Problem 77-1
Source: Problem created by the author of the site.
A ferromagnetic core is shown in Figure 77-01.1 whose path length
average is 60 cm. There is a thin 0.05 cm air gap in the core, which is one piece
in the rest. The cross-sectional area of the core is 36 cm2, the relative permeability of the core is 4800 and the coil wound in the core has 240 turns through which a current of
10 A circulates.
Assume that the spread in the air gap increase the effective area of the cross section by 6%. Given this information, find
a) the total reluctance of the flow path (iron plus air gap).
b) what is the magnetic flux density in the air gap?
Figure 77-01.1
Solution of the Problem 77-1
Item a
To calculate the magnetic reluctance we will use the eq. 77-18, repeated below:
Replacing with numeric values
Rcore = 0.6 m / (4 800 x 4 π x 10-7 x 0.0036 m2 )
Performing the calculation, we have
Rcore = 27 631 A.e / Wb
This is the reluctance value in the ferromagnetic part of the core. Now let's calculate the reluctance at the core's air gap.
The effective area of the core in the air gap is
Aef = 1.06 x 36 = 38.16 cm2. Thus, the reluctance at the air gap,
Rgap, is given by
By numerically replacing the values, we have:
Rgap = 0.005 m / (4 π x 10-7 x 0.003816 m2 )
Performing the calculation, we have
Rgap = 1 042 682 A.e / Wb
Now we can calculate the total reluctance that the core path offers to the magnetic flux, that is:
Rtotal = Rcore + Rgap
After replacing the numerical values and performing the calculation:
Rtotal = 27 631 + 1 042 682 = 1 070 313 A.e / Wb
Item b
From eq. 77-13 we know that F =ΦR and beyond that,
Φ =BA e F =N I. Relating these equations, we find
that N I =BAR . And so, we arrive at the following equation
B = N I / A Rtotal
Substituting the numerical values, we have
B = 240 x 10 / 1 070 313 x 0.003816
Performing the calculation, we found
B = 0.588 T
As the problem requires the determination of the magnetic flux density in the air gap,
then we use the effective area of the air gap.
