When a current of 10 mA is passed through a tangent galvanometer, it exhibits a deflection of 45^ . Determine the number of turns in the coil, given that the horizontal component of the earth's magnetic field is B_H = 3.6 10⁻⁵ T and the coil's radius is 10 cm .

Step-by-step solution

The formula for a tangent galvanometer is given by: B = B_H The magnetic field at the center of the coil is: B = ₀ N I 2 R Equating the two expressions for B : ₀ N I 2 R = B_H Rearranging to solve for the number of turns N : N = 2 R B_H ₀ I Substituting the given values R = 0.1 m , B_H = 3.6 10⁻⁵ T , = 45^ , ₀ = 4 10⁻⁷ T m/A , and I = 10 10⁻³ A : N = 2 0.1 3.6 10⁻⁵ 45^ 4 10⁻⁷ 10 10⁻³ N = 0.72 10⁻⁵ 4 10⁻⁹ N = 7200 4 = 1800 Using 3.14 : N 1800 3.14 573 The closest integer value among the given options is 570 .