A solenoid inductor carries a current of 180 mA. It has a magnetic flux of 19.0 mu Wb per turn and stores 1.40 mJ of energy.How many turns does the inductor have?
the magnetic energy stored in inductor (L), having N turns, is E (mag) 0.5 L I^2 ----------------- given mag. flux with each turn [phi (1)] 19*10^-6 Weber/turn Total flux linked with L [phi (N)] 19*10^-6* N but total flux L I 19*10^-6* N L 19*10^-6* N / I E (mag) 0.5 [19*10^-6* N / I] I^2 E (mag) 0.5 *19*10^-6* N * I N 1.40*10^-3 /[0.5 *19*10^-6* 180*10^-3] N 818.7
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