Give the $SI$ unit of magnetic field from the Biot-Savart law.

Two very thin metallic wires placed along $X$ and $Y$-axis carry equal currents as shown in the figure. $AB$ and $CD$ are lines at $45^\circ$ with the axes,with the origin of the axes at $O$. The magnetic field will be zero on the line:

An electron moving in a circular orbit of radius $r$ makes $n$ rotations per second. The magnetic field produced at the centre has magnitude

Two coils $P$ and $Q$ each of radius $R$ carry currents $I$ and $\sqrt{8} I$ respectively in the same direction. These coils are lying in perpendicular planes such that they have a common centre. The magnitude of the magnetic field at the common centre of the two coils is ($\mu_0 =$ permeability of free space).

$A$ current of $1 \, A$ flows through an infinitely long straight wire. The magnetic field produced at a point $1 \, m$ away from it is:

Two long parallel wires carrying currents $8 \text{ A}$ and $15 \text{ A}$ in opposite directions are placed at a distance of $7 \text{ cm}$ from each other. $A$ point $P$ is equidistant from both the wires such that the lines joining the point $P$ to the wires are perpendicular to each other. The magnetic field at $P$ is $X \times 10^{-6} \text{ T}$. Find $X$. (Given: $\sqrt{2} = 1.4, \mu_0 = 4\pi \times 10^{-7} \text{ SI unit}$)