Magnetic lines of force due to a current in a circular conductor are

$A$ small cube of side $1 \text{ mm}$ is placed at the centre of a circular loop of radius $10 \text{ cm}$ carrying a current of $2 \text{ A}$. The magnetic energy stored inside the cube is $\alpha \times 10^{-14} \text{ J}$. The value of $\alpha$ is . . . . . . . ($\mu_0 = 4\pi \times 10^{-7} \text{ Tm/A}$, $\pi = 3.14$)

Two identical long conducting wires $AOB$ and $COD$ are placed at a right angle to each other,with one above the other such that $O$ is their common point. The wires carry currents $I_1$ and $I_2$,respectively. Point $P$ lies at a distance $d$ from $O$ along a direction perpendicular to the plane containing the wires. The magnetic field at point $P$ will be:

Shown in the figure is a conductor carrying a current $I$. The magnetic field intensity at the point $O$ (common centre of all the three arcs) is

Write the equation for the magnetic field of a current-carrying circular loop at: $(i)$ a point on the axis,and $(ii)$ a point on the plane of the loop at a distance $x$ from the center of the loop.

$A$ particle carrying a charge equal to $100$ times the charge on an electron is rotating one rotation per second in a circular path of radius $0.8 \ m$. The value of magnetic field produced at the centre will be $(\mu_0 = \text{permeability of vacuum})$