Question

A circular current carrying coil has a radius r. Find the distance from the centre of coil, on its axis, where the magnetic induction will be 1/8 th of its value at the centre of coil.

Answer 1

$B_A=\frac{{\mu }_{0}I}{2}\times \frac{r^2}{{(x^2+r^2)}^{\frac{3}{2}}}{B}_0=\frac{{\mu }_{0}I}{2}\times \frac{r^2}{r^3}=\frac{{\mu }_{0}I}{2}\times \frac{1}{r}\therefore 8\times B_A=B_{0}8\times \frac{{\mu }_{0}I}{2}\times \frac{r^2}{{(x^2+r^2)}^{\frac{3}{2}}}=\frac{{\mu }_{0}I}{2}\times \frac{1}{r}8r^3={(x^2+r^2)}^{\frac{3}{2}}64r^6={(x^2+r^2)}^{3}or{x}^2+r^2=4\times r^{2}3r^2=x^{2}x=r\sqrt{3}$