A circular coil of wire consisting of ' n ' turns each of radius 8 cm carries a current of 0.4 A . The magnitude of the magnetic field at the centre of coil is 3.14 10⁻⁴ ~T . The value of ' n ' is [Take ₀=12 56 10⁻⁷ SI unit]

Step-by-step solution

The magnetic field at the center of a circular coil with n turns, each of radius R carrying current I , is B = ₀ n I 2R . Given R = 0.08 m , I = 0.4 A , B = 3.14 10⁻⁴ T , and ₀ = 12.56 10⁻⁷ T m/A , solve for n : n = 2 B R ₀ I = 2 3.14 10⁻⁴ 0.08 12.56 10⁻⁷ 0.4 . Simplify numerator and denominator: 2 3.14 10⁻⁴ 0.08 = 5.024 10⁻⁵ 12.56 10⁻⁷ 0.4 = 5.024 10⁻⁷ Then n = 5.024 10⁻⁵ 5.024 10⁻⁷ = 10^2 = 100 . Final answer: C