From Ampere's circuital law for a long straight wire of circular cross-section carrying a steady current,the variation of magnetic field in the inside and outside region of the wire is :

$A$ long solenoid is fabricated by closely winding a wire of radius $0.5 \, mm$ over a cylindrical frame so that the successive turns nearly touch each other. What is the magnetic field at the centre of the solenoid if it carries a current of $5 \, A$?

$A$ 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 $\pi$)

$A$ long straight wire with a circular cross-section having radius $R$ is carrying a steady current $I$. The current $I$ is uniformly distributed across this cross-section. Then the variation of magnetic field due to current $I$ with distance $r$ $(r < R)$ from its centre will be:

$A$ steady current $I$ flows along an infinitely long hollow cylindrical conductor of radius $R$. This cylinder is placed coaxially inside an infinite solenoid of radius $2R$. The solenoid has $n$ turns per unit length and carries a steady current $I$. Consider a point $P$ at a distance $r$ from the common axis. The correct statement$(s)$ is (are) :
$(A)$ In the region $0 < r < R$,the magnetic field is non-zero.
$(B)$ In the region $R < r < 2R$,the magnetic field is along the common axis.
$(C)$ In the region $R < r < 2R$,the magnetic field is tangential to the circle of radius $r$,centered on the axis.
$(D)$ In the region $r > 2R$,the magnetic field is non-zero.

$A$ current-carrying solenoid is placed vertically and a particle of mass $m$ with charge $Q$ is released from rest. The particle moves along the axis of the solenoid. If $g$ is the acceleration due to gravity,then the acceleration $(a)$ of the charged particle will satisfy: