Example 7.
A magnetic circuit consists of three parts as connected in series as follows:
Part A Part B Air-gap
Length 300mm 120mm 1.0mm
Cross sectional area 450mm 2 300mm 2 350mm 2
Determine the mmf necessary to produce a flux of 0.35mWb in the airgap. Neglect magnetic
leakage and fringing. The magnetic characteristic for parts A and B is given by:
H (A/m) 400 560 800 1280 1800
B (T) 0.7 0.85 1.0 1.15 1.
The Airgap
T
a
B
6
3
×
×
−
−
A m
B
H
3
7
0
= ×
×
−
A
mmf HAg Ag
3 3
= × × ×
−
Part A
H A m
T
a
B
6
3
×
×
−
−
A
mmf HA A
3
= × ×
−
Part B
H A m
T
a
B
6
3
×
×
−
−
kAWb
a
S
o
Ag
7 6
3
× × ×
× ×
− −
−
π
μ
7.5.
A
mmf NI
= ×
( )
Wb
S S S
mmf S
A B Ag
T T
3
×
= + + ×Φ
= ×Φ
T
a
B
6
6
×
×
−
−
Since the cross-sectional area of all the series components is the same. The magnetic flux
density will be the same throughout the whole length of the circuit.
N
Ba F (^) p
7
2 6
0
2
× ×
× ×
= ×
= ×
−
−
π
μ
kg
m
F (^) p mg
Supplementary Example 7.
A magnetic circuit consists of an iron ring of mean circumference of 90cm with a cross
sectional area of 13cm 2 throughout. A current of 1.9A in the magnetizing coil of 200 turns
produces a total flux of 1.2mWb in the iron. Calculate:
7.6.1 The flux density in the iron (0.923T)
7.6.2 The absolute and relative permeability of the iron (2.186mWb/A.m, 1739)
7.6.3 The total reluctance of the circuit. (0.317MA/Wb)
7.6.
T
a
B
4
3
×
×
−
−
7.6.
A m
NI
H
2
×
×
−
7
3
×
×
−
−
π
μ
μμ
r
o r
mWb A
H
B
7.6.
MA Wb
a
S
o r
3 4
2
× × ×
×
− −
−
Supplementary Example 7.
A coil of 450 turns and resistance of 18Ω is wound uniformly on an iron ring of mean
circumference of 60cm and cross-sectional area of 5cm² it is connected to a 24V D.C. supply.
The relative permeability of iron is 800. Calculate:
7.7.1 The magnetic field strength (1000A/m)
7.7.2 The total flux (503μWb)
7.7.3 The total reluctance of the ring. (1.193MA/Wb)
Find:
7.8.1 The reluctance of part A (176.8kA/Wb)
7.8.2 The reluctance of part B (250.5kA/Wb)
7.8.3 The reluctance of two air gaps (3537kA/Wb)
7.8.4 The total reluctance (3960kA/Wb)
7.8.5 The total mmf (2000A)
[ ]
[ ]
m
m
m
m
B
A
2
2
2
2
−
−
−
−
= ×
= + + − − − − ×
= ×
= − + − + ×
7.8.
kA Wb
a
S
o r
A
7 4
2
× × × ×
×
− −
−
7.8.
kA Wb
a
S
o r
B
7 4
2
× × × ×
×
− −
−
7.8.
3537 kA/ Wb
a
S
7 4
3
o
Ag
π× × ×
× ×
μ
− −
−
7.8.
( )
kA Wb
S (^) T SA SB SAg
7.8.
A
mmf NI
= + ×
Supplementary Example 7.
