$A$ long solenoid produces a magnetic field $B$ on its axis. If it is cut into four equal parts and,for the same current,half the number of turns are wound on any one piece,the value of the magnetic field on its axis becomes:

Statement-$1$: Ampere's law can be used to find the magnetic field due to a finite length of a straight current-carrying wire.
Statement-$2$: The magnetic field due to a finite length of a straight current-carrying wire is symmetric about the wire.

The figure shows the cross-sectional view of a hollow cylindrical conductor with inner radius $R$ and outer radius $2R$,carrying a uniformly distributed current $i$ along its axis. The magnetic induction at point $P$ at a distance $3R/2$ from the axis of the cylinder will be

$A$ long wire carrying a current of $18 \,A$ is kept along the axis of a long solenoid of radius $1 \,cm$. The magnetic field due to the solenoid is $8.0 \times 10^{-3} \,T$. The magnitude of the resultant magnetic field at a point $0.6 \,mm$ from the solenoid axis is (Assume $\mu_0 = 4 \pi \times 10^{-7} \,Tm/A$):

$A$ closely wound solenoid $120 \ cm$ long has $4$ layers of windings of $400$ turns each. The diameter of the solenoid is $1.8 \ cm$. If the current carried is $8.0 \ A$,estimate the magnitude of $B$ inside the solenoid near its centre.

$A$ coaxial cable consists of an inner wire of radius $a$ surrounded by an outer shell of inner and outer radii $b$ and $c$ respectively. The inner wire carries an electric current $i_o$,which is distributed uniformly across its cross-sectional area. The outer shell carries an equal current in the opposite direction,also distributed uniformly. What will be the ratio of the magnetic field at a distance $x$ from the axis when $(i)$ $x < a$ and $(ii)$ $a < x < b$?