Question

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

• A. 2 L
• B. L
• C. 4 L
• D. 8 L

Answer 1

Answer: (A)

$L=\frac{{\mu }_0{N}^2\pi r^2}{l}$
Length of wire $=N2\pi r=$ Constant $(=C$, suppose $)$
$\therefore L={\mu }_0{\left(\frac{C}{2\pi r}\right)}^2\frac{\pi r^2}{l}$
$\therefore L\propto \frac{1}{l}$
$\therefore$ Self-inductance will become $2L$.