A circular coil of radius $R$ carries a current $i$ The magnetic field at its centre is $B$. At what

Question

A circular coil of radius $R$ carries a current $i$ The magnetic field at its centre is $B$. At what distance from the centre, on the axis of the coil the magnetic field will be $\frac{B}{8}$\

Professor PC Thomas Classes · Answer

At the center $B=\frac{\mu_{0} n I}{2 r}$ At a distance $x$ from the center $B^{\prime}=\frac{B}{8}$ $$ \begin{equation*} \frac{\mathrm{B}}{8}=\frac{\mu_{0} \mathrm{nIr}^{2}}{2\left(\mathrm{r}^{2}+\mathrm{x}^{2}ight)^{3 / 2}} 	ag{2} \end{equation*} $$ $\frac{(1)}{(2)} 8=\frac{\mu_{0} n r^{2}}{r^{3}}$\ $8 r^{3}=\left(r^{2}+x^{2}ight)^{3 / 2}$\ Taking cube root $2 r=\left(r^{2}+x^{2}ight)^{1 / 2}$\ Square and get $x$\