Question

A circular coil has 100 turns each
of radius 8.0 cm and bears a
current of 0.40A. The magnitude
of magnetic induction at the centre
(A) 3.1 x 10-41
(B) 3.14 x 10-4T
(C) 3.1415 x 10-4 T
(D) 3.14159 10-4T​

Answer 1

Answer:

Number of turns on the circular coil, n = 100

Radius of each turn, r = 8.0 cm = 0.08 m

Current flowing in the coil, I = 0.4 A

Magnitude of the magnetic field at the centre of the coil is given by the relation,

| B | = μ0 2πnI / 4π r

Where,

μ0 = Permeability of free space

= 4π × 10–7 T m A–1

| B | = 4π x 10-7 x 2π x 100 x 0.4 / 4π x 0.08

= 3.14 x 10-4 T

Hence, the magnitude of the magnetic field is 3.14 × 10–4 T