Two inductive coils which are close to each other have a mutual inductance of 0.6 H. Current through one coil is increased from 1 A to 6 A in 0.03 s. The voltage induced in the other coil is:

Correct option is C

​$$\text{The problem provides the mutual inductance } (M), \text{ the initial current } (I_1), \text{ the final current } (I_2), \text{ and the time interval } (\Delta t). \\[10pt]\bullet \ M = 0.6 \text{ H} \\[4pt]\bullet \ I_1 = 1 \text{ A} \\[4pt]\bullet \ I_2 = 6 \text{ A} \\[4pt]\bullet \ \Delta t = 0.03 \text{ s}$$

$$\text{The change in current } (\Delta I) \text{ is calculated as the final current minus the initial current.} \\[8pt]\text{The rate of change of current is } \frac{\Delta I}{\Delta t}. \\[14pt]\Delta I = I_2 - I_1 = 6\,\text{A} - 1\,\text{A} = 5\,\text{A} \\[14pt]\frac{\Delta I}{\Delta t} = \frac{5\,\text{A}}{0.03\,\text{s}}$$

$$\text{The magnitude of the voltage } (V) \text{ induced in the second coil due to the change in current in the first coil is given by the formula} \\[6pt]V = M \left| \frac{dI}{dt} \right| \\[12pt]V = M \times \frac{\Delta I}{\Delta t} \\[14pt]V = 0.6 \, \text{H} \times \frac{5 \, \text{A}}{0.03 \, \text{s}} \\[14pt]V = \frac{3}{0.03} \, \text{V} \\[14pt]V = 100 \, \text{V}$$​​​​