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The, flux . _ , To find the intensity of the field, H , at a point dis- m AsLet Then the flux coming from sume (Fig. 10). pole of strength that the lines of force are perpendicular to the surface. B = flux density at any flat distance. one of the surfaces of the magnet 1 is FIG. 10. 9. tant d from one side of a If I 2m = 2m -> and is #= the area of the pole face is is < = S. then ^ = 2irm. " ELECTRICAL ENGINEERING 16 in Meters. To find the magnetic between the area of a pole Let S be intensity poles (Fig.
This is given solely for the purpose of impressing upon them the fact that inductance is a "constant" of the circuit. problem 3 33 ELECTRICAL ENGINEERING 34 From thus obtained, it independent of the current; the values of constant, and is I/, stant" similar to resistance in a Transposing Eq. seen that L is a " a circuit con- is it is having a non-magnetic core. f. of self-induction. In Fig. f. of self-induction due to the setting in the coil. , e. f. has to overcome the resistance, r, and also = ir + L^ fundamental, and up of flux (15) is general for circuits possessing only resistance and inductance of constant value.
Substituting this value of dx in the formula, 2irml c r sin 3 a , dF = r = ^ da sm 2 a 2irm7 c sin ada. - ai 27rm/ c sin ada = Co. *. /a =" TT H or For very long coils, 4irl, m cos ai = 47r/ c cos ai. cos ai = 1, H = 47T/ C . and The relation between H and the ampere-turns of a coil may be found as follows: Let there be a current of I abamp. in the coil, and let n = number of turns. -turns. -turns per cm. Then = nl -r- = H= Jc , where 47m/ 4ir7 c I = length of coil in centimeters. 1 When the current is in amperes, I 1 NOTE.