A solenoid has the length $1 \mathrm{~m}$ and the area of cross-section $0.02 \mathrm{~m}^2$. If number of…
A solenoid has the length $1 \mathrm{~m}$ and the area of cross-section $0.02 \mathrm{~m}^2$. If number of turns in the solenoid is $\mathbf{5 0 0 0}$ then the self inductance of the solenoid is
$0.2 \pi$ henry
$0.4 \pi$ henry
$0.02 \pi$ henry
$0.04 \pi$ henry
Solution
For a solenoid, coefficient of self induction is given by
$L=\frac{\mu_0 N^2 A}{l}$
Here, Number of turns, $N=5000$, length., $l=1 \mathrm{~m}$,
Area, $A=0.02 \mathrm{~m}^2$
Also $\mu_0=4 \pi \times 10^{-7} \mathrm{Tm} / \mathrm{A}$
So, self inductance of solenoid is
$L=\frac{4 \pi \times 10^{-7} \times(5000)^2 \times 0.02}{1}=0.2 \pi \mathrm{H}$