Question

A galvanometer coil 5 cm$\times$ 2 cm with 200 turns is suspended vertically in a field of $5\times {10}^{-2}T$.The suspension fibre needs a torque of $0.125\times {10}^{-7}N-m$ to twist it through one radian. If $i$ is the strength of the current required to be maintained in the coil when we require a deflection of $6^o$ then find $x$ such that $x=i\times {10}^8$.

Answer 1

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

$=\frac{(0.125\times {10}^{-7})\left(6\right)\left(\pi /180\right)}{200\times 5\times {10}^{-2}\times 5\times 2\times {10}^{-}4}$

$=1.3\times {10}^{-7}A$