Question

A coil of moving coil galvanometer twists through ${90}^o$ when a current of one microampere is passed through it.If the area of the coil is ${10}^{-4}{m}^2$ and it has 100 turns, calculate the magnetic field of the magnet of the galvanometer.Given $k={10}^{-8}N-m/degree$.

Answer 1

$i=(\frac{k}{NBA})\varphi$

$\therefore B=\frac{k\varphi }{NiA}=\frac{\left({10}^{-8}\right)\left(90\right)}{100\times {10}^{-6}\times {10}^{-4}}$$=90J$