In a coil, the current changes from -2 A to +2 A in 0.2 s and induces an emf of 0.1 V. The self inductance of the coil is :

Solution

Physics sir · 3 steps

Current going from −2 A to +2 A is a change of 4 A, not zero, so mind the sign flip.

1. Change in current. The current reverses direction, so Δ I = (+2) - (-2) = 4 A. This happens in Δ t = 0.2 s, so (Δ I)/(Δ t) = 4/0.2 = 20 A s⁻¹.

   |ε| = L (Δ I)/(Δ t)

2. Solve for L. L = 0.1/20 = 5 × 10⁻³ H.
3. So. The self inductance is 5 mH, which is option (d).

The idea to keep. Always find Δ I as final minus initial with signs. A current that reverses changes by the sum of the two magnitudes, and L is just emf divided by the rate of change of current.

Answer from the NTA final answer key. Solution by PrepWiser.