Question

A solenoid of diameter 0.05 m and 500 turns /cm has length 1 m. current 3 A passes through it Calculate magnetic flux

Answer 1

The diameter of solenoid $\left(D\right)=0.05m$

Radius $\left(R\right)=\frac{0.05}{2}=0.025m$

Length of solenoid $L=1m$

$\frac{Numberofturns}{Length}=\frac{N}{L}=\frac{500}{cm}=\frac{50000}{meter}$

Current flowing $\left(I\right)=3A$

$\therefore$ The magnetic field produced by the solenoid $\left(B\right)={\mu }_{0}nl$

$=4n\times {10}^{-7}\times 50000\times 3$

$=4\times 3.14\times 15\times {10}^{-3}$

$=0.188T$

$\therefore$ Magnetic flux is passing through the solenoid $\left(\varphi \right)=NBA=NB\pi r^2$

$=50000\times 0.188\times 3.14\times 0.025\times 0.025$

$=18.49Wb$