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?
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.