Question

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

Answer 1

We know that

$B={\mu }_{0}ni$

$3.14\times {10}^{-2}=4\pi \times {10}^{-7}\times n\times 5$

$\Rightarrow n=\frac{{10}^{-2}}{20\times {10}^{-7}}$

$=5\times {10}^8=5000turns/m$