A ring has a mean length of 66cm and a cross-sectional area of 10cm 2
. The ring is made up
of semi-circular sections of cast iron and cast steel, with each joint having reluctance
equalling an air-gap of 0.2mm. The relative permeability of cast steel and cast iron are 800
and 166 respectively. Neglect fringing and leakage effects. If the flux produced in the air-gap
is 800 x 10-6^ Wb, determine the following:
7.9.1 The flux density in the air-gaps (0.8T)
7.9.2 The reluctance of the air-gaps (318kA/Wb)
7.9.3 The reluctance of cast iron (1.582MA/Wb)
7.9.4 The reluctance of cast steel (328.26kA/Wb)
7.9.5 The total magneto-motive force of the magnetic circuit. (1.783kA)
7.9.
T
a
B
4
6
×
×
−
−
7.9.
kA Wb
a
S
o
Ag
7 4
3
× × ×
×
− −
−
7.9.
kA Wb
a
S
o r
CI
7 4
2
× × × ×
×
− −
−
7.9.
kA Wb
a
S
o r
CS
7 4
2
× × × ×
×
− −
−
7.9.
[ ]
A
S S S
mmf S
Ag CI CS
T T
3 6
= + + × × ×
= + + ×Φ
= ×Φ
−
A m
NI
H
A m
NI
H
out
in
×
×
8.1.
mm
Mean d
= ×
π
π
MA Wb
a
S
o
in
6
7 6
3
= ×
× × ×
×
− −
−
A
I
N
SΦ
6 6
× × ×
−
8.1.
H
di
d L N
- 85 μ
6
×
= ×
−
Supplementary Example 8.
The field winding of a four pole separately excited dc generator consists of four coils connected
in series, each coil being wound with 1200 turns. If a current of 2A produces a magnetic flux of
400 μWb, calculate:
8.2.1 The inductance of the field circuit.
8.2.2 The average value and direction of the induced emf if the field switch is opened at such
a speed that the flux falls to the residual value of 20μWb in 0.01s.
8.2.
mH
di
d L N
6
×
×
= ×
×
−
8.2.
( )
V
dt
di emf L
=− ×
=− ×
Supplementary Example 8.
Two coils, with terminals AB and CD respectively, are inductively coupled. The inductance
measured between terminals AB is 380μH and that between terminals CD is 640μH. With B
joined to C, the inductance measured between terminals AD is 1600μH. Calculate:
8.3.1 Draw the sketch of the circuit for both conditions (Cumulative and Differential coupled)
8.3.2 The mutual inductance of the coils. (Cumulative coupled)
8.3.3 The inductance between terminals AC when B is connected to D.
8.4.
V
dt
di emf L
=− ×
8.4.
mH
I
M N
3
1
2
×
=− ×
−
Supplementary Example 8.
A steel rod of 1cm in diameter and 60cm in length is formed into a closed ring. The ring is
uniformly wound with 500 turns of wire. A D.C. current of 5A is passed through the winding
and produces a flux density of 0.75T. If all the flux links with every turn of the winding,
determine:
8.5.1 The relative permeability of the steel ring (143.2)
8.5.2 The inductance of the coil (5.896mH)
8.5.3 The average value of the induced emf when the interruption of the current causes the
flux in the steel to decay to 20% of its original value in 20ms. (1.178V)
8.5.
A m
NI
H
2
×
×
−
7
6
×
×
−
−
π
μ
μ
μμ
r
o r
Wb A
H
B
8.5.
Wb
d B
Ba
2
2
×
= ×
= ×
mH
I
L N
6
×
= ×
−
8.5.
11. 78 μ Wb
6
2 1
= × ×
Φ = ×Φ
−
A
L
i N
6
2 2
×
×
= ×
−
−
V
dt
i i emf L
3
3
2 1
×
=− ×
−
−
Supplementary Example 8.
A lighting circuit is operated by a relay of which the coil has a resistance of 5Ω and an
inductance of 0.5 H. The relay coil is supplied from a 6V DC source through a push button
switch. The relay operates when the current in the relay coil attains a value of 500mA.
Calculate:
8.6.1 The time constant (100ms)
8.6.2 The maximum current ( 1.20A)
8.6.3 The time interval between the pressing of the push-button and the closing
of lighting circuit (54ms)
