Answer –

We are given –

Number of turns – 15 turns / cm = 1500 turns / m

Number of turns per unit length is given by n = 1500 turns

The solenoid has a small loop of area which is given by A = 2.0 cm² = 2 × 10⁻⁴ m²

It is also given that the current carried by the solenoid changes from 2 A to 4 A.

Therefore, Change in current in the solenoid becomes

di = 4 – 2 = 2 A

and change in time is given by dt = 0.1 s

therefore, according to Faraday’s law, induced emf in the solenoid will be given by –

$e=\frac{d\phi }{dt}\to (1)$

Where,                                                                  

Phi represents Induced flux through the small loop and is equal to BA.

B is the magnetic field and is given by –

$B={{\mu }_{0}}ni$

Where µ₀ is the permitivitty of free space

Hence, equation (1) can be reduced to –

$e=\frac{d}{dt}(BA)$

$e=A{{\mu }_{0}}n\times \left( \frac{di}{dt} \right)$

$e=2\times {{10}^{-4}}\times 4\pi \times {{10}^{-7}}\times 1500\times \frac{2}{0.1}=7.54\times {{10}^{-6}}V$

Therefore, the induced voltage in the loop is 7.54×10^-6V.