Mutual inductance

In mutual inductance there are two coils, current is passed in one coil, as current increases there is change in the flux, and as a result current is induced in the second coil.

Consider coil 1 connected to battery and coil 2 is connected to the galvanometer as shown in the figure.

Mathematically:-

• Φ₍₂₎∝ I₍₁₎
• =>Φ₍₂₎= MI₍₁₎where M = constant of proportionality known as Mutual Inductance.
• Induced emf in coil 2 e=-(dΦ₍₂₎/dt)
• =>e =-d/dt(MI ₍₁₎) where I current flowing in coil (1).
• Therefore e =-d/dt (M I ₍₁₎)

 Mutual Inductance between long co-axial solenoids

• Co-axial solenoids means the centres of both the solenoids are same.
• Radius of the smaller solenoid (1) =r₁. Number of turns in smaller solenoid= N₁.
• Radius of the bigger solenoid (2) = r₂.Number of turns in bigger solenoid= N₂.

Case 1:-

• Current flowing in the bigger solenoid = I₂, as a result magnetic flux Φ₁will be induced in smaller solenoid.
• Therefore N₁Φ₁∝I₂
• =>N₁Φ₁=M₁₂I₂ equation(1)
  • where M₁₂= mutual inductance of 1 w.r.t 2
• Magnetic field due to I₂in (bigger solenoid 2) B =μ₀n₂I₂
• => B =(μ₀n₂I₂)/l_ength
• Total Flux N₁Φ₁=N₁ BA₁
• => =(N₁A₁μ₀N₂I₂)/l_engthequation(2)
• From equation(1) and (2)
• M₁₂I₂= (N₁A₁μ₀N₂I₂)/l_ength
• =>M₁₂= (μ₀N₁ N₂A)/ l_ength equation(a)

Case 2:-

• Current I₁flowing through solenoid (1) this will result in flux solenoid (2)
• Total flux N₂Φ₂= M₂₁I₁ equation(3)
• Also total flux N₂Φ₂= N₂ B₁ A₁
  • where B =magnetic field due to smaller solenoid;
  • B₁=μ₀n₁I₁;
  • =(μ₀N₁I₁)/l_ength
• =>N₂Φ₂= N₂((μ₀N₁I₁)/l_ength)A₁ equation(4)
• Comparing (3) and (4)
• M₂₁I₁= N₂((μ₀N₁I₁)/l_ength)A₁
• M₂₁= (μ₀N₂ N₁A₁)/ l_ength equation(b)
• Comparing (a) and (b)
• ThereforeM₁₂= M₂₁