Addis Ababa Science and Technology University
College of Electrical and Mechanical Engineering
Department of Electromechanical Engineering
Lecture of Introduction to Electrical Machine
(EMEg-3104)
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Example 1.1
The total magnetic flux out of a cylindrical permanent magnet is
found to be 0.032 mWb. If the magnet has a circular cross
section and a diameter of 1 cm, what is the magnetic flux
density at the end of the magnet?
Solution
The total flux
, cross-sectional area of magnet:
Note that this magnetic flux density exists only at the
immediate end of the magnet. As we move away from
the end of the magnet, the magnetic flux spreads out,
and therefore the magnet flux density decreases.
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Example 1.2
The coil below has 1000 turns wound on a cardboard toroid.
The mean (or average) diameter D of the toroid is 10 cm, and
the cross section is 1 cm. The total magnetic flux in the toroid is
3µWb when there is an excitation current of 10 mA in the coil.
a) What is the magnetic flux when the current
is increased to 20 mA?
a) What is the magnetic flux density within
the coil when the current is 20 mA?
Solution
a) If we double the current to 20 mA, then
b) For a toroid. the magnetic flux is assumed to be uniform
across the interior cross-sectional area of the coil.
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Example 1.3 The circuit given below is a magnetic core made of
cast steel. A coil of N turns is wound on it. For a flux of 560 µWb,
calculate the necessary current, neglecting any fringing effects. The
cross-sectional area A is constant.
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Example 1.4 An electromagnet is of the form and dimensions
as shown below. It is made of iron of square section 4 cm side.
A flux of 1.1 mWb is required in the air gap. Neglecting leakage
and fringing, calculate the number of ampere turns required.
Take the relative permeability to be 2000 at this flux density.
Solution
The magnetic circuit of the electromagnet shown in
Figure 1.13 is completed by four parts connected in
series, viz. (i) iron portion C (ii) air gap, (iii) iron portion
D, and (iv) air gap. Total ampere turns required for this
magnetic circuit, FT = ampere turns required for iron
portion C , FC + ampere turns required for air gap, Fag
+ ampere turns for iron portion D, FD + ampere turns of
air gap, Fag
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• The flux that is produced by the
MMF in the center leg exists in the
center leg and then divides into two
parts, one going in the path afe and
the other in the path bcd. If we
assume for simplicitv that afe = bcd,
the flux is distributed evenly between
the two paths.
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Example 1.5 The following dimensions are given
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