A circular current carrying coil has radius $\mathrm{R}$. At what distance from the centre of the coil on…
A circular current carrying coil has radius $\mathrm{R}$. At what distance from the centre of the coil on the axis, the magnetic induction will become $\frac{1}{8}$ th of its value at the centre of the coil?
$\frac{2 \mathrm{R}}{\sqrt{3}}$
$\mathrm{R} \sqrt{3}$
$\frac{\mathrm{R}}{2 \sqrt{3}}$
$\frac{\mathrm{R}}{\sqrt{3}}$
Solution
By using \(\frac{B_{\text {centre }}}{B_{\text {axis }}}=\left(1+\frac{x^2}{r^2}\right)^{3 / 2}\), given \(r=R\) and \(B_{\text {axis }}=\frac{1}{8} B_{\text {centre }}\)
\(\begin{aligned}
& \Rightarrow 8=\left(1+\frac{x^2}{R^2}\right)^{3 / 2} \\
& \Rightarrow(2)^2=\left\{\left(1+\frac{x^2}{R^2}\right)^{1 / 2}\right\}^3: \\
& \Rightarrow 2=\left(1+\frac{x^2}{R^2}\right)^{1 / 2}: \\
& \Rightarrow 4=1+\frac{x^2}{R^2} \Rightarrow x=\sqrt{3} R
\end{aligned}\)