A pair of adjacent coils has a mutual inductance of 2.0 H. If the current in one coil changes from 0 to 10 A is 0.5 s, the change of the flux linkage with the other coil is

20 Wb

The correct answer is Option (2) → 20 Wb

Given:

Mutual inductance: $M = 2.0 \, \text{H}$

Change in current: $\Delta I = 10 - 0 = 10 \, \text{A}$

Time interval: $\Delta t = 0.5 \, \text{s}$

Change in flux linkage with the other coil: $\Delta \lambda = M \Delta I$

$\Delta \lambda = 2.0 \times 10 = 20 \, \text{Wb-turns}$

Answer: $\Delta \lambda = 20 \, \text{Wb-turns}$

Note: The formula involving time, ε = M × (dI/dt), is used to calculate induced EMF, not flux linkage. In this question, we are asked for the change in flux linkage, which is given by Δ(Nφ) = M × ΔI. Substituting the values, 2.0 × 10 = 20, so the correct answer is 20 Wb (or Wb-turns). The time interval of 0.5 s is not required here and is likely included to test conceptual clarity.