An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is:

$B_{X}$ and $B_{Y}$ are the magnetic fields at the centre of two coils $X$ and $Y$ respectively,each carrying equal current. If coil $X$ has $200$ turns and $20 \ cm$ radius and coil $Y$ has $400$ turns and $20 \ cm$ radius,the ratio of $B_{X}$ and $B_{Y}$ is:

Two concentric thin circular rings of radii $50 \text{ cm}$ and $40 \text{ cm}$,each carry a current of $3.5 \text{ A}$ in opposite directions. If the two rings are coplanar,the net magnetic field due to the two rings at their centre is

For a circular coil of radius $R$ and $N$ turns carrying current $I$,the magnitude of the magnetic field at a point on its axis at a distance $x$ from its centre is given by,
$B=\frac{\mu_{0} I R^{2} N}{2\left(x^{2}+R^{2}\right)^{3 / 2}}$
$(a)$ Show that this reduces to the familiar result for field at the centre of the coil.
$(b)$ Consider two parallel co-axial circular coils of equal radius $R$ and number of turns $N,$ carrying equal currents in the same direction,and separated by a distance $R$. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to $R,$ and is given by,
$B=0.72 \frac{\mu_{0} N I}{R}, \quad \text { approximately }$

$A$ thin circular frame of radius $a$ is made of insulating material. $A$ square loop is constructed within it. If the loop carries current $I$,then the magnetic induction at the geometrical centre $O$ will be:

The magnitude of the magnetic field at the center of an equilateral triangular loop of side $1 \ m$ which is carrying a current of $10 \ A$ is $.... \mu T$.