Two different coils have self inductance $8mH$ and $2mH$. The current in both coils are increased at the same constant rate. The ratio of the induced emf in the coil is:
A. $4:1$
B. $1:4$
C. $1:2$
D. $2:1$

Hint-Relation between self inductance and electro-motive force (emf’s) is
 $L=-\frac{e}{\left( \frac{\Delta i}{\Delta t} \right)}$
Where $L=$ Self inductance, $e=$ Electromotive force and $\frac{\Delta i}{\Delta t}=$ Rate of change of current

 Complete step-by-step solution:
Self induction:- The phenomenon of electromagnetic induction in which, on changing the current in a coil, an opposing induced emf is set up in that every coil is self induction.
$L=-\frac{e}{\left( \frac{\Delta i}{\Delta t} \right)}$
The S.I unit of self induction is henry (H).
Let us consider two coils, have self induction respectively ${{L}_{1}}$ and ${{L}_{2}}$
${{L}_{1}}=8mH=8\times {{10}^{-3}}H$
${{L}_{2}}=2mH=2\times {{10}^{-3}}H$
Current is increased in both the coils at the same rate. It means that $\left( \frac{\Delta i}{\Delta t} \right)$ is the same for both.
For first coil
${{L}_{1}}=8mH=8\times {{10}^{-3}}H$
Use the formula
${{e}_{1}}=-{{L}_{1}}\left( \frac{\Delta i}{\Delta t} \right)$
${{e}_{1}}=-8\times {{10}^{-3}}\left( \frac{\Delta i}{\Delta t} \right)$ …………….(1)
For second coil
${{L}_{2}}=2\times {{10}^{-3}}H$
${{e}_{2}}=-{{L}_{2}}\left( \frac{\Delta i}{\Delta t} \right)$
${{e}_{2}}=-2\times {{10}^{-3}}\left( \frac{\Delta i}{\Delta t} \right)$ …………..(2)
The equation (1) divided by equation (2)
$\frac{{{e}_{1}}}{{{e}_{2}}}=\frac{-8\times {{10}^{-3}}\times \left( \frac{\Delta i}{\Delta t} \right)}{-2\times {{10}^{-3}}\times \left( \frac{\Delta i}{\Delta t} \right)}$
$\frac{{{e}_{1}}}{{{e}_{2}}}=\frac{4}{1}$
${{e}_{1}}:{{e}_{2}}=4:1$

Note:
 The value of $\left( \frac{\Delta i}{\Delta t} \right)$ does not take different for different coils because in this question the rate of current increases is constant.