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.