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Q.
A circular coil carrying current 'I' has radius 'R' and magnetic field at the centre is 'B' . At what distance from the centre along the axis of the same coil, the magnetic field will be $\frac{B}{8}$ ?
According to the question, magnetic field at the centre of circular coil is given by
$B_{\text {centre }}=\frac{\mu_{0} N i}{2 R} \,\,\,\,\,\,\, ...(i)$
$B_{\text {axis }}=\frac{\mu_{0}}{4 \pi} \frac{2 \pi N i R^{2}}{\left(x^{2}+R^{2}\right)^{3 / 2}}$
$\frac{B}{8}=\frac{\mu_{0} N i R^{2}}{2\left(x^{2}+R^{2}\right)^{3 / 2}}$
From Eq. (i), we get
$\frac{\mu_{0} N i}{8 \times 2 R}=\frac{\mu_{0} N i R^{2}}{2\left(x^{2}+R^{2}\right)^{3 / 2}} $
$\frac{1}{8 R}=\frac{1}{\left(x^{2}+R^{2}\right)^{3 / 2}}$
$\frac{1}{8 R^{3}}=\frac{1}{\left(x^{2}+R^{2}\right)^{3 / 2}}$
$8 R^{3}=\left(x^{2}+R^{2}\right)^{3 /2}$ (on taking cube root)
$2 R=\left(x^{2}+R^{2}\right)^{1/2} $ (on taking square root)
$ 4 R^{2} =\left(x^{2}+R^{2}\right) $
$4 R^{2}-R^{2}=x^{2} $
$ 3 R^{2} =x^{2}$
$ x =R \sqrt{3} $