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(C) The magnetic field at the centre of a circular coil of radius $R$ carrying current $i$ is given by $B = \frac{\mu_{0} i}{2 R}$.The magnetic field

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$\frac{B}{\sqrt{8}}$

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(C) The magnetic field at the centre of a circular coil of radius $R$ carrying current $i$ is given by $B = \frac{\mu_{0} i}{2 R}$.The magnetic field at a point on the axis of the coil at a distance $x$ from the centre is given by $B_{axis} = \frac{\mu_{0} i R^{2}}{2(R^{2} + x^{2})^{3/2}}$.Given $x = R$,we substitute this into the formula:$B_{axis} = \frac{\mu_{0} i R^{2}}{2(R^{2} + R^{2})^{3/2}}$$B_{axis} = \frac{\mu_{0} i R^{2}}{2(2R^{2})^{3/2}}$$B_{axis} = \frac{\mu_{0} i R^{2}}{2(2^{3/2} R^{3})}$$B_{axis} = \frac{\mu_{0} i}{2 R \cdot 2^{3/2}}$Since $2^{3/2} = \sqrt{8}$,we have:$B_{axis} = \frac{B}{\sqrt{8}}$.

The magnetic field at the centre of a circular coil of radius $R$ due to current $i$ flowing through it is $B$. The magnetic field at a point along the axis at a distance $R$ from the centre is

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