A tangent galvanometer shown 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 xx 10^(-5) T` and radius of the coil is `10cm`, find the number of turns in the coil.

Explanation

To find the number of turns in the coil of the tangent galvanometer, we can use the formula that relates the magnetic field created by the coil to the horizontal component of the Earth's magnetic field. The formula is given by:

\[ B_H \tan \theta = \frac{\mu_0 n I}{2R} \]

Where:
- \( B_H \) is the horizontal component of the Earth's magnetic field.
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