Question

In a tangent galvanometer, a current of $0.1A$ produces a deflection of ${30}^{\circ }$. The current (in ampere) required to produce a deflection of ${60}^{\circ }$ is

Answer 1

Tangent galvanometer

for a tangent galvanometer, if deflection $=\theta$, the current reading is given by $i=k\tan \theta$

where , $k=\frac{2r{B}_H}{{\mu }_{0}n}=$ reduction factor of

galvanometer $=$ constant

$(n=\text{no. of turns in the coil,}r=\text{radius of the coil})$

$i\propto \tan \theta$

$\therefore \frac{i_1}{i_2}=\frac{\tan {\theta }_1}{tan{\theta }_2}$

$\Rightarrow \frac{0.1A}{i_2}=\frac{\tan {30}^{\circ }}{\tan {60}^{\circ }}$

$\Rightarrow i_2=0.1\times \sqrt{3}\times \sqrt{3}$

$=0.3A$