A long solenoid with 15 turns per \(\mathrm{cm}\) has a small loop of area \(2.0 \mathrm{~cm}^2\) placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 Ato \(4.0 \mathrm{~A}\) in \(0.1 \mathrm{~s}\), the induced emf in the loop while the current is changing is nearly [Take \(\pi=3.14\) ]

The magnetic field produced inside the solenoid is \(B=\mu_0 n I\)
If \(A\) is the area of the loop placed inside the solenoid, then the magnetic flux linked with the loop is given by,
\(\phi=B A=\mu_0 n I A\)
If \(e\) is the induced e.m.f. produced due to change in current through the solenoid, then
\(e=-\frac{d \phi}{d t}=-\frac{d}{d t}\left[\mu_0 n I A\right]=-\mu_0 \times n \times A \times \frac{d I}{d t}\)
Given: Number of turns per unit length of the solenoid,
\(n=15\) turns \(\mathrm{cm}^{-1}=1500\) turns \(\mathrm{m}^{-1}\)
Given: \(A=2 \mathrm{~cm}^2=2 \times 10^{-4} \mathrm{~m}^2\) and \(\frac{d I}{d t}=\frac{4-2}{0.1}=20 \mathrm{As}^{-1}\)
\(\therefore e=-4 \pi \times 10^{-7} \times 1500 \times 2 \times 10^{-4} \times 20=\) \(-7.54 \times 10^{-6} \mathrm{~V}\).