If a long hollow copper pipe carries a current,then the magnetic field is produced:

The magnetic field $B$ within a solenoid having $n$ turns per meter length and carrying a current of $i$ amperes is given by:

$A$ long solenoid with $ 40 $ turns per cm carries a current of $ 1 \,A $. The magnetic energy stored per unit volume is $ J m^{-3} $. (in $\pi$)

$A$ toroid is:

$A$ $25\,cm$ long solenoid has radius $2\,cm$ and $500$ total number of turns. It carries a current of $15\,A.$ If it is equivalent to a magnet of the same size and magnetization $\vec M$ (magnetic moment per unit volume),then $\left| {\vec M} \right|$ is

$A$ solenoid has a core of a material with relative permeability $400$. The windings of the solenoid are insulated from the core and carry a current of $1 \text{ A}$. If the number of turns is $1000 \text{ per metre}$,the magnetic field $(B)$ is . . . . . . $\text{T}$. (Given: $\mu_0 = 4\pi \times 10^{-7} \text{ SI units}$)