Question

The magnetic field $B$ inside a long solenoid, carrying a current of $5.00A$ , is $3.14\times {10}^{-2}T$ . Find the number of turns per unit length of the solenoid.

Answer 1

We know the magnetic field in long solenoid given as,

$B={\mu }_0\frac{N}{l}i$

By substituting the given values in above equation we get

$3.14\times {10}^{-2}T=1.25\times {10}^{-6}\frac{Tm}{A}\cdot \frac{N}{l}5A$

$\frac{N}{l}=\frac{3.14\times {10}^{-2}}{1.25\times 5\times {10}^{-6}m}$

Hence the number of turns per unit length is$5024$