$A$ 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

The correct Biot-Savart law in vector form is

Current through $ABC$ and $A^{\prime} B^{\prime} C^{\prime}$ is $I$. What is the magnetic field at $P$? $BP = PB^{\prime} = r$. (Here $C^{\prime} B^{\prime} PBC$ are collinear).

The resistances of three parts of a circular loop are as shown in the figure. The magnetic field at the centre $O$ is:

The radius of a circular current-carrying coil is $R$. At what distance from the centre of the coil on its axis,the intensity of the magnetic field will be $\frac{1}{2 \sqrt{2}}$ times that at the centre?

$A$ non-conducting disc of radius $R$ has a surface charge density which varies with distance from the centre as $\sigma(r) = \sigma_0 \left[1 + \sqrt{\frac{r}{R}}\right]$, where $\sigma_0$ is a constant. The disc rotates about its axis with angular velocity $\omega$. If $B$ is the magnitude of magnetic induction at the centre, then $\frac{B}{\mu_0 \sigma_0 \omega R}$ will be