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

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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 of radius $r$ 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 $x$ from the centre is given by:$B_a = \frac{\mu_0 I r^2}{2(r^2 + x^2)^{3/2}}$Given $x = r$,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,taking the ratio $B_c : B_a$:$\frac{B_c}{B_a} = \frac{\frac{\mu_0 I}{2r}}{\frac{\mu_0 I}{4\sqrt{2}r}} = \frac{4\sqrt{2}r}{2r} = 2\sqrt{2}$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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