Question

Calculate the inductance of an air core solenoid containing $300$ turns if the length of the solenoid is $25.0\text{cm}$ and its cross-sectional area is $4.00{\text{cm}}^2$.

• A. $1.80\times {10}^{-4}\text{H}$
• B. $1.81\times {10}^{-2}\text{H}$
• C. $3.62\times {10}^{-4}\text{H}$
• D. None of the above

Answer 1

The correct option is A $1.80\times {10}^{-4}\text{H}$

Given:
$N=300;l=25\text{cm};A=4{\text{cm}}^2$

The inductance of a solenoid is,

$L=\frac{N\varphi }{i}=\frac{N\left(BA\right)}{i}$

$\Rightarrow L=\frac{{\mu }_0{N}^{2}A}{l}$

Substituting the values, we have

$L=\frac{(4\pi \times {10}^{-7})\left(300{)}^2\right(4.00\times {10}^{-4})}{(25.0\times {10}^{-2})}$

$\therefore L=1.80\times {10}^{-4}\text{H}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{A}\right)$ is the correct answer.