$A$ solenoid with an air core has length $48 \pi \ cm$,area of cross section $12 \ cm^2$ and $1200$ turns. The self-inductance of the solenoid is: (in $mH$)

An $e.m.f.$ of $12$ $V$ is induced in a coil when the current in it changes at the rate of $48$ $A/min$. The self-inductance of the coil is:

Consider a current in a circuit falls from $6.0 \,A$ to $1.0 \,A$ in $0.2 \,s$. If an average emf of $150 \,V$ is induced by the circuit,then the self inductance of the circuit is (in $\,H$)

If a current of $10 \, A$ flows in one second through a coil,and the induced $e.m.f.$ is $10 \, V$,then the self-inductance of the coil is:

$A$ graph of magnetic flux ( $\phi$ ) versus current $(I)$ is shown for four inductors $A, B, C, D$. The smaller value of self-inductance is for inductor:

Why is self-induced $emf$ known as back $emf$?