The field normal to the plane of a coil of ‘n’ turns and radius ‘r’ which carries a current ‘i’ is measured on the axis of the coil at a small distance ‘h’ from the centre of the coil. This is smaller than the field at the centre by the fraction.

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$\frac{3}{2} \frac{h^{2}}{r^{2}}$

$\frac{2}{3} \frac{h^{2}}{r^{2}}$

$\frac{3}{2} \frac{r^{2}}{h^{2}}$

$\frac{2}{3} \frac{r^{2}}{h^{3}}$

answer is A.

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Detailed Solution

$\begin{array}{l} B_{a} = \frac{μ_{0} {nir}^{2}}{2 {[r^{2} + h^{2}]}^{\frac{3}{2}}} = \frac{μ_{0} {nir}^{2}}{2 [r^{2} {(1 + \frac{h^{2}}{r^{2}})}^{\frac{3}{2}}} \\ B_{a} = \frac{μ_{0} ni}{2 r} {[1 + \frac{h^{2}}{r^{2}}]}^{\frac{- 3}{2}} \\ (∵ B_{c} = \frac{μ_{0} ni}{2 r}); \frac{B_{a}}{B_{c}} = 1 - \frac{3}{2} \frac{h^{2}}{r^{2}}; \frac{3}{2} \frac{h^{2}}{r^{2}} = \frac{B_{c} - B_{a}}{B_{c}} \end{array}$