For a long current-carrying solenoid,the magnetic field inside is $0.6 \ T$. The magnetic energy per unit volume is . . . . . . .

$A$ solenoid of $1000$ turns is wound uniformly on a glass tube $4 \text{ m}$ long and $0.3 \text{ m}$ in diameter. The magnetic intensity at the centre of the solenoid when a current of $4 \text{ A}$ flows through it is

The region inside a current-carrying toroid is filled with a material (Niobium) having magnetic susceptibility $\chi = 2.6 \times 10^{-5}$. The percentage increase in the magnetic field in the presence of Niobium over that without it is:

$A$ 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:

Consider two idealized systems: $(i)$ a parallel plate capacitor with large plates and small separation,and $(ii)$ a long solenoid of length $L \gg R$,where $R$ is the radius of the cross-section. In $(i)$,$E$ is ideally treated as a constant between the plates and zero outside. In $(ii)$,the magnetic field is constant inside the solenoid and zero outside. These idealized assumptions,however,contradict fundamental laws as follows:

$A$ closely wound solenoid of $1200$ turns and area of cross-section $5 \,cm^2$ carries a current. If the magnetic moment of the solenoid is $1.2 \,J \,T^{-1}$, then the current through the solenoid is (in $\,A$)