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(B) The magnetic field $B$ on the axis of a current-carrying circular coil is given by the formula:$B = \frac{\mu_0 n I r^2}{2(x^2 + r^2)^{3/2}}$where

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$\frac{\mu_0 n I}{16 r}$

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(B) The magnetic field $B$ on the axis of a current-carrying circular coil is given by the formula:$B = \frac{\mu_0 n I r^2}{2(x^2 + r^2)^{3/2}}$where $I$ is the current,$n$ is the number of turns,$r$ is the radius of the coil,and $x$ is the distance of the point from the center.Given $x = \sqrt{3} r$,we substitute this into the formula:$B = \frac{\mu_0 n I r^2}{2((\sqrt{3} r)^2 + r^2)^{3/2}}$$B = \frac{\mu_0 n I r^2}{2(3r^2 + r^2)^{3/2}}$$B = \frac{\mu_0 n I r^2}{2(4r^2)^{3/2}}$$B = \frac{\mu_0 n I r^2}{2(8r^3)}$$B = \frac{\mu_0 n I}{16r}$

$A$ circular coil of wire of radius $r$ has $n$ turns and carries a current $I$. The magnetic induction $B$ at a point on the axis of the coil at a distance $\sqrt{3} r$ from its centre is

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