The field normal to the plane of wire of n turns and radius r which carries a current i is measured on the axis 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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$\frac{3}{2} \frac{h^{2}}{r^{2}}$

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

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

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

answer is C.

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Detailed Solution

The fields at axis and centre respectively are given by
$B_{axis} = \frac{μ_{0}}{4 π} \times \frac{2 {πir}^{2}}{{(r^{2} + h^{2})}^{3 / 2}}$
and $B_{centre} = \frac{μ_{0}}{4 π} \times \frac{2 πi}{r}$
$∴ \frac{B_{axis}}{B_{centre}} = \frac{r^{3}}{{(r^{2} + h^{2})}^{3 / 2}} = \frac{r^{3}}{r^{3} {[1 + \frac{h^{2}}{r^{2}}]}^{3 / 2}}$
$= {[1 + \frac{h^{2}}{r^{2}}]}^{- 3 / 2} = (1 - \frac{3 h^{2}}{2 r^{2}})$
or $(1 - \frac{B_{axis}}{B_{centre}}) = \frac{3}{2} \frac{h^{2}}{r^{2}}$
or $\frac{B_{centre} - B_{axis}}{B_{centre}} = \frac{3}{2} \frac{h^{2}}{r^{2}}$