1. Tardigrade
  2. Question
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  4. A wire of fixed length is wound in such a way that it forms a solenoid of length ' l ' and radius ' r '. Its self-inductance is found to be L. Now if same wire is wound in such a way that it forms a solenoid of length l / 2 and radius r / 2, then the self-inductance will be

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Q. A wire of fixed length is wound in such a way that it forms a solenoid of length ' $l$ ' and radius ' $r$ '. Its self-inductance is found to be $L$. Now if same wire is wound in such a way that it forms a solenoid of length $l / 2$ and radius $r / 2$, then the self-inductance will be

Electromagnetic Induction

Solution:

$L=\frac{\mu_{0} N^{2} \pi r^{2}}{l}$
Length of wire $=N 2 \pi r=$ 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 $ Self-inductance will become $2 L$.