A small coil of 10 turns is placed inside a solenoid of length 20 cm and 240 turns carry a current of 10π A The area of small coil is 2.5 cm2 and resistance 4.8 Ω then the current reduces to zero in 25 ms, the value of average induced current is :

Values given for coil :Number of turns of small coil = 10Area of coil= 2.5 cm2 = 2.5 x 10-4m2∆t=25 x 10-3sValues given for solenoidI = 20 x 10-2 m, N = 240I=10π amp, n=24020 x 10-2=12 x 102According to Faraday's law, induced emf E=-dϕdtInduced emf in coil :I=-dϕdtR=(NBA)/ dtR=μ0nINA∆tR  =10 x 4π x 10-7 x 12 x 102 x 10/π x 2.5 x 10-425 x 10-3 x 4.8 =4 x 12 x 10-7 x 102 x 10 x 10-410-3 x 4.8=10-4 I=0.1 mA