Question

A long solenoid has $1000turns$. When a current of $4A$ flows through it, the magnetic flux linked with each turn of the solenoid is $4\times {10}^{-3}Wb$. What is the self-inductance of the solenoid?

Answer 1

Step 1: GIven data

A long solenoid has $\left(N\right)=1000turns$

Current $I=4A$

Magnetic flux ${\varphi }_{\circ }=4\times {10}^{-3}Wb$

Step 2: Formula used

$\varphi =LI$

This formula belongs to the self-inductance of the solenoid.

Here, $\varphi$=magnetic flux

$L$= self-inductance

$I$=current

Step 3: Calculating self-inductance

So, we have flux linked with each turn

${\varphi }_{\circ }=4\times {10}^{-3}Wb$

Total flux linked =$1000\times 4\times {10}^{-3}$

Now, on the basis of the formula

$$\begin{gathered} \varphi =LI \\ L=\frac{\varphi }{I} \end{gathered}$$

Now on putting the values

$L=\frac{4\times {10}^{-3}}{4}$

So, $L=1H$

Hence, the self-inductance of the solenoid is $L=1H$.