Question

Calculate the magnetic field due to a circular coil of 500 turns and of diameter 0.1 m, carrying a current of 7 A, at a point on the axis of the coil at a distance of 0.12 m.

Answer 1

Given,

$n=500turn$

$r=\frac{d}{2}=\frac{0.1}{2}=0.05m$

$I=7A$

$a=0.12m$

Magnetic field, $B=\frac{{\mu }_{0}nI{r}^2}{2(r^2+a^2{)}^{3/2}}$

$B=\frac{4\pi \times {10}^{-7}\times 500\times 7\times 0.05\times 0.05}{2({0.05}^2+{0.12}^2{)}^{3/2}}$

$B=2.5\times {10}^{-3}T=2.5mT$