In an ionised sodium atom,an electron is moving in a circular path of radius $r$ with angular velocity $\omega$. The magnetic induction in $Wb/m^2$ produced at the nucleus will be

The magnetic field due to a current-carrying square loop of side $a$ at a point located symmetrically at a distance of $a/2$ from its centre (as shown in the figure) is:

The magnetic field due to a current-carrying circular coil on its axis at a distance of $\sqrt{2} \,d$ from the centre of the coil is $B$. If $d$ is the diameter of the coil,then the magnetic field at the centre of the coil is: (in $B$)

$A$ square coil of side $a$ carries a current $I$. The magnetic field at the centre of the coil is

$A$ long curved conductor carries a current $I$. $A$ small current element of length $dl$ on the wire induces a magnetic field at a point away from the current element. If the position vector between the current element and the point is $\vec{r}$,making an angle $\theta$ with the current element,then the induced magnetic field density $d\vec{B}$ at the point is $(\mu_0 = \text{permeability of free space})$:

$A$ loop carrying current $I$ has the shape of a regular polygon of $n$ sides. If $R$ is the distance from the centre to any vertex,then the magnitude of the magnetic induction vector $B$ at the centre of the loop is