Class 12 Physics Electromagnetic Induction Self and Mutual Inductance

Self and Mutual Inductance

  • Consider a coil (1) which is connected to a battery and there is a key.
  • There is another coil (2) which is connected to galvanometer.
  • When the key is closed, the current flows through coil (1) there is a change in the flux of coil (1) because of which self-induced emf will be produced.
  • Because of current flow in coil(1) there will be change in the magnetic flux of coil(2) because of which induced emf will be produced in coil(2).
  • As a result mutual induction takes place in coil(2).
  • This shows both self-inductance and mutual inductance takes place simultaneously.
  • Therefore flux linked to coil 1:-N1Φ1 =L1I1 +M12I2
  • Induced emf in coil (1) e= -L1(dI1/dt) - M12(dI2/dt)
    • Where(-)ive sign shows it opposes the growth of current.


Problem:- Two coils have mutual inductance of 1.5 H. If the current in the primary circuit is raised to 5A in 1 millisecond after closing the circuit, what is the emf induced in the secondary?


M=1.5 H

Ii=0 A, If=5A, dt =1ms =10-3sec

e2 =-M (dIi/dt)

=1.5 x ((If-Ii)/dt)

=1.5 x ((5-0)/10-3)


Problem:- Two concentric circular coils, one of small radius r1 and the other of large radius r2, such that r1<< r2, are placed co-axially with centres coinciding. Obtain the mutual inductance of the



 Let a current I2 flow through the outer circular coil. The field at the centre of the coil is

B2 = μ0I2 / 2r2. Since the other co-axially placed coil has a very small radius, B2 may be considered constant over its cross-sectional area. Hence,

Φ1 = πr12 B2

=I2 0 π r12)/ (2r2)

= M12I2

Thus, M12 = (μ0 π r12)/ (2r2)

Using equation M12=M21= (μ0 π r12)/ (2r2)

Note that we calculated M12 from an approximate value of Φ1, assuming the magnetic field B2 to be uniform over the area π r12. However, we can accept this value because r1 << r2.

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