$A$ solenoid has a length of $1 \,m$ and a cross-sectional area of $0.02 \,m^2$. If the number of turns in the solenoid is $5000$, then the self-inductance of the solenoid is: (in $\pi \,H$)

$A$ long solenoid of $N$ turns has a self-inductance $L$ and area of cross-section $A$. When a current $i$ flows through the solenoid,the magnetic field inside it has magnitude $B$. The current $i$ is equal to:

Consider a long solenoid of length $l$ and radius $r$. If $n$ is the number of turns per unit length and $\mu_0$ is the permeability of free space, the inductance of the solenoid is:

$A$ 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).

The self-inductance $L$ of a solenoid of length $l$ and area of cross-section $A$,with a fixed number of turns $N$,increases as:

In an inductor,the current $I$ varies with time $t$ as $I = 5 \,A + 16 \,(A/s) \,t$. If the induced emf in the inductor is $5 \,mV$,the self-inductance of the inductor is .............. $\times 10^{-4} \,H$.