The self-inductance of an air-cored solenoid of length $40 \ cm$,diameter $7 \ cm$ having $200$ turns is nearly: (in $\mu H$)

$A$ 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 $\,H$)

$A$ long solenoid has $1000$ turns. When a current of $4\, A$ flows through it,the magnetic flux linked with each turn of the solenoid is $4 \times 10^{-3}\, Wb$. The self-inductance of the solenoid is......$H$

The total number of turns and cross-section area in a solenoid are fixed. However,its length $L$ is varied by adjusting the separation between windings. The inductance of the solenoid will be proportional to

Current in a circuit falls from $5.0 \, A$ to $0.0 \, A$ in $0.1 \, s$. If an average $emf$ of $200 \, V$ is induced,give an estimate of the self-inductance of the circuit in $H$.

Regarding self-inductance:
$A:$ The self-inductance of the coil depends on its geometry.
$B:$ Self-inductance does not depend on the permeability of the medium.
$C:$ Self-induced e.m.f. opposes any change in the current in a circuit.
$D:$ Self-inductance is the electromagnetic analogue of mass in mechanics.
$E:$ Work needs to be done against self-induced e.m.f. in establishing the current.
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