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.