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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.

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Solution

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,
BHtan θ=μ0 In2rn=BHtan θ×2rμ0In=3.6×10-5×2×1×10-14π×10-7×10-2n=0.57332×103=573

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