An air-cored solenoid with length \(30 \mathrm{~cm}\), area of cross-section \(25 \mathrm{~cm}^2\) and number of turns 500 carries a current of \(2.5 \mathrm{~A}\). The current is suddenly switched off for a brief time of \(10^{-3} \mathrm{~s}\). How much is the (nearly) average back e.m.f. induced across the ends of the open switch in the circuit?
(Ignore the variation of magnetic field near the ends of solenoid)

Field at the centre of the solenoid with \(N\) turns over length \(l\) is :
\(
B=\mu_0\left(\frac{N}{l}\right) i
\)
The self-flux associated with the coil with area \(A\) is: \(\Phi=N(B A)\) We define self-inductance \(L\) as, \(\Phi=L i\)
Therefore, self-inductance of the solenoid is given by,
\(
L=\frac{\Phi}{i}=\frac{N(B A)}{i}=\frac{N\left[\mu_0\left(\frac{N}{l}\right) i A\right]}{i}=\frac{\mu_0 N^2 A}{l}
\)
Back emf,
\(e=L\left(\frac{d i}{d t}\right)=\left(\frac{\mu_0 N^2 A}{l}\right) \frac{d i}{d t}=\) \(\frac{4 \pi \times 10^{-7} \times 500 \times 500 \times 25 \times 10^{-4}}{0.3} \times \frac{2.5}{10^{-3}} \mathrm{~V}\)
\(\therefore e=6.5 \mathrm{~V}\)