A circular coil of wire consisting of 100 turns each of radius 9 cm carries a current of 0.4 A. The magnitude of the magnetic field at the centre of coil is [ μ 0 = 12.56 × 10 − 7 SI Units ]

Step-by-step solution

Given, number of turns, n = 100, radius of coil, r = 9 c m = 9 × 10 - 2 m and current flow in the coil, I =0.4 A The magnitude of the magnetic field at the centre of a coil of n turns is given by B = μ 0 n l 2 r …. (i) Substituting given values in Eq. (i), we get ⇒ B = 1256 × 10 - 7 × 100 × 0.4 2 × 10 - 1 × 9 × 10 - 2 = 2.79 × 10 - 4 T