T.O. 33B11
340
Example:
What would the effective diameter be for a cylindrical part 10 inches long that has a 2 inch
outside diameter with a 0.125 inch wall thickness (see Figure 326).
Figure 326. Calculating Effective Diameter
To find A_{t}, calculate the crosssectional area of the outside diameter of the part as follows:
A_{t} = pr^{2}
A_{t} = p(1)^{2}
A_{t} = 3.1416 sq. inches
To find A_{h}, calculate the crosssectional area for the inside diameter of the part (the part’s hollow portion)
as follows:
A_{h} = pr^{2}
A_{h} = p(0.875)^{2}
A_{h} = 2.40 sq. inches
Insert the results for A_{t} and A_{h} into the formula to find D_{eff}.
D_{eff}
=

2^{31416}
2
3
1416
1
2
.
.40
.
D_{eff} = 0.97 inch
To calculate the current required to longitudinally magnetize the part in the above example, use the formula
from paragraph 3.3.12.6.3.1 (for the part in the bottom of a 12 inch diameter coil with 5 turns),
except replace D with D_{eff.} (0.97):
I
KD
NL
=
I
=
45000
0
97
5
10
.
I = 873 amperes

