Mutual Inductance

In mutual inductance there are 2 coils, current is passed in one coil, as current increases there is change in the flux, and as aresult current is induced in the second coil.

Consider coil 1 connected to battery and coil 2 is connected to the galvanometer.

• When the key is pressed attached to the coil 1 the current starts flowing, when the current starts increasing flux linked also starts increasing.
• Because of the increase in the flux linked with the coil1, the flux of coil 2 also increases.
• There is change in the flux of the coil 2 as a result emf is induced in the coil 2.
• Because of the induced emf induced current will be there in coil2.
• This induced current opposes the increase of the current in coil 1.

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₎)

• M (mutual inductance) depends on:-

1. Geometry of both coils.
2. Distance between coils.
3. Orientation of coils.

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 Φ₁ will 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 I₂ in (bigger solenoid 2) B =μ₀n₂I₂
• => B =(μ₀n₂I₂)/l_ength
• Total Flux N₁Φ₁ =N₁ BA₁
• => =(N₁A₁μ₀N₂I₂)/l_ength equation(2)
• From equation(1) and (2)
• M₁₂I₂ = (N₁A₁μ₀N₂I₂)/l_ength
• =>M₁₂ = (μ₀N₁ N₂A)/ l_ength equation(a)

Case 2:-

• Current I₁ flowing 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

Problem:- A solenoid of length 50cm with 20 turns per cm and area of cross-section 40cm² completely surrounds another co-axial solenoid of the same length, area of cross section 25cm² with 25 turns per cm. Calculate the mutual inductance of the system?

Answer:- Let the outer solenoid be 1 and inner solenoid be 2.

Length l₁ =50cm=50x10^-2m

n₁ =2000 turns /m

A₁ = 40 x10^-4 m²

Length l₂=50x10^-2m

n₂ =2500 turns/m

A₂=25x10^-4 m²

M₁₂=M₂₁

M₁₂= (μ₀N₁N₂A₂)/l_ength =μ₀n₁n₂A₂l_ength

=4x3.14x10^-7x2000x2500x50x10^-2x25x10^-4

=7.85x10^-3 H

Problem:- (a)  A toroidal solenoid with an air core has an average radius of 15cm, area of cross-section 2cm² and 1200 turns. Obtain the self-inductance of the toroid. Ignore field variations across the cross-section of toroid.

(b) A second coil of 300 turns is wound closely on the toroid above. If the current in the primary coil is increased from 0 to 2A in 0.05s.Obtain the induced emf in the second coil.

Answer:- (a)  Radius r₁=15x10^-2m

Area A₁=2x10^-4m²

N₁=1200

B=μ₀nI= (μ₀N₁I)/ (2πr₁)

Total flux: -N₁xBxA₁=N₁x(μ₀xN₁I/2πr₁) x A₁

Also Total flux =LI

LI =μ₀xA₁x N₁²I/ (2πr₁) 

= (4πx10^-7x (1200)²x 2x10^-4)/ (2πx15x10^-2)

=2.304x10^-4 H

The self-inductance of the toroid is 2.304x10^-4 H.

(b) A₁=2x10^-4m²

r₁=15x10^-2m

N₁=1200

N₂ =300

(dI₁/ dt) = (I_f –I_i)/dt = (2-0)/ (0.05) =40A/s

Total flux in 2:- Φ₂ =N₂xB₁xA₁

=N₂(μ₀n₁I₁)/A₁  equation (1)

Also,Φ₂ =MI₁ equation (2)

=> M=N₂xμ₀xn₁xA₁

e₂ =M (dI₁/dt) =μ₀N₂n₁A₁x40

=4πx10^-7x 300x (N₁/2πr₁) x2x10^-4x40

=0.023V

The emf induced in the second coil is 0.023V.