The field normal to the plane of a wire 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
Diffcult
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(a) Field at the centre ${B_1} = \frac{{{\mu _0}}}{{4\pi }} \times \frac{{2\pi in}}{r} = \frac{{{\mu _0}}}{2}.\frac{{ni}}{r}$
Field at a distance $h$ from the centre
${B_2} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi ni{r^2}}}{{{{({r^2} + {h^2})}^{3/2}}}} = \frac{{{\mu _0}}}{2}.\frac{{ni{r^2}}}{{{r^3}{{\left( {1 + \frac{{{h^2}}}{{{r^2}}}} \right)}^{3/2}}}}$
$ = {B_1}{\left( {1 + \frac{{{h^2}}}{{{r^2}}}} \right)^{ - 3/2}} = {B_1}\left( {1 - \frac{3}{2}.\frac{{{h^2}}}{{{r^2}}}} \right)$(By binomial theorem)
Hence $B_2$ is less than $B_1$ by a fraction $ = \frac{3}{2}\frac{{{h^2}}}{{{r^2}}}$
Field at a distance $h$ from the centre
${B_2} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi ni{r^2}}}{{{{({r^2} + {h^2})}^{3/2}}}} = \frac{{{\mu _0}}}{2}.\frac{{ni{r^2}}}{{{r^3}{{\left( {1 + \frac{{{h^2}}}{{{r^2}}}} \right)}^{3/2}}}}$
$ = {B_1}{\left( {1 + \frac{{{h^2}}}{{{r^2}}}} \right)^{ - 3/2}} = {B_1}\left( {1 - \frac{3}{2}.\frac{{{h^2}}}{{{r^2}}}} \right)$(By binomial theorem)
Hence $B_2$ is less than $B_1$ by a fraction $ = \frac{3}{2}\frac{{{h^2}}}{{{r^2}}}$
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