Physics - Magnetism Question with Solution | TestHub

The magnetic field along the axis of a circular coil of $n$ turns is given by

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

Given $x=2\sqrt{2}r$, we have

$x^2=(2\sqrt{2}r{)}^2=8r^2$

Substituting this into the formula, we get

$B=\frac{{\mu }_{0}nI{r}^2}{2(r^2+8r^2{)}^{3/2}}$

$B=\frac{{\mu }_{0}nI{r}^2}{2(9r^2{)}^{3/2}}$

$B=\frac{{\mu }_{0}nI{r}^2}{2(27r^3)}$

$B=\frac{{\mu }_{0}nI}{54r}$

Answer: $\frac{{\mu }_0\mathrm{nI}}{54\mathrm{r}}$