Question

An air-cored solenoid with length 30 cm, area of cross-section 25 cm² 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.

Answer 1

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

Area of cross-section, A = 25 cm² = 25 × 10⁻⁴ m²

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.