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 $\sqrt{3}$)

$A$ circular loop and an infinitely long straight conductor carry equal currents,as shown in the figure. The net magnetic field at the centre of the loop is $B_1$,when the current in the loop is clockwise and $B_2$ when the current in the loop is anti-clockwise. Then $\frac{B_1}{B_2}$ is

The magnetic field normal to the plane of a wire coil of $n$ turns and radius $r$ carrying a current $i$ is measured on the axis of the coil at a small distance $h$ from the centre. By what fraction is this field smaller than the field at the centre?

$A$ cell is connected between the points $A$ and $C$ of a circular conductor $ABCD$ of center $O$ with angle $AOC = 60^o$. If $B_1$ and $B_2$ are the magnitudes of the magnetic fields at $O$ due to the currents in $ABC$ and $ADC$ respectively,the ratio $B_1/B_2$ is

$A$ non-conducting thin disc of radius $R$ rotates about its axis with an angular velocity $\omega$. The surface charge density on the disc varies with the distance $r$ from the center as $\sigma(r)=\sigma_0\left[1+\left(\frac{r}{R}\right)^\beta\right]$, where $\sigma_0$ and $\beta$ are constants. If the magnetic induction at the center is $B=\left(\frac{9}{10}\right) \mu_0 \sigma_0 \omega R$, the value of $\beta$ is

$A$ long straight wire of radius $a$ carries a steady current $i$. The current is uniformly distributed across its cross-section. The ratio of the magnetic field at $a/2$ and $2a$ is