First slide

Inductance

Question

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

Moderate

A

120 mH
B

100 mH
C

140 mH
D

160 mH

Solution

$L = μ_{0} n^{2} \times l A = μ_{0} {(\frac{N}{l})}^{2} \times l A = \frac{μ_{0} N^{2} A}{l}$
$∴ \frac{L^{1}}{L} = {(\frac{N^{1}}{N})}^{2} \times (\frac{l}{l_{1}}) = {(2)}^{2} \times (2) = 8 \Rightarrow L^{1} = 120 m H$

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