Question

For a solenoid keeping the turn density constant its length makes halved and its cross section radius is doubled then the inductance of the solenoid increased by

Answer 1

New inductance, $L'=8L$

Explanation:

The inductance of a solenoid is given by :

$L=\dfrac{\mu_oN^2A}{l}$

New length, l' = l/2

Radius, r' = 2r

New inductance of a solenoid is given by :

$L'=\dfrac{\mu_oN^2\pi r'^2}{l'}$

$L'=\dfrac{\mu_oN^2\pi r(2r)^2}{l/2}$

$L'=8\dfrac{\mu_oN^2A}{l}$

$L'=8L$

So, the inductance of a solenoid becomes 8 times of the initial value of inductance. Hence, this is the required solution.

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Topic : Inductance of the solenoid

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