Question

$A$ $200$-turn solenoid having a length of $25\, cm$ and a diameter of $10 \,cm$ carries a current of $0.30 \,A$. Calculate the magnitude of the magnetic field $\vec{B}$ (in mT) inside the solenoid

Answer 1

The magnetic field at the centre of the solenoid is given by
$B=\frac{\mu_{0} ni}{2} \left(cos\,\theta_{1}-cos\,\theta_{2}\right) $
Here $cos\,\theta_{1}=0.93, cos\,\theta_{2}=cos\left(180^{\circ}-\theta_{1}\right)=-cos\,\theta_{1}$

$\therefore B=\frac{4\pi\times10^{-7}\times\left(\frac{200}{0.25}\right)\times0.3\times\left(2\times0.93\right)}{2}$
$=0.3\,mT$