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 B_H = 3.6 × 10⁻⁵ T and radius of the coil is 10 cm, find the number of turns in the coil.

Answer 1

Given:
Horizontal component of Earth's magnetic field, B_H = 3.6 × 10⁻⁵ T
Deflection shown by the tangent galvanometer, θ = 45°
Current through the galvanometer, I = 10 mA = 10⁻² A
Radius of the coil, r = 10 cm = 0.1 m
Number of turns in the coil, n = ?
We know,
$$\begin{gathered} B_{\mathrm{H}}\tan \theta =\frac{{\mathrm{\mu }}_0\mathit{}In}{\mathit{2}r} \\ \Rightarrow n=\frac{B_{\mathrm{H}}\tan \theta \times 2r}{{\mathrm{\mu }}_0\mathrm{I}} \\ \Rightarrow n=\frac{3.6\times {10}^{-5}\times 2\times 1\times {10}^{-1}}{4\mathrm{\pi }\times {10}^{-7}\times {10}^{-2}} \\ \Rightarrow n=0.57332\times {10}^3=573 \end{gathered}$$