What will be the resultant magnetic field at the origin due to four infinite length wires? If each wire produces a magnetic field '$B$' at the origin.

For a wire, as shown in the figure, carrying a current of $10 \,A$, find the magnetic induction field at the point $O$. (Given: $\mu_0 = 4 \pi \times 10^{-7} \,H/m$)

Two wires $A$ and $B$ are of lengths $40 \ cm$ and $30 \ cm$. $A$ is bent into a circle of radius $r$ and $B$ into an arc of radius $r$. $A$ current $i_1$ is passed through $A$ and $i_2$ through $B$. To have the same magnetic inductions at the centre, the ratio of $i_1: i_2$ is

$A$ current of $0.1\, A$ circulates around a coil of $100$ turns and having a radius equal to $5\, cm$. The magnetic field set up at the centre of the coil is $(\mu_0 = 4\pi \times 10^{-7} \, Wb/A \cdot m)$

Two circular coils $X$ (smaller) and $Y$ (bigger), each having a single turn, carry equal currents in the same direction and subtend the same angle at point '$O$' along the axis of the coil. The distances of the centers of coil to point '$O$' are $d$ and $(d/2)$ for coil $Y$ and $X$ respectively. The radii of coils $Y$ and $X$ are $(2r)$ and $(r)$ respectively. The magnetic induction due to the bigger coil at point '$O$' is $B_y$ and that due to smaller coil $X$ at point '$O$' is $B_x$. $(d \gg r)$ The relation between $B_x$ and $B_y$ is

The magnetic induction at the centre of a current-carrying circular coil of radius $8 \ cm$ is $6 \sqrt{6}$ times the magnetic induction at a point on its axis. Then the distance of the point from the centre of the coil in $cm$ is $(\sqrt{5} = 2.236)$.