Number of turns on the solenoid – 15 turn / cm = 1500 turn / m

Number of turns per unit length, n = 1500 turns

The solenoid has a small loop of the area, $A=2.0c{m}^2=2\times {10}^{-4}{m}^2$

The current carried by the solenoid changes from 2 A to 4 A.

Therefore, Change in current in the solenoid, di = 4 – 2 = 2 A

Change in time, dt = 0.1 s

Induced emf in the solenoid is given by Faraday's law as:

$\mathrm{ \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; e }= \frac{d\varphi }{dt} \dots \left(1\right)$

Where,

$\varphi$ = Induced flux through the small loop  = BA                                 . . . (2)

B = Magnetic fieldnullni

null = Permeability of free space $$$=4\pi \times {10}^{-7}H/m$

Hence, equation (1) can be reduced to:

$$$\begin{array}{l}\mathrm{e}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{BA}\right)\\ ={\mathrm{A\mu }}_0\mathrm{n}\times \left(\frac{\mathrm{di}}{\mathrm{dt}}\right)\\ =2\times {10}^{-4}\times 4\mathrm{\pi }\times {10}^{-7}\times 1500\times \frac{2}{0.1}\\ =7.54\times {10}^{-6}\mathrm{V}\end{array}$

Hence, the induced voltage in the loop is 7.54 ×${\mathrm{}}^{}$${10}^{-6}$V.