JEE Main 2024 Physics question
The current in an inductor is given by I = (3t + 8) where t is in second. The magnitude of induced emf produced in the inductor is 12 mV. The self- inductance of the inductor mH.
Answer: 4
Solution
ε = L dI/dt = 3L ⇒ L = 12 mV/3 = 4 mH.