Class 12 Physics Electromagnetic Induction Mutual Inductance

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.


  • 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)/length
  • Total Flux N1Φ1 =N1 BA1
  • => =(N1A1μ0N2I2)/length equation(2)
  • From equation(1) and (2)
  • M12I2 = (N1A1μ0N2I2)/length
  • =>M12 = (μ0N1 N2A)/ length 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;
    • B10n1I1;
    • =(μ0N1I1)/length
  • =>N2Φ2 = N2((μ0N1I1)/length)A1 equation(4)
  • Comparing (3) and (4)
  • M21I1 = N2((μ0N1I1)/length)A1
  • M21 = (μ0N2 N1A1)/ length 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= (μ0N1N2A2)/length0n1n2A2length


=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


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



N2 =300

(dI1/ dt) = (If –Ii)/dt = (2-0)/ (0.05) =40A/s

Total flux in 2:- Φ2 =N2xB1xA1

=N20n1I1)/A1  equation (1)

Also,Φ2 =MI1 equation (2)

=> M=N20xn1xA1

e2 =M (dI1/dt) =μ0N2n1A1x40

=4πx10-7x 300x (N1/2πr1) x2x10-4x40


The emf induced in the second coil is 0.023V.

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