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.

#### Solution

Length of the solenoid, l = 30 cm = 0.3 m

Area of cross-section, A = 25 cm^{2} = 25 × 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" = ("d"phi)/("dt")` .........(1)

Where,

`"d"phi` = Change in flux = NAB …........(2)

Where,

B = Magnetic field strength = `mu_0("NI")/"l"` ......(3)

Where,

`mu_0` = Permeability of free space = 4π × 10^{−7} T m A^{−1}

Using equations (2) and (3) in equation (1), we get

e = `(mu_0"N"^2"IA")/("lt")`

= `(4pi xx 10^-7 xx (500)^2 xx 2.5 xx 25 xx 10^-4)/(0.3 xx 10^-3)`

= 6.5 V

Hence, the average back emf induced in the solenoid is 6.5 V.