The resistances of three parts of a circular loop are as shown in the figure. The magnetic field at the centre $O$ is:

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,
$B=0.72 \frac{\mu_{0} N I}{R}, \quad \text { approximately }$

Two circular coils $1$ and $2$ are made from the same wire. The radius of the first coil is twice that of the second coil. What is the ratio of potential difference applied across them $V_1 / V_2$,so that the magnetic field at their centre is the same?

$A$ 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?

In the following figure, the magnitude of the magnetic field at point '$O$' will be

Write the formula for the magnetic field due to a circular current-carrying loop having $N$ turns and $R$ radius at a point on the axis of the loop at a distance $x$ from the center.