$A$ long solenoid has $500$ turns. When a current of $2 \,A$ is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3} \,Wb$. The self-inductance of the solenoid is: (in $\,H$)

$A$ toroid is a long coil of wire ($N$ turns) wound over a circular core. The coefficient of self-induction of the toroid is [The magnetic field in it is uniform and $R >> r$,where $r=$ radius of wire,$R=$ radius of coil] ($\mu_0=$ permeability of free space).

When the current through a solenoid increases at a constant rate, the induced current

The self-inductance of a coil having $500$ turns is $50\,mH$. The magnetic flux through the cross-sectional area of the coil when the current through it is $8\,mA$ is found to be:

The self-inductance of a coil is $L$. Keeping the length and area the same,the number of turns in the coil is increased to four times. The self-inductance of the coil will now be

The ratio of the number of turns per unit length of two solenoids $A$ and $B$ is $1: 3$ and the lengths of $A$ and $B$ are in the ratio $1: 2$. If the two solenoids have the same cross-sectional area,the ratio of the self-inductances of the solenoids $A$ and $B$ is