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self inductance , mutual inductance and coeffecient of coupling
Gandhinagar Institute Of
Element Of Electrical Engineering
Active Learning Assignment
When the switch is closed, the battery (source
emf) starts pushing electrons around the circuit.
The current starts to rise, creating an increasing
magnetic flux through the circuit.
This increasing flux creates an induced emf in
The induced emf will create a flux to oppose the
The direction of this induced emf will be opposite the source emf.
This results in a gradual increase in the current rather than an instantaneous one.
The induced emf is called the self-induced emf or back emf.
•When an electric current is passed
through an insulated conducting coil, it
gives rise to a magnetic field in the coil
so that the coil itself behaves like a
•The magnetic flux produced by the
current in the coil is linked with the coil
As the strength of the current in the
coil is changed, the flux linked with the coil
also changes. Under such circumstances an
emf is induced in the coil too. Such emf is
called a self-induced emf and this
phenomenon is known as self-induction.
Direction of the Current
Mutual induction is the phenomenon
of production of induced emf in one
coil due to a change of current in the
At any instant ,
Magnetic Flux linked with the secondary coil ∝ current in the primary coil.
i.e. ᵩ ∝ I
The proportionality constant M is called th mutual inductance or coeffecient of mutual
induction of the two coils.
Any change in the current I sets up and induced emf in the secondary coil which is given
= -M −𝑑I
Consider two long co-axial solenoids S1 and S2, with S2 wound over S1 .
Let l= length of each solenoid
r1, r2=radii of the two solenoids
𝐴 = 𝜋r1
= area of cross section of inner solenoid S1
N1, N2 = number of turns in the two solenoids
First we pass a time varying current I2 through S2 . The magnet field setup inside S2 due to I2
B2 = µ0 n2 I2
where n2 = N2 /l =number of turns per unit length of S2.
Total magnetic flux linked with the inner solenoid S1 is
Φ1 = B2 A N1 = µ0 n2 I2 A N1
∴ Mutual inductance of coil 1 w.r.t coil 2 is
M12 = Φ1 / I2 = µ0 n2 A N1 = µ0 N2 A N1 / l
We now consider the flux linked with the outer solenoid S2 due to the
current I1 in the inner solenoid S1 . The field B1 due to I1 is constant
inside S1 but zero in the annular region between the two solenoids.
B1 = µ0 n1 I1
Where n1 = N1 /l = the number of turns per unit length of S1
The total flux linked with the outer solenoid S2 is
Φ2 = B1 A N2 = µ0 n1 I1 A N2 = µ0 N2 A N1 I1 / l
∴ Mutual inductance of coil 2 w.r.t coil 1 is
M21 = Φ2 / I1 = µ0 N2 A N1 / l
M12 = M21 = M
∴M= µ0 N2 A N1 / l
Thus mutual inductance of two coils is the property of their
combination. It does not matter which one of them functions as
primary or secondary coil.
1. Number of Turns : larger the number of turns , larger the
2. Larger the common cross-sectional area of two solenoids ,
larger will be their mutual inductance.
3. Larger the distance b/w two solenoids , smaller will be the
magnetic flux linked with the secondary coil due to current
in the primary coil. Hence smaller will be the value of M.
Multiplying equations (15) and (17),
But N1Φ1/i1 = Self induced of coil 1 = L1
N2Φ2/i2 = Self induced of coil 2 = L2
... M2 = k1k2L1L2
... M = √(k1k2) √(L1L2)
Let k = √(k1k2)
... M = k √(L1L2) ............(18)
where k is called coefficient of coupling.
... k = M/(√(L1L2)) .........(19)