Question

A tangent galvanometer shows a deflection of 45° when 10 mA of current is passed through it. If the horizontal component of the earth's magnetic field is BH = 3.6 × 10−5 T and radius of the coil is 10 cm, find the number of turns in the coil.

Answer 1

Number of turns inside the coil is 573 turns

Explanation:

Given data in the question  

Horizontal part of a magnetic field on Earth, $B_{H}=3.6 \times 10-5 T$

The tangent galvanometer shows deflections, = 45°

Current via galvanometer, I = 10 mA =$10^{-2} A$

Coil Radius , r = 10 cm = 0.1 m

Number of turns inside the coil, n = ?

We know that

$B_{H} \tan \theta=\frac{\mu_{0} I n}{2 r}$

           $h=\frac{B_{H} \tan \theta \times 2 r}{\mu_{0} I}$

          $n=\frac{\left(3.6 \times 10^{-5} \times 2 \times 1 \times 10^{-1}\right)}{\left(4 \pi \times 10^{-7} \times 10^{-2}\right)}$

         $n=0.57332 \times 10^{3}$

             = 573 turns