The magnitude of the magnetic field (in $SI$ units) at the centre of a hexagonal coil of side $10 \, cm$, having $50$ turns and carrying a current $I$ (Ampere), in units of $\frac{\mu_{0} I}{\pi}$ is: (in $\sqrt{3}$)

$A$ current loop consists of two identical semicircular parts each of radius $R,$ one lying in the $x-y$ plane and the other in the $x-z$ plane. If the current in the loop is $i,$ the resultant magnetic field due to the two semicircular parts at their common centre is

State the Biot-Savart law.

$A$ coil having $N$ turns is wound tightly in the form of a spiral with inner and outer radii $a$ and $b$ respectively. When a current $I$ passes through the coil,the magnetic field at the centre is

When a helium nucleus makes a full rotation of a circle of radius $0.8 \,m$ in $2.5 \,s$, then the value of magnetic field $B$ at the centre of the circle will be

As shown in the figure,a long straight conductor with a semicircular arc of radius $R = \frac{\pi}{10} \, m$ is carrying a current $I = 3 \, A$. The magnitude of the magnetic field at the center $O$ of the arc is $........... \mu T$. (The permeability of the vacuum $\mu_0 = 4 \pi \times 10^{-7} \, T \cdot m/A$)