Question

Suppose you wish to make a solenoid whose self-inductance is 1.2 mH. The inductor is to have a cross-sectional area of 2.00 10-3 m2 and a length of 0.053 m. How many turns of wire are needed?

Answer 1

Giver, Self Inductae = 102mH -3.2 Cross sec Area = 2X10 M Length = 0.053 m turns = ? we know L=NNA L = Inductance = 0.0012 H A = 2x10 m L = 0.053 m Na ?

IM = 2716 N=407107 put values, we get D.0012 - I XD XNX2X10 O. 053 0.0012 = 4 * 3.14 X 10' xnx2x103 0.053 - 6 pol2 = 12.5% 12x16 15 Or 053 0.0012 = 25.12*10 XN 00053 020012 = 473.96 x 18 xn? 1020103 = 273.96X10" XN? N2 - 102x153 473.96X15 10 3 12 = 102x107 473.96 NZ 2053 2104

N = 2.53 * 104

N = 159

Number of turns would be 159

hope this helps , regards