In the figure shown,there are two semicircles of radii $r_1$ and $r_2$ in which a current $i$ is flowing. The magnetic induction at the centre $O$ will be

$A$ 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 carries a current of $\pi \, A$. The magnetic field due to it will be $5 \times 10^{-5} \, Wb/m^2$ at what distance from the wire? $[\mu_0 = \text{permeability of free space}]$

$A$ particle having a charge $100 e$ is revolving in a circular path of radius $0.8 \ m$ with $1 \ r.p.s$. The magnetic field produced at the center of the circle in $SI$ unit is ($\mu_0$ is the permeability of vacuum,$e = 1.6 \times 10^{-19} \ C$).

An infinitely long wire carrying $1 \ A$ current in the $+z$ direction is placed at $(1 \ cm, 1 \ cm)$. Another wire carrying $1 \ A$ in the $+x$ direction is placed at $y=1 \ cm$. If the magnetic field due to this configuration at the origin is $B$. Let $B_0$ be the magnitude of the field if only the wire at $(1 \ cm, 1 \ cm)$ was present,then $\frac{B}{B_0}$ is

$A$ 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