A long solenoid has 1500 turns. When a current of $3.5 \mathrm{~A}$ flows through it, the magnetic flux…
A long solenoid has 1500 turns. When a current of $3.5 \mathrm{~A}$ flows through it, the magnetic flux linked with each turn of solenoid is $2.8 \times 10^{-3}$ weber. The self-inductance of solenoid is
$1.2 \mathrm{H}$
$2.4 \mathrm{H}$
$3.6 \mathrm{H}$
$6 \mathrm{H}$
Solution
Flux linked with each turn of the solenoid is, $\phi=2.8 \times 10^{-3} \mathrm{~Wb}$
$\therefore \quad$ Total magnetic flux of the solenoid is,
$\begin{aligned}
& \phi_{\mathrm{net}}=\mathrm{N} \phi \\
& \phi_{\mathrm{net}}=1500 \times 2.8 \times 10^{-3}=4.2 \mathrm{~Wb}
\end{aligned}$
Flux can also be given by, $\phi=\mathrm{LI}$
$\therefore \quad$ The self-inductance of the solenoid is,
$\begin{aligned}
& \mathrm{L}=\frac{\phi}{\mathrm{I}}=\frac{4.2}{3.5} \\
& \mathrm{~L}=1.2 \mathrm{H}
\end{aligned}$