# The radius of a circular current carrying coil is R. At what distance from the centre of the coil on its axis, the intensity of magnetic field will be 1/2root2 times that at the centre?

Please find below the solution to the asked query:

$magneticfielsalongtheaxisofacircularcurrentcarryingcoilis:\frac{{\mu}_{o}i{R}^{2}}{2({x}^{2}+{r}^{2}{)}^{3/2}}\phantom{\rule{0ex}{0ex}}magneticfieldatthecenter=\frac{{\mu}_{o}i}{2R}\phantom{\rule{0ex}{0ex}}given\frac{{\mu}_{o}i{R}^{2}}{2({x}^{2}+{R}^{2}{)}^{3/2}}=\frac{1}{2\sqrt{2}}\frac{{\mu}_{o}i}{2R}\phantom{\rule{0ex}{0ex}}or,\frac{{R}^{2}}{({x}^{2}+{R}^{2}{)}^{3/2}}=\frac{1}{2\sqrt{2}R}\phantom{\rule{0ex}{0ex}}or2\sqrt{2}{R}^{3}=({x}^{2}+{R}^{2}{)}^{3/2}\phantom{\rule{0ex}{0ex}}or,8{R}^{6}=({x}^{2}+{R}^{2}{)}^{3}\phantom{\rule{0ex}{0ex}}or,2{R}^{2}={x}^{2}+{R}^{2}\phantom{\rule{0ex}{0ex}}orx=\pm \sqrt{2{R}^{2}-{R}^{2}}=\pm R\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}answer$

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