Two circular coils X and Y,having an equal nu | vedclass

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Question

(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 I R^2}{2(x^2 + R^2)

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The magnetic field on the axis of a circular coil of radius $R$ at a distance $x$ from its center is given by $B = \frac{\mu_0 I R^2}{2(x^2 + R^2)^{3/2}}$.For the smaller coil $X$: radius is $r$,distance from $O$ is $x = d/2$. Thus,$B_X = \frac{\mu_0 I r^2}{2((d/2)^2 + r^2)^{3/2}} = \frac{\mu_0 I r^2}{2(\frac{d^2}{4} + r^2)^{3/2}} = \frac{\mu_0 I r^2}{2(\frac{d^2 + 4r^2}{4})^{3/2}} = \frac{\mu_0 I r^2}{2 \cdot \frac{1}{8} (d^2 + 4r^2)^{3/2}} = \frac{4 \mu_0 I r^2}{(d^2 + 4r^2)^{3/2}}$.For the bigger coil $Y$: radius is $2r$,distance from $O$ is $x = d$. Thus,$B_Y = \frac{\mu_0 I (2r)^2}{2(d^2 + (2r)^2)^{3/2}} = \frac{4 \mu_0 I r^2}{2(d^2 + 4r^2)^{3/2}} = \frac{2 \mu_0 I r^2}{(d^2 + 4r^2)^{3/2}}$.Taking the ratio: $\frac{B_Y}{B_X} = \frac{2 \mu_0 I r^2 / (d^2 + 4r^2)^{3/2}}{4 \mu_0 I r^2 / (d^2 + 4r^2)^{3/2}} = \frac{2}{4} = \frac{1}{2}$.

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