First slide

Inductance

Question

A coil of wire of radius r has 600 turns and a self inductance of 108 mH. The self inductance of a similar coil of 500 turns will be

Easy

A

75 mH
B

108 mH
C

90 mH
D

108 x (5/6)³ mH

Solution

The inductances of two coils ate given by
$\begin{array}{l} L_{1} = \frac{μ_{0} N_{1}^{2} πr}{2} and L_{2} = \frac{μ_{0} N_{2}^{2} πr}{2} \\ Now \frac{L_{2}}{L_{1}} = {(\frac{N_{2}}{N_{1}})}^{2} or L_{2} = L_{1} {(\frac{N_{2}}{N_{1}})}^{2} \\ ∴ L_{2} = 108 \times {(\frac{500}{600})}^{2} = 75 mH . \end{array}$

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