Question

Coil 1 has L₁ = 37 mH and N₁ = 122 turns. Coil 2 has L₂ = 50 mH and N₂ = 206 turns. The coils are fixed in place; their mutual inductance M is 4.3 mH. A 9.4 mA current in coil 1 is changing at the rate of 6.7 A/s. (a) What magnetic flux Φ₁₂ links coil 1, and (b) what self-induced emf appears in that coil? (c) What magnetic flux Φ₂₁ links coil 2, and (d) what mutually induced emf appears in that coil?

Answer 1

For coil 1

L₁ = 37 mH= 37*10^-3 H

N₁ = 122 turns

For Coil 2

L₂ = 50 mH = 50*10^-3 H

N₂ = 206 turns

mutual inductance M  = 4.3 mH = 4.3*10^-3 H

i1 = 9.4 mA = 9.4*10^-3 A

Part(a)

magnetic flux Φ12 Links to coil 1

Φ12 = L1i1/N1 = (37*10^-3)( 9.4*10^-3)/ 122

Φ12 = 2.85*10^-6 wb or 2.85 wb

Part(b)

self-induced emf

E1 = L1*di/dt

E1 = ( 37*10^-3 ) *6.7

E1 = 0.2479 V

Part(c)

magnetic flux Φ21 links coil 2

Φ21 = M*i1 /N2

Φ21 = (4.3*10^-3)*(9.4*10^-3) / 206

Φ21 = 1.96*10^-7 wb or 0.96 wb

Part(d)

mutually induced emf

E12 = M*di/dt

E12 = (4.3*10^-3)*6.7

E12 = 0.028 v

Good Luck !!!