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

$A$ circular coil of wire consisting of $100$ turns each of radius $9 \ cm$ carries a current of $0.4 \ A$. The magnitude of the magnetic field at the centre of the coil is $[\mu_0 = 12.56 \times 10^{-7} \text{ SI Units}]$.

Two long parallel wires carrying currents $I_1 = 4 \ A$ and $I_2 = 3 \ A$ in opposite directions are placed at a distance of $d = 5 \ cm$ from each other. $A$ point $P$ is at equidistance from both the wires such that the lines joining the point $P$ to the wires are perpendicular to each other. The magnitude of the magnetic field at point $P$ is ( $\mu_0 = 4 \pi \times 10^{-7} \ T \cdot m/A$ ).

The magnetic field normal to the plane of a wire coil of $n$ turns and radius $r$ carrying a current $i$ is measured on the axis of the coil at a small distance $h$ from the centre. By what fraction is this field smaller than the field at the centre?

$A$ current of $I$ $A$ is passed through a straight wire of length $2.0$ $m$. The magnetic field at a point in air at a distance of $3$ $m$ from either end of the wire and lying on the axis of the wire will be:

Four identical long solenoids $A, B, C$ and $D$ are connected to each other as shown in the figure. If the magnetic field at the center of $A$ is $3\, T$,the field at the center of $C$ would be........... $T$ (Assume that the magnetic field is confined within the volume of the respective solenoid).