Flux in coil A = 0.05 mWb = 5 × 10−5 wb = φA No. of turns NA = NB = 1000 Flux linkage in a coil with B = Flux linkage in coil A × 80/100 = 0.8 × 5 × 10−5 = 4 × 10−5 wb =0.04 mwb
The self-inductance is given as L = μN2A/I L ∝ N2 where N is the number of turns of the solenoid A is the area of each turn of the coil l is the length of the solenoid and μ is the permeability constant L1/L2 = N21/N22 0.5/0.5 = N21/N22 N1/N2 = 1
EXAMIANS