AP EAMCET202123 Aug 2021Evening ShiftPhysicsElectromagnetic InductionActual
The current in an inductor of self-inductance L=40 mH is to be increased uniformly from 2 ~A to 12 ~A in 8 ~ms . The emf induced in the inductor during this process is
Options
- A50 ~V
- B0.4 ~V
- C40 ~V
- D100 ~V
Correct answer
A. 50 ~V
Step-by-step solution
Given, Self-inductance of inductor, L=40 mH Initial current, I₁=2 ~A Final current, I₂=12 ~A Time interval, d t=8 ~ms Using expression for induced emf, =-L d i d t | |=L d i d t =40 10⁻³ ( 12-2 8 10⁻³ )=50 ~V