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The flux density of the air gap in Fig. 11.35(b) is given by
where, for our purposes,
Bg � � �g
�
� g �� core
and A g � A core
(11.13)
For most practical applications, the permeability of air is taken to be
equal to that of free space. The magnetizing force of the air gap is then
determined by
Hg � �� m
(11.14)
and the mmf drop across the air gap is equal to H gl g.Anequation for
H g is as follows:
Hg � � Bg
Bg � ����7
4p � 10
m o
H g � (7.96 � 10 5 )B g
and (At/m) (11.15)
EXAMPLE 11.6 Find the value of I required to establish a magnetic
flux of � �0.75 � 10 �4 Wb in the series magnetic circuit of Fig.
11.36.
I
N = 200 turns
I
f
e
A g
B g
o
All cast steel
�
�
l cdefab = 100 × 10 –3 m
l bc = 2 × 10 –3 m
�
a
b
c
FIG. 11.36
Relay for Example 11.6.
d
Area (throughout)
= 1.5 × 10 –4 m 2
Air gap
� = 0.75 × 10 –4 Wb
Solution: An equivalent magnetic circuit and its electric circuit
analogy are shown in Fig. 11.37.
The flux density for each section is
0.75 � 10
B � � �0.5 T
�4 �
Wb
� ��
A
1.5 � 10 �4 m 2
E
�
�
I
� core
(a)
R cdefab
(b)
AIR GAPS ⏐⏐⏐ 453
� gap
R bc
FIG. 11.37
(a) Magnetic circuit equivalent and
(b) electric circuit analogy for the relay of
Fig. 11.36.