$A$ uniform circular wire loop is connected to the terminals of a battery. The magnetic field induction at the centre due to $ABC$ portion of the wire will be (length of $ABC = l_1$, length of $ADC = l_2$):

For a circular coil of radius $R$ and $N$ turns carrying current $I$,the magnitude of the magnetic field at a point on its axis at a distance $x$ from its centre is given by,
$B=\frac{\mu_{0} I R^{2} N}{2\left(x^{2}+R^{2}\right)^{3 / 2}}$
$(a)$ Show that this reduces to the familiar result for field at the centre of the coil.
$(b)$ Consider two parallel co-axial circular coils of equal radius $R$ and number of turns $N,$ carrying equal currents in the same direction,and separated by a distance $R$. Show that the field on the axis around the mid-point between the coils is uniform over a distance that is small as compared to $R,$ and is given by,
$B=0.72 \frac{\mu_{0} N I}{R}, \quad \text { approximately }$

$PQRS$ is a square loop made of a uniform conducting wire. The current enters the loop at $P$ and leaves at $S$. Then the magnetic field will be

$A$ long straight rod of diameter $4 \text{ mm}$ carries a steady current '$i$'. The current is uniformly distributed across its cross-section. The ratio of the magnetic fields at distances $1 \text{ mm}$ and $4 \text{ mm}$ from the axis of the rod is

Explain the experiment which demonstrates the magnetic field produced due to a straight long current-carrying wire.

$A$ coil of $50$ turns and $4$ cm radius carries $2$ $A$ current. The magnetic field at its centre is ....... $mT$.