$A$ direct current $I$ flows along the length of an infinitely long straight thin-walled pipe. Then the magnetic field is

$A$ long solenoid carrying a current produces a magnetic field $B$ along its axis. If the number of turns per cm are tripled and the current is made $\left(\frac{1}{4}\right)^{th}$,then the new value of the magnetic field will be:

$A$ 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 $(ii)$ $a < x < b$?

$A$ toroid is a long coil of wire wound over a circular core. If $r$ and $R$ are the radii of the coil and toroid respectively,the coefficient of self-induction of the toroid is (The magnetic field in it is uniform and $R >> r$). ($N =$ number of turns of the coil and $\mu_{0} =$ permeability of free space)

$A$ closely wound solenoid $80\; cm$ long has $5$ 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.

$A$ winding wire which is used to frame a solenoid can bear a maximum $10\, A$ current. If the length of the solenoid is $80\, cm$ and its cross-sectional radius is $3\, cm$,then the required length of the winding wire is $(B = 0.2\, T)$.