The magnetic field intensity $(H)$ at the centre of a long solenoid carrying a current of $2 \ A$ is found to be $1000 \ A/m$. The number of turns per centimeter of the solenoid is: (Use $\mu_0 = 4 \pi \times 10^{-7} \ T \ m \ A^{-1}$)

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

$A$ long solenoid carrying a current produces a magnetic field $B$ along its axis. If the current is tripled and the number of turns per cm is halved, then the new value of the magnetic field will be:

Inductance per unit length near the middle of a long solenoid is ($\mu_0=$ permeability of free space,$n=$ number of turns per unit length,$d=$ the diameter of the solenoid).

Calculate the axial magnetic field of a finite solenoid.

Verify Ampere's law for the magnetic field of a point dipole with dipole moment $\vec{M} = M\hat{k}$. Take $C$ as the closed curve running clockwise along the quarter circle of radius $R$ and center at the origin,in the first quadrant of the $x-z$ plane.