AP EAMCET20227 Jul 2022Morning ShiftPhysicsElectromagnetic InductionActual
A long solenoid having 100 turns per cm carries a current of 4 A. At the centre of it is placed a coil of 200 turns of cross-sectional area 25 ~cm ^2 having its axis parallel to the field produced by the solenoid. When the direction of the current in the solenoid is reversed with in 0.04 ~s , the induced emf in the coil is
Options
- A0.2 ~V
- B0.4 ~V
- C0.002 ~V
- D0.016 ~V
Correct answer
B. 0.4 ~V
Step-by-step solution
Due to solenoid a magnetic field exists around coil placed at centre Magnetic field intensity produced by solenoid, B= ₀ n I Flux linked with coil is _B=N . B A Where, N= Number of turns in coil A= Area of coil As current in solenoid is reversed, change in flux is _B=2 N B A This change in flux produces an emf in coil given by Induced emf e= Rate of change of flux e= 2 N B A t = 2 N ( ₀ n I ) A t Here N=200, A=25 ~cm ^2=25 10⁻⁴ ~m ^2I= 4 A , n=100 tuns / cm =100 100 turns / m ₀=4 10⁻⁷ ~T mA ⁻¹ and t=0.04 ~s So indu