The current in a coil changes from 1 mA to 5 mA in 4 millisecond. If the coefficient of self - induction of the coil (L) is 10 mH then find the magnitude of self-induced emf is:
a. 10mV
b. 5mV
c. 2.5mV
d. 1mV

Hint: When there is change in the current then an induced potential difference is generated. For finding the emf of self-inductance, we use the formula
\[\varepsilon = - L\dfrac{{di}}{{dt}}\]
Where
$\varepsilon $= Emf of self-inductance
$L$ = Coefficient of self-inductance
$\dfrac{{di}}{{dt}}$= Rate of change of current in the coil
In the above question, 1 mA and 5 mA are the initial and final values of current respectively. The total time taken to change the current is 4 millisecond. Hence the rate of change of current is the ratio of difference of current and total time.

Step by step solution:
The induced emf is generated by two methods self-inductance as well as mutual-inductance. In this question self-induction gives rise to induced current and hence induced emf.
Self-induced emf is given by,
\[\varepsilon = - L\dfrac{{di}}{{dt}}\]

According to question,
$\dfrac{{di}}{{dt}} = \dfrac{{1 - 5}}{4}A{s^{ - 1}} \Leftrightarrow \dfrac{{di}}{{dt}} = - 1A{s^{ - 1}}$

On substituting the values of L and $\dfrac{{di}}{{dt}}$ we get,
$\varepsilon = - \left( {10mH} \right)\left( { - 1A{s^{ - 1}}} \right)$
$\varepsilon = 10mV$

Hence the induced emf through the self-inductance is 10 mV.
Therefore the correct option is A.

Note: The induced emf can be generated by self-induction as well as mutual induction. Self-induction is defined as the inductance of voltage by changing the current in the self coil. In the question both current and time are in the same unit so their respective units cancel out to become both in S.I. unit. The negative sign indicates that the self-induced emf always opposes the change in the flux.