The ratio of the magnetic field at the centre of a current-carrying circular wire and the magnetic field at the centre of a square coil made from the same length of wire will be

$A$ current of $1\,A$ is flowing through the sides of an equilateral triangle of side $4.5\times10^{-2}\,m$. The magnetic field at the centre of the triangle will be

$A$ current of $1 \,A$ is flowing along the sides of an equilateral triangle of side $a = 4.5 \times 10^{-2} \,m$. The magnetic field at the centroid of the triangle is $(\mu_0 = 4 \pi \times 10^{-7} \,T \cdot m/A)$.

$A$ circular arc of radius $r$ carrying current $I$ subtends an angle $\frac{\pi}{16}$ at its centre. The radius of the metal wire is uniform. The magnetic induction at the centre of the circular arc is [where $\mu_0$ is the permeability of free space].

$A$ 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 $\mu\text{T}$)

The magnetic induction at the centre $O$ in the figure shown is