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(C) The magnetic field on the axis of a circular coil of radius $R$ at a distance $x$ from its center is given by $B = \frac{\mu_0 N i R^2}{2(x^2 + R^

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$\frac{B_Y}{B_X} = \frac{1}{2}$

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(C) The magnetic field on the axis of a circular coil of radius $R$ at a distance $x$ from its center is given by $B = \frac{\mu_0 N i R^2}{2(x^2 + R^2)^{3/2}}$.For the smaller coil $X$,radius is $r$ and distance from $O$ is $x = d/2$. Thus,$B_X = \frac{\mu_0 N i r^2}{2((d/2)^2 + r^2)^{3/2}} = \frac{\mu_0 N i r^2}{2(\frac{d^2+4r^2}{4})^{3/2}} = \frac{4\mu_0 N i r^2}{2(d^2+4r^2)^{3/2}}$.For the bigger coil $Y$,radius is $2r$ and distance from $O$ is $x = d$. Thus,$B_Y = \frac{\mu_0 N i (2r)^2}{2(d^2 + (2r)^2)^{3/2}} = \frac{4\mu_0 N i r^2}{2(d^2+4r^2)^{3/2}}$.Wait,re-evaluating the geometry: The coils subtend the same solid angle,meaning the ratio of radius to distance is constant. For $X$,$r/(d/2) = 2r/d$. For $Y$,$2r/d$. This is consistent.Recalculating $B_X$: $B_X = \frac{\mu_0 N i r^2}{2(d^2/4 + r^2)^{3/2}} = \frac{\mu_0 N i r^2}{2(\frac{d^2+4r^2}{4})^{3/2}} = \frac{8\mu_0 N i r^2}{2(d^2+4r^2)^{3/2}}$.Recalculating $B_Y$: $B_Y = \frac{\mu_0 N i (2r)^2}{2(d^2 + 4r^2)^{3/2}} = \frac{4\mu_0 N i r^2}{2(d^2+4r^2)^{3/2}}$.Therefore,$\frac{B_Y}{B_X} = \frac{4}{8} = \frac{1}{2}$.

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