$A$ long wire carries a steady current. It is bent into a circle of one turn and the magnetic field at the centre of the coil is $B$. It is then bent into a circular loop of $n$ turns. The magnetic field at the centre of the coil for the same current will be

The current through $ABC$ and $A'B'C'$ is $I$. What is the magnetic field at $P$? Given $BP = PB' = r$ (Here $C', B', P, B, C$ are collinear).

$A$ long wire carries a steady current. It is bent into a coil of one turn such that the magnetic induction at the centre is $B$. If the same wire is bent to form a coil of smaller radius with $n$ turns,then the new magnetic induction $B^{\prime}$ at the centre is:

$A$ dielectric circular disc of radius $R$ carries a uniform surface charge density $\sigma$. If it rotates about its axis with angular velocity $\omega$, the magnetic field at the center of the disc is:

Consider the circular loop carrying current $i$ as shown in the figure. The magnetic field at the central point $O$ is

An infinitely long wire,located on the $z$-axis,carries a current $I$ along the $+z$-direction and produces the magnetic field $\vec{B}$. The magnitude of the line integral $\int \vec{B} \cdot d\vec{l}$ along a straight line from the point $(-\sqrt{3} a, a, 0)$ to $(a, a, 0)$ is given by [$\mu_0$ is the magnetic permeability of free space.]