Question

The radius of the coil of a tangent galvanometer, which has $10$ turns, is $0.1m.$ The current (in ampere) required to produce a deflection of ${60}^{\circ }$ is $(B_H=4\times {10}^{-5}T)$

Answer 1

Tangent law

in tangent galvanometer from the tangent law

$B=B_H\tan \theta$

Where, $\theta$ is angle of deflection, $B=$ magnetic field of coil (at coil’s center ) ,

$B_H=$ horizontal magnetic field of earth

$\Rightarrow \frac{{\mu }_{\circ }ni}{2r}=B_H\tan \theta$

Where, $\theta$ is angle of deflection, $B=$ magnetic field of coil (at coil’s center ) , $B_H=$ horizontal magnetic field of earth

$\Rightarrow \frac{{\mu }_{\circ }ni}{2r}=B_H\tan \theta$

Where, $n$ is number of turns in coil , $i=$ current

$\Rightarrow i=\frac{2r{B}_H\tan \theta }{{\mu }_{\circ }n}=\frac{2\times 0.1m\times 4\times {10}^{-5}T}{10\times 4\pi \times {10}^{-7}}$

$=1.1A$