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
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 188.8.131.52.2 to
determine cross-sectional area ratio between part and coil.)
Substituting the known values and doing the calculations gives:
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
Table 3-2. Typical Coil-Shot Current for a Five-Turn Coil with Part in Bottom of Coil.
Part Length in
Part Diameter in
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 = 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.