Two circular coils $X$ and $Y$, having equal number of turns, carry equal currents in the same sense and subtend same solid angle at point $O$. If the smaller coil, $X$ is midway between $O$ and $Y$, then if we represent the magnetic induction due to bigger coil $Y$ at $O$ as $BY$ and that due to smaller coil $X$ at $O$ as $BX$ , then
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(c) Magnetic field at $O$ due to bigger coil $Y$, is
${B_Y} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi i{{(2r)}^2}}}{{{{\{ {d^2} + {{(2r)}^2}\} }^{3/2}}}} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{8\pi i{r^2}}}{{{{({d^2} + 4{r^2})}^{3/2}}}}$
Magnetic field at $O$ due to smaller coil $X$ is
${B_X} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi i{r^2}}}{{{{\left\{ {{{\left( {\frac{d}{2}} \right)}^2} + {r^2}} \right\}}^{3/2}}}} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{16\pi i{r^2}}}{{{{({d^2} + 4{r^2})}^{3/2}}}}$
$==>$ $\;\frac{{{B_Y}}}{{{B_X}}} = \frac{1}{2}$
${B_Y} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi i{{(2r)}^2}}}{{{{\{ {d^2} + {{(2r)}^2}\} }^{3/2}}}} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{8\pi i{r^2}}}{{{{({d^2} + 4{r^2})}^{3/2}}}}$
Magnetic field at $O$ due to smaller coil $X$ is
${B_X} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\pi i{r^2}}}{{{{\left\{ {{{\left( {\frac{d}{2}} \right)}^2} + {r^2}} \right\}}^{3/2}}}} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{16\pi i{r^2}}}{{{{({d^2} + 4{r^2})}^{3/2}}}}$
$==>$ $\;\frac{{{B_Y}}}{{{B_X}}} = \frac{1}{2}$
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