Question

An air-cored solenoid with length 30 cm, area of cross-section 25 cm2and number of turns 500, carries a current of 2.5 A. The current issuddenly switched off in a brief time of 10–3 s. How much is the averageback 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

The given length of the solenoid, l=30 cm or 0.3 m.

Cross section area of the solenoid, A=25  cm 2  or 25× 10 −4   m 2 .

Number of turns, N=500.

Current in the solenoid, I=2.5 A.

Time for which the current flows, t= 10 −3  s.

The formula for average back emf is,

e= dϕ dt (1)

Change in flux is given by, dϕ=NAB and magnetic field strength is B= μ 0 NI l .

Substitute the values in equation (1).

e= μ 0 N 2 IA lt = 4π× 10 −7 × ( 500 ) 2 ×2.5×25× 10 −4 0.3× 10 −3 = 1.96× 10 −3 0.3× 10 −3 =6.5 V

Thus, the average back emf in the solenoid is 6.5 V.