The self and mutual-inductance of linear conductors . ate the expression for the mag-netic force between the wires due to unit current, //= 2/^.Thus, id Ar=i I ?^=2/iog^ f Multiplying this by two (for both wires) and adding the term dueto the magnetic field within the wires we have the result given by(14). If the end effect is large, as when the wires are relatively farapart, use the expression for the self-inductance of a rectanglebelow (24); or, better, add to the value of (14) the self-inductance ofAB + CD using equation (10) in which /= 2AB. 4. MUTUAL INDUCTANCE OF TWO PARALLEL WIRES BY NE
The self and mutual-inductance of linear conductors . ate the expression for the mag-netic force between the wires due to unit current, //= 2/^.Thus, id Ar=i I ?^=2/iog^ f Multiplying this by two (for both wires) and adding the term dueto the magnetic field within the wires we have the result given by(14). If the end effect is large, as when the wires are relatively farapart, use the expression for the self-inductance of a rectanglebelow (24); or, better, add to the value of (14) the self-inductance ofAB + CD using equation (10) in which /= 2AB. 4. MUTUAL INDUCTANCE OF TWO PARALLEL WIRES BY NEUMANNS FORMULA. Neumanns formula for the mutual inductance of any two cir-cuits is M-- r Tcos e ds ds , . In this case e — o and cos e=i, r—Jd^^i^y — Uf^ and the integra-tion is along both lines. M= /^?^fxx^ The quantity in the brackets is the mutual inductance of the lineAB and unit length of CD at a point distant b from the lower end,Fig. 4a. Now making b variable and calling it j/, and integratingalong CD from o to /, we have 3o8. Bulletin of the Bi^reau of Standards. [ Vol. 4, No. 2. Fig. 4a. M=2
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