Question

Calculate the magnetic field of induction, due to circular coil of $400$ turns and radius $0.05m$ carrying a current of $5A$, at a point on the axis of the coil at a distance $0.1m$.

Answer 1

Given,

$n=400$ turns

$I=5A$

$r=0.05m$

$x=0.1m$

The magnetic field at point x on the axis of the circular coil if given by

$B=\frac{{\mu }_{0}nI{r}^2}{2(r^2+x^2{)}^{2/3}}$

$B=\frac{4\pi \times {10}^{-7}400\times 5\times (0.05{)}^2}{2\left[\right(0.05{)}^2+(0.1{)}^2{]}^{3/2}}$

$B=2.24mT$