Inductance

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A wire of fixed length is wound on a solenoid of length $l$ and radius r. Its self inductance is found to be L. Now if same wire is wound on a solenoid of length $\frac{l}{2}$ and radius $\frac{r}{2}$ then its self-inductance will be

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Solution

$\begin{array}{l} L = μ_{0} n^{2} A l = μ_{0} . \frac{N^{2}}{l^{2}} π r^{2} l = \frac{μ_{0} {(N r)}^{2} . π}{l} \\ N o w l' = L e n g t h o f w i r e = 2 π r N \\ ∴ N r = \frac{l^{'}}{2 π} = C = c o n s \tan t \Rightarrow L = \frac{μ_{0} C^{2} . π}{l} \\ ∴ L \propto \frac{1}{l} \\ ∴ \frac{L^{1}}{L} = \frac{l}{l / 2} = 2 \Rightarrow L^{'} = 2 L \end{array}$

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Inductance

Remember concepts with our Masterclasses.

A wire of fixed length is wound on a solenoid of length l and radius r. Its self inductance is found to be L. Now if same wire is wound on a solenoid of length l2 and radius r2 then its self-inductance will be

L=μ0n2 Al=μ0.N2l2 πr2l=μ0Nr2. πlNow l' = Length of wire = 2πrN∴ Nr=l'2π =C=constant ⇒ L=μ0C2.πl∴ L ∝ 1l ∴L1L=ll/2=2 ⇒ L'=2L

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A wire of fixed length is wound on a solenoid of length $l$ and radius r. Its self inductance is found to be L. Now if same wire is wound on a solenoid of length $\frac{l}{2}$ and radius $\frac{r}{2}$ then its self-inductance will be