T.O. 33B-1-13-38Where: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 coilExample:Determine the current required to longitudinally magnetize a steel part 10 inches long with adiameter of 2 inches using a 12 inch diameter coil having 5 turns. (See paragraph 3.3.12.6.2 todetermine cross-sectional area ratio between part and coil.)Substituting the known values and doing the calculations gives:I=450002510I = 1800 amperesTable 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 coilwall.Table 3-2. Typical Coil-Shot Current for a Five-Turn Coil with Part in Bottom of Coil.Part Length inInches (L)Part Diameter inInches (D) L/D RatioAmpere-TurnsRequiredAmperes Required12 3 4 11,250 2,25012 2 6 7,500 1,50016 2 8 5.625 1,12510 1 10 4,500 90018 1 ½ 12 3,750 75014 1 14 3,214 6433.3.12.6.3.2Formula for Part in Center of Coil. This formula SHALL be used when the cross-sectional area of part is greater thanone-tenth and less than one-half of the cross-sectional area of the coil(s).IKRNL D=-( (/) )6 5Where:I = current through the coil amperes)K = 43,000 (Constant) (ampere-turns)R = Radius of coil, (inches)N = Number of coil turnsL = Length of part (inches)D = Diameter of part (inches)The term 6(L/D)-5 is called the effective permeability.

Integrated Publishing, Inc.

6230 Stone Rd, Unit QPort Richey, FL34668

Phone For Parts Inquiries: (727) 493-0744
Google +