Question

A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A. What is the magnitude of the magnetic field B at the centre of the coil?

Answer 1

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,
                                       \(|B|= \frac{μ_0}{4π}.\frac{2πnl}{r}\)
Where,
\(μ_0\)= Permeability of free space
    \(= 4π × 10^{–7} T m A^{–1}\)
So, 
                 \(|B| =\frac{4π × 10^{–7}}{4π} × \frac{2π × 100 × 0.4}{r}\)
                       \(= 3.14 × 10^{-4}\,T\)
Hence, the magnitude of the magnetic field is\(= 3.14 × 10^{-4}\,T.\)