The magnetic field at the centre of a circular coil of radius r, due to current I flowing through it, is B. The magnetic field at a point along the axis at a distance $\frac{r}{2}$ from the centre is :

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B/2

${(\frac{2}{\sqrt{3}})}^{3} B$

${(\frac{2}{\sqrt{5}})}^{3} B$

answer is C.

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Detailed Solution

$\begin{aligned} B_{C} = \frac{μ_{0} I}{2 r}, B_{a} = \frac{μ_{0} {Ir}^{2}}{2 {(x^{2} + r^{2})}^{3 / 2}} \\ At x = \frac{r}{2} \\ B_{a} = \frac{μ_{0} {Ir}^{2}}{2 {(\frac{r^{2}}{4} + r^{2})}^{3 / 2}} \\ = \frac{μ_{0} {Ir}^{2}}{2 {(\frac{5}{4} r^{2})}^{3 / 2}} = \frac{μ_{0} I}{2 r} {(\frac{4}{5})}^{3 / 2} \\ = \frac{μ_{0} I}{2 r} {(\frac{2}{\sqrt{5}})}^{3} \end{aligned}$