$A$ 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 of $n$ turns,mean radius $R$ and cross-sectional radius $a$ carries current $I$. It is placed on a horizontal table taken as $xy$-plane. Its magnetic moment $m$ is:

$A$ long cylindrical conductor with a large cross-section carries an electric current distributed uniformly over its cross-section. The magnetic field due to this current is:

$A$ 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)

$A$ 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:

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