Question

A magnetic flux of $8\times {10}^{-4}\text{Wb}$ is linked with each turn of a $200$ turns coil when there is an electric current of $4\text{A}$ in it. Calculate the self inductance of the coil.

• A. $10\text{mH}$
• B. $20\text{mH}$
• C. $40\text{mH}$
• D. $50\text{mH}$

Answer 1

The correct option is C $40\text{mH}$

Given, $\varphi =8\times {10}^{-4}\text{Wb};N=200;i=4\text{A}$

We know that for solenoid, net flux linked with the coil is,

$N\varphi \propto i$

$N\varphi =Li$

$\Rightarrow 200\times 8\times {10}^{-4}=L\times 4$

$\Rightarrow L=\frac{16\times {10}^{-2}}{4}$

$\therefore L=4\times {10}^{-2}\text{H}=40\text{mH}$

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