$A$ charge $q$ coulomb moves in a circle at $n$ revolutions per second and the radius of the circle is $r$ metre; then the magnetic field at the centre of the circle is

$A$ current-carrying wire in its neighborhood produces:

$A$ 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

The correct Biot-Savart law in vector form is

When the current flowing in a circular coil is doubled and the number of turns of the coil in it is halved,the magnetic field at its centre will become

Two identical coils having the same number of turns and carrying equal current have a common centre,and their planes are at right angles to each other. What is the ratio of the magnitude of the resultant magnetic field at the centre to the magnetic field due to one of the coils at the centre?