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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 the self inductance will be :
Option: 1
$2L$

Option: 2
$L$

Option: 3
$4L$

Option: 4
$8L$

Answers (1)

best_answer

Co efficient of self induction -

$\phi \, \alpha \, I\Rightarrow N\phi \, \alpha\: I$

$N\phi \,=L\, I$

$L=\frac{N\phi }{I}\,$

- wherein

$N\phi =$ Number of flux linkage with coil.

$L=\frac{\mu _{0}N^{2}\pi r^{2}}{l }$

Lenght of wire $=N2\pi r=Constant (=C, suppose )$

$\therefore L=\mu _{0}\left ( \frac{C}{2\pi r} \right )^{2}\frac{\pi r^{2}}{l }$

$\therefore L\propto \frac{1}{l }$

$\therefore$ inductance wil be become $2L$

Posted by

avinash.dongre

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