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

Explanation

Initial magnetic flux through a given solenoid is,
`Phi_1=LI`
`=(mu_0N^2A)/l I(because L=(mu_0N^2A)/l)`
`=((4pixx10^(-7))(500)^2(25xx10^(-4))(2.5))/(0.3)`
`therefore Phi_1=6.542xx10^(-3)` Wb
Finally,
`Phi_2=0 (because I=0)`
Average induced emf,
`lt epsilon gt = -(DeltaPhi)/(Deltat)`
`therefore ltepsilongt=-((Phi_2-Phi_1)/(Deltat))`
`therefore lt epsilon gt =(Phi_1-Phi_2)/(Deltat)`
`=(6.542xx10^(-3)-0)/10^(-3)`
`therefore lt epsilon gt` = 6.542 V