**Question:**

The magnetic intensity $\mathrm{H}$ at the centre of a long solenoid carrying a current of $2.0 \mathrm{~A}$, is found to be $1500 \mathrm{~A} \mathrm{~m}^{-1}$. Find the number of turns per centimetre of the solenoid.

**Solution:**

Here given in question, Current in the solenoid, $I=2$ A Magnetic intensity at the centre of long solenoid, $H=1500 \mathrm{Am}^{-1}$ Magnetic field produced by a solenoid(B) is given by $B=\mu_{0} n i . . .(T)$ where, $n=$ number of turns per unit length $i=$ electric current through the solenoid Also, the relation between magnetic field strength(B) and magnetic intensity( $\mathrm{H})$ is given by $H=$ $\frac{B}{u 0}$

..(2) From equations (1) and (2), we get: $H=n i \Rightarrow 1500 \mathrm{~A} / \mathrm{m}=n \times 2 \Rightarrow n=7.5 \mathrm{turns} / \mathrm{cm}$ The number of turns per centimetre of the solenoid is $7.5$.