T.O. 33B-1-13-39Example:Determine the current needed to longitudinally magnetize a l2 inch long part with a diameter of4 inches and using a 5 turn, 12 inch diameter coil. (See paragraph 3.3.12.6.2 to determine thecross-sectional area ratio between the part and the coil). If the part contains hollow portions, Dshould be replaced with D_{eff} (see paragraph 3.3.12.6.3.4).Substituting known values gives:I=-430006561245( ( /) )I = 3969 amperes3.3.12.6.3.3 FormulaforCableWraporHighFill-FactorCoils.When using a cable wrap or when the cross-sectional area of the part is greater than one-half of the cross-sectional areaof the coil, the following formula SHALL be used for estimating the current required to longitudinally magnetize a partcentered in the coil. If the part has hollow portions, replace D with D_{eff}, in the formula (see paragraph 3.3.12.6.3.4).IKNLD=+((/ ) )2Where:I = Current through the coil (amperes)K = 35,000, a constant (ampere-turns)N = Number of coil or cable turnsL = Length of the part (inches)D = Diameter of the part (inches)Example:Determine the required current to longitudinally magnetize a part 12 inches long with a 4 inchdiameter using the cable wrap technique with a 3 turn wrap.Substituting known values gives:I=+3500031242(( /) )I = 35000/3(12/4 + 2)I = 2333 amperes3.3.12.6.3.4 FormulaforHollowPartsorPartsHavingHollowPortions.If a part has hollow portions, replace the diameter (D) with the effective diameter (D_{eff}), which is calculated using:DA Aefft h=-212pWhere:D_{eff} = Effective Diameter (inches)A_{t} = Total cross-sectional area of part (square inches)A_{h} = Area of part hollow sections of part (square inches)p= 3.1416

Integrated Publishing, Inc.

6230 Stone Rd, Unit QPort Richey, FL34668

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