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Q.

Induction of a solenoid is 15 mH. Now length of the solenoid is halved, number of turns doubled and area of cross section remains the same. Then the new inductance of the solenoid will be

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a

120 mH

b

100 mH

c

140 mH

d

160 mH

answer is B.

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Detailed Solution

L=μ0n2×lA=μ0(Nl)2×lA=μ0N2Al∴ L1L=(N1N)2×(ll1)=(2)2×(2)=8⇒L1=120 mH
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