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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^

Vedclass · Accepted Answer

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

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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,$\frac{B_Y}{B_X} = \frac{4}{8} = \frac{1}{2}$.

List-$I$	List-$II$
$(A)$ Permeability of free space	$(I) \ [M L^2 T^{-2}]$
$(B)$ Magnetic field	$(II) \ [M T^{-2} A^{-1}]$
$(C)$ Magnetic moment	$(III) \ [M L T^{-2} A^{-2}]$
$(D)$ Torsional constant	$(IV) \ [L^2 A]$

List-$I$	List-$II$
$a$. Fleming's left-hand rule	$e$. Direction of induced current
$b$. Fleming's right-hand rule	$f$. South pole
$c$. Clockwise current	$g$. North pole
$d$. Anticlockwise current	$h$. Direction of force

Similar Questions

Two insulated rings,one of slightly smaller diameter than the other,are suspended along their common diameter as shown. Initially,the planes of the rings are mutually perpendicular. When a steady current is set up in each of them:

Match the following:
List-$I$ List-$II$
$a$. Fleming's left-hand rule $e$. Direction of induced current
$b$. Fleming's right-hand rule $f$. South pole
$c$. Clockwise current $g$. North pole
$d$. Anticlockwise current $h$. Direction of force

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Which of the following statements is correct?

Two infinitely long wires each carrying current $I$ along the same direction are made into the geometry as shown in the figure below. The magnetic field at the point $P$ is

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