The magnetic induction at the centre of a current carrying circular coil of radius $10\, cm$ is $5\sqrt 5 \,times$ the magnetic induction at a point on its axis. The distance of the point from the centre of the coil (in $cm$) is
Diffcult
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Magnetic field at any point on the axis of current carrying coil at a distance $x$ from the centre is given by :
$\mathrm{B}_{\mathrm{a}}=\frac{\mu_{0} \mathrm{nir}^{2}}{2\left(\mathrm{r}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}$
At the centre,
${\mathrm{x}=0}$
$\therefore $ ${\mathrm{B}_{\mathrm{e}}=\frac{\mu_{0} \mathrm{nir}^{2}}{2 \mathrm{r}^{3}}=\frac{\mu_{0} \mathrm{ni}}{2 \mathrm{r}}}$
$\therefore $ ${\frac{\mathrm{B}_{\mathrm{c}}}{\mathrm{B}_{\mathrm{a}}}=\frac{\mu_{0} \mathrm{ni}}{2 \mathrm{r}} \times \frac{2\left(\mathrm{r}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}{\mu_{0} \mathrm{nir}^{2}}}$
Given : $\quad \mathrm{B}_{\mathrm{c}}=5 \sqrt{5} \mathrm{B}_{\mathrm{a}}$
$\therefore $ $5 \sqrt{5}=\frac{2\left(\mathrm{r}^{2}+\mathrm{x}^{2}\right)^{3 / 2}}{\mathrm{r}^{3}}=\frac{\left(100+\mathrm{x}^{2}\right)^{3 / 2}}{1000}$
Squaring on both sides.
${125=\frac{\left(100+x^{2}\right)^{3}}{10^{6}}} $
or ${125 \times 10^{6}=\left(100+x^{2}\right)^{3}}$
$100+x^{2}=5 \times 10^{2}=500$
$\therefore \quad \mathrm{x}=20 \,\mathrm{cm}$
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