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(B) The magnetic field on the axis of a circular coil of radius $R$ at a distance $r$ from the centre is given by $B_1 = \frac{\mu_0 I R^2}{2(R^2 + r^

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$\frac{1}{8}$

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(B) The magnetic field on the axis of a circular coil of radius $R$ at a distance $r$ from the centre is given by $B_1 = \frac{\mu_0 I R^2}{2(R^2 + r^2)^{3/2}}$.Given $r = R \sqrt{3}$,we substitute this into the formula:$B_1 = \frac{\mu_0 I R^2}{2(R^2 + (R \sqrt{3})^2)^{3/2}} = \frac{\mu_0 I R^2}{2(R^2 + 3R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(4R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(8R^3)} = \frac{\mu_0 I}{16R}$.The magnetic field at the centre of the coil is $B_2 = \frac{\mu_0 I}{2R}$.Taking the ratio $\frac{B_1}{B_2} = \frac{\mu_0 I / 16R}{\mu_0 I / 2R} = \frac{2}{16} = \frac{1}{8}$.

If $B_1$ is the magnetic field induction at a point on the axis of a circular coil of radius $R$ situated at a distance $R \sqrt{3}$ and $B_2$ is the magnetic field at the centre of the coil,then the ratio of $\frac{B_1}{B_2}$ is equal to

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