The field due to a wire of n turns and radius r which carries a current J is measured on the axis of the coil at a small distance h from the center of the coil. This is smaller than the field at the center by the fraction:

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

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

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

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

answer is A.

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Detailed Solution

The magnetic field on the axis of a coil carrying current d having n turns, radius r and at a distance h from the centre of the coil, is given by

$B = \frac{μ_{0}}{4 π} \times \frac{2 π N I r^{2}}{{(r^{2} + h^{2})}^{3 / 2}} T h e f i e l d a t t h e c e n t r e i s g i v e n b y B_{C} = \frac{μ_{0}}{4 π} \times \frac{2 π N I}{r} ∴ \frac{B}{B_{C}} = \frac{r^{3}}{{(r^{2} + h^{2})}^{3 / 2}} = \frac{r^{3}}{r^{3} {[1 + \frac{h^{2}}{r^{2}}]}^{3 / 2}} = [1 - \frac{3}{2} \frac{h^{2}}{r^{2}}] We have to find: f = \frac{B_{C} - B}{B_{C}} = 1 = \frac{B}{B_{C}} = \frac{3}{2} \frac{h^{2}}{r^{2}}$