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 e.m.f. induced across the ends of the open switch in the cuicuit ? Ignore the variation in magnetic field near the ends of the solenoid.

Explanation

Length of the solenoid, l = 30 cm = 0.3 m
Area of cross-section, A = 25 `cm^(2) = 25 xx 10^(-4) m^(2)`
Number of turns on the solenoid, N = 500
Current in the solenoid, I = 2.5 A
Current flows for time,` t = 10^(-3)` s
Average back emf, `e=(dphi)/(dt)" "...(1)`
Where,
`dphi`=Change in flex
= NAB … (2)
Where
B = Magnetic field strength
`=mu_(0)(NI)/(l) " "(3)`
Where
`mu_(0)` = Permeability of free space `= 4pi xx 10^(-7) T m A^(-1)`
Using equations (2) and (3) in equation (1), we get
`e=(mu_(0)N^(2)IA)/("lt")`
`=(4pixx10^(-7)xx(500)^(2)xx2.5 xx 25xx 10^(-4))/(0.3xx10^(-3))=6.5 V`
Hence, the average back emf induced in the solenoid is 6.5 V.