Two coils of 100 turns and 200 turns have self inductances 25 mH and 4

1.	Two coils of 100 turns and 200 turns have self inductances 25 mH and 40 mH, respectively. Their mutual inductance is 3 mH. If a 6 mA current in the first coil is changing at the rate of 4 A/s, calculate (a) 2 that links the first coil (b) self induced emf in the first coil (c) Φ21 that links the second coil (d) mutually induced emf in the second coil.
Answer» Data : N₁= 100, N₂ = 200, L₁ = 25 mH, L₂ = 40 mH, I₁ = 6 mA, dI /dt = 4 A/s (a) The flux per unit turn in coil 1, Φ₂₁ = \(\cfrac{L_1I_1}{N_1}\) = \(\cfrac{(25\times10^{-3})(6\times10^{-3})}{100}\) = 1.5 × 10^-6 Wb =1.5 μ Wb (b) The magnitude of the self induced emf in coil 1 is L₁ = \(\cfrac{dI_1}{dt}\) = (25 × 10^-3)(4) = 0.1 V (c) The flux per unit turn in coil 2, Φ₂₁ = \(\cfrac{MI_1}{N_2}\) = \(\cfrac{(3\times10^{-3})(6\times10^{-3})}{200}\) = 90 × 10^-9 Wb = 90 nWb (d) The mutually induced emf in coil 2 is e₂₁ = M \(\cfrac{dI_1}{dt}\) = (3 × 10^-3)(4) = 12 × 10^-3 V = 12 mV

Two coils of 100 turns and 200 turns have self inductances 25 mH and 40 mH, respectively. Their mutual inductance is 3 mH. If a 6 mA current in the first coil is changing at the rate of 4 A/s, calculate (a) 2 that links the first coil (b) self induced emf in the first coil (c) Φ21 that links the second coil (d) mutually induced emf in the second coil.

Answer»

Data : N₁= 100, N₂ = 200, L₁ = 25 mH, L₂ = 40 mH,

I₁ = 6 mA, dI /dt = 4 A/s

(a) The flux per unit turn in coil 1,

Φ₂₁ = \(\cfrac{L_1I_1}{N_1}\) = \(\cfrac{(25\times10^{-3})(6\times10^{-3})}{100}\)

= 1.5 × 10^-6 Wb =1.5 μ Wb

(b) The magnitude of the self induced emf in coil 1 is

L₁ = \(\cfrac{dI_1}{dt}\) = (25 × 10^-3)(4) = 0.1 V

(c) The flux per unit turn in coil 2,

Φ₂₁ = \(\cfrac{MI_1}{N_2}\) = \(\cfrac{(3\times10^{-3})(6\times10^{-3})}{200}\)

= 90 × 10^-9 Wb = 90 nWb

(d) The mutually induced emf in coil 2 is

e₂₁ = M \(\cfrac{dI_1}{dt}\) = (3 × 10^-3)(4) = 12 × 10^-3 V

= 12 mV