The current in a coil decreases from $1 \, A$ to $0.2 \, A$ in $10 \, s$. Calculate the coefficient of self-inductance if the induced $e.m.f.$ is $0.4 \, V$. (in $, H$)

The current in an inductor is given by $I = (3t + 8)$ $A$,where $t$ is in seconds. The magnitude of the induced emf produced in the inductor is $12 \ mV$. The self-inductance of the inductor is . . . . . . $mH$.

$A$ current of $0.5 \ A$ is passed through the winding of a long solenoid having $400$ turns. The magnetic flux linked with each turn is $3 \times 10^{-3} \ Wb$. The self-inductance of the solenoid is: (in $H$)

Consider a solenoid carrying current supplied by a $DC$ source with a constant $emf$ containing an iron core inside it. When the core is pulled out of the solenoid,the change in current will:

Current in a coil changes from $4 \,A$ to zero in $0.1 \,s$ and the emf induced is $100 \,V$. The self inductance of the coil is (in $\,H$)

$A$ coil of $Cu$ wire (radius $r$,self-inductance $L$) is bent into two concentric turns,each having a radius of $r/2$. What is the new self-inductance?