Question

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

Answer 1

Answer: 0.3

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 $cos{\theta }_1=0.93,cos{\theta }_2=cos\left(180^{\circ }-{\theta }_1\right)=-cos{\theta }_1$

$\therefore B=\frac{4\pi \times 10^{-7}\times \left(\frac{200}{0.25}\right)\times 0.3\times \left(2\times 0.93\right)}{2}$
$=0.3mT$