Two circular coils $X$ and $Y$, having equal number of turns, carry equal currents in the same sence and subtend same solid angle at point $O$. If the smaller coil $X$ is midway between $O$ and $Y$, and If we represent the magnetic induction due to bigger coil $Y$ at $O$ as $B_Y$ and that due to smaller coil $X$ at $O$ as $B_X$, then $\frac{{{B_Y}}}{{{B_X}}}$ is
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Magnetic field at $\mathrm{O}$ due to bigger coil $\mathrm{Y},$ is
$\mathrm{B}_{\mathrm{Y}}=\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \pi \mathrm{i}(2 \mathrm{r})^{2}}{\left\{\mathrm{d}^{2}+(2 \mathrm{r})^{2}\right\}^{3 / 2}}=\frac{\mu_{0}}{4 \pi} \cdot \frac{8 \pi \mathrm{ir}^{2}}{\left(\mathrm{d}^{2}+4 \mathrm{r}^{2}\right)^{3 / 2}}$
Magnetic field at $\mathrm{O}$ due to smaller coil $\mathrm{X}$ is
$\mathrm{B}_{\mathrm{x}}=\frac{\mu_{0}}{4 \pi} \cdot \frac{2 \pi \mathrm{ir}^{2}}{\left\{\left(\frac{\mathrm{d}}{2}\right)^{2}+\mathrm{r}^{2}\right\}^{3 / 2}}=\frac{\mu_{0}}{4 \pi} \cdot \frac{16 \pi \mathrm{ir}^{2}}{\left(\mathrm{d}^{2}+4 \mathrm{r}^{2}\right)^{3 / 2}}$
$\Rightarrow \quad \frac{\mathrm{B}_{\mathrm{Y}}}{\mathrm{B}_{\mathrm{X}}}=\frac{1}{2}$
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