T.O. 33B-1-1
3-38
Where:
I = Current through coil (amperes)
K = 45,000 (a constant, ampere-turns)
L = Length of the part (inches)
D = Diameter of the part (inches)
N = Number of turns in coil
Example:
Determine the current required to longitudinally magnetize a steel part 10 inches long with a
diameter of 2 inches using a 12 inch diameter coil having 5 turns. (See paragraph 3.3.12.6.2 to
determine cross-sectional area ratio between part and coil.)
Substituting the known values and doing the calculations gives:
I
=
45000
2
5
10
I = 1800 amperes
Table 3-2 gives typical currents for a five turn coil with the parts lying in the bottom of the coil or held next to the coil
wall.
Table 3-2. Typical Coil-Shot Current for a Five-Turn Coil with Part in Bottom of Coil.
Part Length in
Inches (L)
Part Diameter in
Inches (D)
L/D Ratio
Ampere-Turns
Required
Amperes Required
12
3
4
11,250
2,250
12
2
6
7,500
1,500
16
2
8
5.625
1,125
10
1
10
4,500
900
18
1 ½
12
3,750
750
14
1
14
3,214
643
3.3.12.6.3.2
Formula for Part in Center of Coil. This formula SHALL be used when the cross-sectional area of part is greater than
one-tenth and less than one-half of the cross-sectional area of the coil(s).
I
KR
N
L D
=
-
( (
/
)
)
6
5
Where:
I = current through the coil amperes)
K = 43,000 (Constant) (ampere-turns)
R = Radius of coil, (inches)
N = Number of coil turns
L = Length of part (inches)
D = Diameter of part (inches)
The term 6(L/D)-5 is called the effective permeability.
