Question

How many turns are required to make a solenoid of length .150m and cross sectional area .00325m^2 with a self inducatance of 12.5 mH?

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Answer #1

self inducatnce L= mu_o*N^2*A*l

given that A = 0.00325 m^2

l = 0.150 m

mu_o = 4*pi*10^-7 H/m

No.of turns is N = sqrt( L/(mu_o*A*l)= sqrt((12.5*10^-3)/(4*3.142*10^-7*0.00325*0.15)) = 4517 turns

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