the length of thin wire required to manufacture a solenoid of inductance L and length l,(if cross section diameter is considered less than its lenght)is

1) {(pie)Ll/2(nu not)}1/2

2) {4(pie)Ll/(nu not)}1/2

3) {2(pie)Ll/(nu not)}1/2

4) {(pie)Li/(nu not)}1/2

The inductance of  a solenoid is given by,
L=μ0N2Al
And the length of the wire needed to make it,
L1=2πNr
and A=πr2
So,
L=μ04π4π2N2r2lL=μ04πL12lL1=4πμ0Ll12
2) is the right answer.

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