The self induction of a solenoid 0.1 m in
diameter, 2.0 m long and of 1200 turns is
(1) 7.11 mH
(2) 8.11 mH
(3) 6.11 mH
(4) 5.11 mH
Answers
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Answer:1) 7.11 mH
Explanation: Self inductance,
L is given by,
L= mew*N^2A/l
Therefore
L=4pie*10^-7*(1200)^2*pie*(0.05)^2/2
Therefore L=7.11 mH
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