Question

What is the self inductance of a solenoid of length $31.4$cm, area of cross-section ${10}^{-3}{m}^2$ and total number of turns ${10}^3$?

• A. $4$mH
• B. $4$H
• C. $40$H
• D. $0.4$H

Answer 1

The correct option is A $4$mH

Given,

Area of cross-section of coil $A={10}^{-3}{m}^2$
The length of the solenoid $l=31.4cm=31.4\times {10}^{-2}m$
and the number of turns in the coil $n={10}^3$

Magnetic flux linked with the coil

$\varphi =Li$
$BA=Li$
${\mu }_{0}niA=Li$

$L={\mu }_{0}N/lA$

$L=\frac{4\pi \times {10}^{-7}\times {10}^3\times {10}^{-3}}{31.4\times {10}^{-2}}$

$=4mH$