Question

A tangent galvanometer has a coil of $50$ turns and a radius of $20$ cm. The horizontal component of the earth’s magnetic field is $B_H=3\times {10}^{-5}T$. Find the current (in ampere) which gives a deflection of ${45}^{\circ }$. (Take $\pi =3$)

Answer 1

Tangent galvanometer

For a tangent galvanometer , if deflection $=\theta$ , the current reading is given by

$i=k\tan \theta$

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

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

$i=\left(\frac{2r{B}_H}{{\mu }_{\circ }N}\right)\tan \theta$

$\therefore i=\frac{\left(2\right)\left(0.2m\right)(3\times {10}^{-5}T)}{4\pi \times {10}^{-7}\times 50}\tan {45}^{\circ }$

$=0.2A$