The magnetic field at the centre of a current-carrying circular coil is

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.

$A$ current $I=5 \text{ A}$ flows along a thin wire shaped as shown in the figure. The radius of the curved part of the wire is $R=100 \text{ mm}$, and the angle $2\phi=90^{\circ}$. The magnitude of the magnetic field at point $O$ is approximately:
$\left[\text{Use, } \frac{\mu_0}{4\pi}=10^{-7} \text{ T m A}^{-1}\right]$ (in $\mu\text{T}$)

$A$ circular current-carrying coil has radius $R$. At what distance from the centre of the coil on the axis,the magnetic induction will become $\frac{1}{8}$ th of its value at the centre of the coil?

Two long straight parallel conductors $A$ and $B$ carrying currents $4.5 \ A$ and $8 \ A$ respectively are separated by $25 \ cm$ in air. The resultant magnetic field at a point $P$ which is at a distance of $15 \ cm$ from conductor $A$ and $10 \ cm$ from conductor $B$ is:

Two circular coils $P$ and $Q$ of $100$ turns each have the same radius of $\pi \text{ cm}$. The currents in $P$ and $Q$ are $1 \text{ A}$ and $2 \text{ A}$ respectively. $P$ and $Q$ are placed with their planes mutually perpendicular and their centers coinciding. The resultant magnetic field induction at the center of the coils is $\sqrt{x} \text{ mT}$,where $x = \_\_\_$.
$\left[\text{Use } \mu_0 = 4\pi \times 10^{-7} \text{ T m A}^{-1}\right]$