MHT CET202618 April 2026Evening ShiftPhysicsElectromagnetic InductionActual
Two coils A and B have 180 and 360 turns respectively. The current of 1A flows through both the coils. Due to current of 1A in coil A, flux per turn of 0.8 10⁻³ Wb is linked with coil A. Due to current of 1A in coil B, flux per turn of 1 10⁻³ Wb is linked with coil B. The self inductance of coil A is L_A and the self inductance of coil B is L_B . The ratio L_A to L_B is
Options
- A1 5
- B2 5
- C3 2
- D5 2
Correct answer
B. 2 5
Step-by-step solution
The self-inductance of a coil is given by L = N I , where N is the number of turns, is the flux per turn, and I is the current. For coil A: L_A = N_A _A I_A = 180 0.8 10⁻³ 1 = 144 10⁻³ H For coil B: L_B = N_B _B I_B = 360 1 10⁻³ 1 = 360 10⁻³ H The ratio of their self-inductances is: L_A L_B = 144 10⁻³ 360 10⁻³ = 144 360 = 2 5 Answer: 2 5