## N

it is found that L0 = 4.5/(3)2 = 0.5 /¿h. Applying this value of Lc to the chart, we find that a coil 2Yz inches in diameter should be about 4.2 inches long to give the proper inductance. The total number of turns necessary is 4.2 N, or (4.2) (3) = 12.6 turns. The length of the coil for other diameters can readily be found from the chart, and the total number of turns determined by multiplying the length by N, the number of turns per inch.

The chart can be used equally well to find the inductance of a coil of known specifications. For example, it is desired to determine the inductance of a coil V/l inches in diameter wound with 90 turns of No. 24 D.C.C. wire. The wire tables show that this size of wire has a winding factor of 33.6 turns per inch, Data by courtesy of Electronics

### Flo. 11

from which the length of the coil is found to be 90/33.6 = 2.68 inches. From the chart, L0 is found to be about 0.29 ph for a 2'1/^-inch coil 2.68 inches long. Because L = L„N2, the inductance of the coil is (0.29) (33.6)2, or 327 fih (approximately) .