When current flowing in a coil changes from 3A to 2A is one millisecond, 5volt emf is induced in it. The self-inductance of the coil will be:
A) Zero
B) 5kh
C) 5h
D) 5mh

Hint: Self-inductance is the property of a coil or a circuit to oppose the change in the current through which it is generated. It is also known as back emf. Almost in every circuit there would be a back emf generated in the coil. Apply the formula for inductance and solve for self-inductance.

Complete step by step solution:
Find out the self-inductance:
${V_L} = - L\dfrac{{di}}{{dt}}$ ;
${V_L}$= Induced voltage in volts;
$L$= Inductance;
$di$= Change in current;
$dt$= Change in time;
Put in the given values:
${V_L} = - L\left( {\dfrac{{2 - 3}}{{{{10}^{ - 3}}}}} \right)$;
$ \Rightarrow {V_L} = - L\left( {\dfrac{{ - 1}}{{{{10}^{ - 3}}}}} \right)$;
The voltage induced is 5 volts (${V_L}$= 5 volts), put it in the above equation:
$ \Rightarrow 5 = L\left( {\dfrac{1}{{{{10}^{ - 3}}}}} \right)$;
Write the above equation in terms of L:
$ \Rightarrow 5 \times {10^{ - 3}} = L$;
$ \Rightarrow L = 5mH$;

Final answer is option D. The self-inductance of the coil will be 5mH.

Note: Here, there are more than two formulas for self-inductance. The first formula for self-inductance is in relation with the magnetic flux, the inductance of the coil is proportional to the negative of the magnetic flux passing per turn in the coil and is inversely proportional to the difference in time. The second formula is in relation with the voltage emf, inductance, current and time. The voltage induced emf is proportional to the difference in the current times the inductance and is inversely proportional to the difference in time. We have to use the second formula for voltage induced emf.