A circular coil of wire consisting of 100 turns, each of radius 8.0 cm carries a current of 0.40 A.
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 at the centre of the coil?

Solution:

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,

$|\mathbf{B}|=\frac{\mu_{0}}{4 \pi} \frac{2 \pi n I}{r}$

Where,

$\mu_{0}=$ Permeability of free space

$=4 \pi \times 10^{-7} \mathrm{Tm} \mathrm{A}^{-1}$

$|\mathbf{B}|=\frac{4 \pi \times 10^{-7}}{4 \pi} \times \frac{2 \pi \times 100 \times 0.4}{0.08}$

=3.14 \times 10^{-4} \mathrm{~T}

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