When a current of 4 ~A changes to 8 ~A in 0.6 ~s in a primary coil it induces an e. m. f. of 50 mV in secondary coil. The mutual inductance between two coils is

Step-by-step solution

Change in current in primary coil, di ₁=(8-4) A =4 ~A Time taken, dt =0.6 ~s Induced e. m. f in secondary coil, E ₂=50 10⁻³ ~V So, mutual inductance is, M= E₂ di ₁ / dt = E ₂ dt di ₁ = 50 10⁻³ ~V 0.6 ~s 4 ~A =7.5 10⁻³ H =7.5 mH