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A tangent galvanometer has a coil of 50 turns and a radius of 20 cm. The horizontal component of earth's magnetic field is B_H = 3 $\times$ 10^-5 T. What will be the current which gives a deflection of 45⁰?

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0.19 A

0.29 A

0.09 A

0.39 A

answer is C.

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Detailed Solution

Given, $B_{H} = 3 \times 10^{- 5} T, θ = 45 °, n = 50$

and $r = 20 cm = 20 \times 10^{- 2} m$

As, current, $I = \frac{2 r B_{H}}{μ_{0} n} \tan θ$

$= \frac{2 \times 20 \times 10^{- 2} \times 3 \times 10^{- 5} \times \tan 45 °}{4 π \times 10^{- 7} \times 50}$

$= \frac{120 \times 10^{- 7}}{4 π \times 10^{- 7} \times 50} = \frac{3}{5 π} = \frac{3}{5 \times 3.14} = \frac{3}{15.7} = 0.19$ A

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