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(B) The magnetic induction at the centre of a circular current-carrying coil is given by $B_{C} = \frac{\mu_{0} I}{2 R}$.The magnetic induction at a d

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$1: 8$

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(B) The magnetic induction at the centre of a circular current-carrying coil is given by $B_{C} = \frac{\mu_{0} I}{2 R}$.The magnetic induction at a distance $r$ from the centre along the axis is given by $B_{A} = \frac{\mu_{0} I R^{2}}{2(R^{2} + r^{2})^{3/2}}$.Given that $r = \sqrt{3} R$,we substitute this into the formula for $B_{A}$:$B_{A} = \frac{\mu_{0} I R^{2}}{2(R^{2} + (\sqrt{3} R)^{2})^{3/2}} = \frac{\mu_{0} I R^{2}}{2(R^{2} + 3 R^{2})^{3/2}} = \frac{\mu_{0} I R^{2}}{2(4 R^{2})^{3/2}}$.Simplifying the denominator: $(4 R^{2})^{3/2} = (2^{2} R^{2})^{3/2} = (2 R)^{3} = 8 R^{3}$.So,$B_{A} = \frac{\mu_{0} I R^{2}}{2(8 R^{3})} = \frac{\mu_{0} I}{16 R}$.Now,calculating the ratio $B_{A} : B_{C}$:$\frac{B_{A}}{B_{C}} = \frac{\frac{\mu_{0} I}{16 R}}{\frac{\mu_{0} I}{2 R}} = \frac{2 R}{16 R} = \frac{1}{8}$.Thus,the ratio is $1: 8$.

$A$ circular current-carrying coil has radius $R$. The magnetic induction at the centre of the coil is $B_{C}$. The magnetic induction of the coil at a distance $\sqrt{3} R$ from the centre along the axis is $B_{A}$. The ratio $B_{A}: B_{C}$ is

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