When equal current is passed through two coils,equal magnetic field is produced at their centres. If the ratio of the number of turns in the coils is $8: 15$,then the ratio of their radii will be

The magnetic field at the point of intersection of diagonals of a square wire loop of side $L$ carrying a current $I$ is

$A$ and $B$ are two concentric circular conductors with center $O$,carrying currents $i_1$ and $i_2$ as shown in the figure. If the ratio of their radii is $1:2$ and the ratio of the magnetic flux densities at $O$ due to $A$ and $B$ is $1:3$,then the value of $i_1/i_2$ is:

$A$ helium nucleus makes a full rotation in a circle of radius $0.8 \ m$ in $2 \ s$. The value of the magnetic field $B$ at the centre of the circle will be

$A$ long straight rod of diameter $4 \text{ mm}$ carries a steady current '$i$'. The current is uniformly distributed across its cross-section. The ratio of the magnetic fields at distances $1 \text{ mm}$ and $4 \text{ mm}$ from the axis of the rod 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