Physics - Magnetism Question with Solution | TestHub

Magnetic Induction of Current carrying coil at its centre: ${\mathrm{B}}_{\mathrm{c}}=\frac{{\mu }_{\mathrm{o}}\mathrm{I}}{2\mathrm{R}}$ Magnetic Induction of Current carrying coil at distance $\mathrm{r}$ : ${\mathrm{B}}_{\mathrm{A}}=\frac{{\mu }_0{\mathrm{IR}}^2}{2{\left({\mathrm{R}}^2+{\mathrm{r}}^2\right)}^{3/2}}$ Given $r=\sqrt{3}R$ $\therefore B_A=\frac{{\mu }_0{I}^2}{2{\left(R^2+(\sqrt{3}R{)}^2\right)}^{3/2}}$ $\therefore$ Ratio of ${\mathrm{B}}_{\mathrm{A}}$ to ${\mathrm{B}}_{\mathrm{C}}$ is $\frac{{\mathrm{B}}_{\mathrm{A}}}{{\mathrm{B}}_{\mathrm{C}}}=\frac{\frac{{\mu }_0{\mathrm{IR}}^2}{2{\left({\mathrm{R}}^2+(\sqrt{3}\mathrm{R}{)}^2\right)}^{3/2}}}{\frac{{\mu }_0\mathrm{I}}{2\mathrm{R}}}=\frac{{\mu }_0{\mathrm{IR}}^2}{2{\left(4{\mathrm{R}}^2\right)}^{3/2}}\frac{2\mathrm{R}}{{\mu }_0\mathrm{I}}$ $\frac{{\mathrm{B}}_{\mathrm{A}}}{{\mathrm{B}}_{\mathrm{C}}}=\frac{1}{8}$