In a coil, the current changes form –2 A to +2A in 0.2 s and induces an emf of 0.1 V. The self-inductance of the coil is :

Calculation: 

(Emf)_induced = \(-\mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}\)

In magnitude form, 

⇒|Emf_ind| = \(\left|(-) \mathrm{L} \frac{\mathrm{di}}{\mathrm{dt}}\right|\)

⇒ 0.1 = \(\frac{(\mathrm{L})[+2-(-2)]}{0.2}\)

⇒ \(\mathrm{L}=\frac{0.1 \times 0.2}{4}=5 \mathrm{mH}\)

∴ The Correct answer is Option (1): 5 mH