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Q.

A wire of fixed length is wound into a solenoid of length 'l' and radius 'r'. Its self inductance is found to be L. Now if same wire is wound into a solenoid of length $\frac{l}{2}$ and radius $\frac{r}{2}$ . Then, the self inductance will be:

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a

L

b

2 L

c

8 L

d

4 L

answer is A.

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Detailed Solution

$L = μ_{0} n^{2} lA = \frac{μ_{0} N^{2} A}{l} = \frac{μ_{0} N^{2} {πr}^{2}}{l}$
Since, the same length of wire is wound over solenoid of length $\frac{l}{2}$ and radius $\frac{r}{2}$ therefore, number of turns $(l_{wire} = [2 πr] N)$ will becomes 2 times larger.
$\begin{array}{l} ∴ \frac{L_{2}}{L_{1}} = {(\frac{N_{2}}{N_{1}})}^{2} \times {(\frac{r_{2}}{r_{1}})}^{2} \times (\frac{l_{1}}{l_{2}}) = (2)^{2} \times {(\frac{1}{2})}^{2} \times (2) = 2 \\ ∴ L_{2} = 2 L_{1} = 2 L (∴ L_{1} = L) \end{array}$

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