$B_{X}$ 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:

The magnetic induction at point $O$ for the current-carrying wire shown in the figure is:

The magnitude of magnetic induction at the mid-point $O$ due to the current arrangement as shown in the figure will be:

$A$ long wire lies along the $X$-axis and carries a current of $40 \, A$ in the positive $x$-direction. $A$ second long wire is perpendicular to the $xy$-plane, passes through the point $(3.0 \, m) \hat{j}$, and carries a current along the positive $z$-direction. If the magnitude of the resultant magnetic field at the point $(2.0 \, m) \hat{j}$ is $R=5 \times 10^{-6} \, T$, then the current in the second wire is (Permeability of free space, $\mu_0=4 \pi \times 10^{-7} \, T \cdot m/A$) (in $A$)

An element $\Delta l = \Delta x \hat{i}$ is placed at the origin and carries a current $I = 10 \,A$. The magnetic field on the $y$-axis at a distance of $0.5 \,m$ from the element of length $\Delta x = 1 \,cm$ is:

$A$ current $i$ is flowing in a straight conductor of length $L.$ The magnetic induction at a point on its axis at a distance $\frac{L}{4}$ from its centre will be