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

n = 100, I = 0.40 A, r = 8.0 cm = 8.0 x 10^-2 m
B = \(\frac{\mu_{0} \mathrm{nI}}{2 \mathrm{r}}\)
= \(\frac{4 \pi \times 10^{-7} \times 100 \times 0.4}{2 \times 8 \times 10^{-2}}\)
= π x 10^-4 T