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

$A$ coil and a bulb are connected in series with a $12 \,V$ direct current $(DC)$ source. $A$ soft iron core is now inserted into the coil. Then,

$A$ long solenoid has $1500$ turns. When a current of $3.5 \, A$ flows through it, the magnetic flux linked with each turn of the solenoid is $2.8 \times 10^{-3} \, Wb$. The self-inductance of the solenoid is (in $ \, H$)

Two coils $A$ and $B$ have $180$ and $360$ turns respectively. $A$ current of $1 \text{ A}$ flows through both the coils. Due to a current of $1 \text{ A}$ in coil $A$, a flux per turn of $0.8 \times 10^{-3} \text{ Wb}$ is linked with coil $A$. Due to a current of $1 \text{ A}$ in coil $B$, a flux per turn of $1 \times 10^{-3} \text{ Wb}$ is linked with coil $B$. The self-inductance of coil $A$ is $L_A$ and the self-inductance of coil $B$ is $L_B$. The ratio $L_A$ to $L_B$ is:

The following figure represents two bulbs $B_1$ and $B_2$, a resistor $R$, and an inductor $L$. When the switch $S$ is turned off, which of the following statements is true?

What will be the self-inductance of a coil of $100$ turns if a current of $5 \ A$ produces a magnetic flux of $5 \times 10^{-5} \ Wb$?