$A$ tightly wound $100$ turns coil of radius $10 \,cm$ carries a current of $7 \,A$. The magnitude of the magnetic field at the centre of the coil is (Take permeability of free space as $4 \pi \times 10^{-7} \,SI$ units):

$A$ linear small part of wire $PQ$ is situated on the $y$-axis from $y = -a/2$ to $y = +a/2$, and a current $I$ is flowing through it. The magnetic field produced due to wire $PQ$ at the point $y = +a$ will be:

$A$ loop $ABCDA$,carrying current $I=12 \ A$,is placed in a plane,consists of two semi-circular segments of radius $R_1=6 \pi \ m$ and $R_2=4 \pi \ m$. The magnitude of the resultant magnetic field at center $O$ is $k \times 10^{-7} \ T$. The value of $k$ is . . . . . . . (Given $\mu_0=4 \pi \times 10^{-7} \ Tm \ A^{-1}$)

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).

Two circular coils $P$ and $Q$ are made from similar wire,but the radius of $Q$ is twice that of $P$. What should be the value of the potential difference across them so that the magnetic induction at their centres is the same?

The ratio of the magnetic field at the centre of a current-carrying coil of radius $r$ to the magnetic field at a distance $r$ from the centre of the coil on its axis is $\sqrt{x}: 1$. The value of $x$ is $............$