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(C) The magnetic field at the centre of a circular current-carrying coil is given by:$B_c = \frac{\mu_0 I}{2r} \quad \dots(i)$The magnetic field on th

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$2\sqrt{2} : 1$

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(C) The magnetic field at the centre of a circular current-carrying coil is given by:$B_c = \frac{\mu_0 I}{2r} \quad \dots(i)$The magnetic field on the axis of a circular coil at a distance $d$ from the centre is given by:$B_a = \frac{\mu_0 I r^2}{2(r^2 + d^2)^{3/2}}$Given $d = r$,we substitute this into the formula:$B_a = \frac{\mu_0 I r^2}{2(r^2 + r^2)^{3/2}} = \frac{\mu_0 I r^2}{2(2r^2)^{3/2}} = \frac{\mu_0 I r^2}{2(2\sqrt{2} r^3)} = \frac{\mu_0 I}{4\sqrt{2} r} \quad \dots(ii)$Now,calculating the ratio $B_c : B_a$:$\frac{B_c}{B_a} = \left( \frac{\mu_0 I}{2r} \right) \div \left( \frac{\mu_0 I}{4\sqrt{2} r} \right) = \frac{4\sqrt{2} r}{2r} = 2\sqrt{2}$Therefore,$B_c : B_a = 2\sqrt{2} : 1$.

The magnetic field at the centre of a circular current-carrying conductor of radius $r$ is $B_c$. The magnetic field on its axis at a distance $r$ from the centre is $B_a$. The value of $B_c : B_a$ will be:

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