At what distance from a long straight wire carrying a current of $12 \, A$ will the magnetic field be equal to $3 \times 10^{-5} \, Wb/m^2$?

$A$ circular arc of radius $r$ carrying current $I$ subtends an angle $\frac{\pi}{16}$ at its centre. The radius of the metal wire is uniform. The magnetic induction at the centre of the circular arc is [where $\mu_0$ is the permeability of free space].

To produce a uniform magnetic field directed parallel to a diameter of a cylindrical region,one can use the saddle coils illustrated in the figure. The loops are wrapped over a somewhat flattened tube. Assume the straight sections of wire are very long. The end view of the tube shows how the windings are applied. The overall current distribution is the superposition of two overlapping circular cylinders of uniformly distributed current,one toward you and one away from you. The current density $J$ is the same for each cylinder. The position of the axis of one cylinder is described by a position vector $\vec{a}$ relative to the other cylinder. The magnetic field inside the hollow tube is:

$A$ steady current flows in a long wire. It is bent into a circular loop of one turn and the magnetic field at the centre of the coil is $B$. If the same wire is bent into a circular loop of $n$ turns,the magnetic field at the centre of the coil is

Two infinite length wires carry currents $8 \ A$ and $6 \ A$ respectively and are placed along the $X$ and $Y$ axes respectively. The magnetic field at a point $P(0, 0, d)$ will be:

$A$ circular coil of wire consisting of $100$ turns,each of radius $8.0 \; cm$,carries a current of $0.40 \; A$. What is the magnitude of the magnetic field $B$ at the centre of the coil?