Q116 / 120 · VITEEE 2014 · Physics
A circular current carrying coil has a radius $\mathrm{R}$. The distance from the centre of the coil on the axis, where the magnetic induction will be $\frac{1}{8}$ th to its value at the centre of the coil is
- A$\frac{R}{\sqrt{3}}$
- Correct answerB$R \sqrt{3}$
- C$2 \sqrt{3} R$
- D$\frac{2}{\sqrt{3}} R$
Solution
Here the ratio, $\frac{\mathrm{B}_{\mathrm{Centre}}}{\mathrm{B}_{\mathrm{axis}}}=\left(1+\frac{\mathrm{x}^{2}}{\mathrm{R}^{2}}\right)^{3 / 2}$
Also, $B_{\text {axis }}=\frac{1}{8} B_{\text {centre }}$
$\begin{array}{l}
\frac{8}{1}=\left(1+\frac{x^{2}}{R^{2}}\right)^{3 / 2} \Rightarrow 4=1+\frac{x^{2}}{R^{2}} \\
3=\frac{x^{2}}{R^{2}} \Rightarrow X^{2}=3 R^{2} \\
\text { or, } x=\sqrt{3} R
\end{array}$
Hence, the correct answer is option: 'B'.