NEET-XII-Physics
36: Permanent Magnets
- #16A 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.Ans : Given:
Horizontal component of Earth's magnetic field, BH = 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,
`` {B}_{\,\mathrm{\,H\,}}\,\mathrm{\,tan\,}\theta =\frac{{\,\mathrm{\,\mu \,}}_{0}In}{\mathit{2}r}``
`` \Rightarrow n=\frac{{B}_{\,\mathrm{\,H\,}}\,\mathrm{\,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``
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