A circular current carrying coil has radius $R$. The magnetic induction at the centre of the coil is…
A circular current carrying coil has radius $R$. The magnetic induction at the centre of the coil is $\mathrm{B}_{\mathrm{C}}$. The magnetic induction of the coil at a distance $\sqrt{3} R$ from the centre along the axis is $\mathrm{B}_{\mathrm{A}}$. The ratio $\mathrm{B}_{\mathrm{A}}: \mathrm{B}_{\mathrm{C}}$ is
$1: 3$
$1: 8$
$8: 1$
$27: 1$
Solution
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 $r=\sqrt{3} R$
$\therefore \quad B_A=\frac{\mu_0 I R^2}{2\left(R^2+(\sqrt{3} R)^2\right)^{3 / 2}}$
$\therefore \quad$ Ratio of $\mathrm{B}_{\mathrm{A}}$ to $\mathrm{B}_{\mathrm{C}}$ is
$\frac{\mathrm{B}_{\mathrm{A}}}{\mathrm{B}_{\mathrm{C}}}=\frac{\frac{\mu_0 \mathrm{IR}^2}{2\left(\mathrm{R}^2+(\sqrt{3} \mathrm{R})^2\right)^{3 / 2}}}{\frac{\mu_0 \mathrm{I}}{2 \mathrm{R}}}=\frac{\mu_0 \mathrm{IR}^2}{2\left(4 \mathrm{R}^2\right)^{3 / 2}} \frac{2 \mathrm{R}}{\mu_0 \mathrm{I}}$
$\frac{\mathrm{B}_{\mathrm{A}}}{\mathrm{B}_{\mathrm{C}}}=\frac{1}{8}$
.