22.5 MUTUAL INDUCTION AND MUTUAL INDUCTANCE
22.5.1 Mutual Induction
 |
| Figure 22.64 Magnetic field lines produced by current $I_1$ in coil 1 pass through coil 2, showing magnetic coupling. A change in current in coil 1 causes a change in magnetic flux in coil 2. |
Consider two coils which are placed close to each other. A current I1 flows through coil 1. Hence, a magnetic field is produced, whose magnetic field lines pass through coil 2, as shown in Figure 22.64. Hence, the two coils are magnetically coupled, even though they are not physically linked to each other. If current I1 changes with time, then the magnetic flux through coil 2 will change with time simultaneously. Due to the change of flux linkage through coil 2, an emf E2 is induced in coil 2. This is in accordance to Faraday’s law of electromagnetic induction. In other words, a change of current in one coil leads to the production of an induced emf in a second coil which is magnetically linked to the first coil. This process of producing an induced emf in one circuit or circuit component X due the change of current in another circuit or circuit component Y which is magnetically coupled to X is known as mutual induction.
22.5.2 Emf Induced in Each Magnetically Coupled Coil
Refer to coil 1 and coil 2 mentioned above. According to Faraday’s law of electromagnetic induction, the induced emf E2 produced in coil 2 due to the change of current I1 in coil 1 is
Emf induced in coil 2 (E2) ∝ − rate of change of current in coil 1 (dI1/dt)
E2 ∝ − dI1/dt
or,
E2 = −M21 dI1/dt
where M21 is a constant known as the mutual inductance of the two coils.
Next, suppose there is no current in coil 1 but there is a current I2 flowing in coil 2. If the current changes with time, mutual induction occurs and an emf E1 will be induced in coil 1. If the rate of change of current I2 is dI2/dt, then
E1 ∝ − dI2/dt
or,
E1 = −M12 dI2/dt
where M12 is the mutual inductance of the two coils. If the geometric properties like sizes, shapes and relative positions of the coils are not changed, then we can show that
M21 = M12 = M
Hence, we have
E1 = −M (dI2/dt) and E2 = −M (dI1/dt)
Note: We need to remember that
(a) dI1/dt in coil 1 induces E2 in coil 2
(b) dI2/dt in coil 2 induces E1 in coil 1
Figure 22.65 shows how the emf induced in one coil is related to the rate of change of current in the other coil.
 |
Figure 22.65 The diagram shows that a changing current in one coil produces an induced emf in the other coil. Specifically, a rate of change of current in coil 1 induces an emf in coil 2, and similarly, a changing current in coil 2 induces an emf in coil 1, demonstrating the principle of mutual induction. |
22.5.3 Mutual Inductance
1. Definition
The mutual inductance between two magnetically coupled circuit components is defined as the property of the circuit component which can induce an emf in the neighbouring magnetically coupled component.
2. Unit
Mutual inductance is also measured in henry.
3. Mutual inductance between two magnetically coupled coaxial coils
 |
| Figure 22.66 Two coils are arranged with their axes coinciding, forming a coaxial system. When the current $I_1$ in coil 1 changes with time, it produces a changing magnetic field that passes through coil 2. This changing magnetic flux induces an emf $I_2$ in coil 2 according to Faraday’s law of electromagnetic induction. |
Figure 22.66 shows two coils whose axes coincide with each other. A current I1 which flows through coil 1 changes with time. The emf E2 induced in coil 2 due to the current changing in coil 1 is given by
E2 = −M (dI1/dt)
According to Faraday’s law of electromagnetic induction, we have
E2 = −N2 (dΦ2/dt)
where N2, Φ2 are the number of turns of coil 2 and the magnetic flux through coil 2 respectively. Hence, we have
M (dI1/dt) = N2 (dΦ2/dt)
Since M and N2 are constant, we have
M I1 = N2 Φ2
or,
M = (N2 Φ2) / I1
Similarly, we can show that
M I2 = N1 Φ1
or,
M = (N1 Φ1) / I2
Notice that the magnetic flux Φ2 through coil 2 is produced by the current I1 in coil 1. If the geometrical properties of the coils are not changed, then we have
Φ2 ∝ I1
Φ2 = kI1
where k is a constant which is determined by the geometrical properties of the coils. We have
M = (N2Φ2) / I1
= (N2kI1) / I1 = N2k
This expression indicates the mutual inductance of the magnetically coupled coaxial coils is determined by the geometrical properties of the coils.
EXAMPLE 22.35
Two coils, X and Y, are magnetically coupled. The emf induced in coil Y is 2.5 V when the current flowing through coil X changes at the rate of 5.0 A s−1. Determine
- the mutual inductance of the coils.
- the emf induced in coil X if there is a current flowing through coil Y which changes at the rate of 1.5 A s−1.
Answer
(a)
|Ey| = M (dIx/dt)
2.5 = M(5.0)
M = 0.50 H
(b)
|Ex| = M (dIy/dt)
= (0.50)(1.5) = 0.75 V
EXAMPLE 22.36
Two coils, X and Y, are magnetically coupled. Coil Y has 50 turns. A current of 2.5 A flowing through coil X produces a magnetic field so that magnetic flux of 35 mWb passes through coil Y.
- Determine the mutual inductance of the coils.
- The current of 2.5 A which flows through coil X changes to 4.5 A in 2.0 ms. Estimate the average emf induced in coil Y in the time interval of 2.0 ms.
Answer
(a)
MI1 = NΦy
M(2.5) = (50)(35 × 10−3)
M = 0.70 H
(b)
|Ey| = M (dIx/dt)
= M (ΔIx/Δt)
= (0.70) ((4.5 − 2.5) / (2.0 × 10−3)) = 700 V
EXAMPLE 22.37
Two coils, X and Y, have mutual inductance of 550 mH. Determine the rate of change of magnetic flux linkage through coil Y at the instant when the current flowing through coil X changes at the rate of 5.5 A s−1.
Answer
At the instant when the current flowing through coil X changes at the rate of 5.5 A s−1, an emf is induced in coil Y, which is given by
|Ey| = M dIx/dt
= (0.550)(5.5) = 3.03 V
According to Faraday’s law of electromagnetic induction, we have
|Ey| = Ny dΦy/dt
= rate of change of flux linkage through coil Y
Hence, the rate of change of flux linkage through coil Y is 3.0 Wb s−1.
EXAMPLE 22.38
 |
| Figure 22.67 A long ideal air-core solenoid is connected in series with a 12 V battery and a 20 Ω resistor. A secondary coil is tightly wound around the central portion of the solenoid, allowing magnetic coupling so that changes in current in the solenoid can induce an emf in the coil. |
Explain the terms mutual induction and mutual inductance of a combination of two magnetically linked coils.
A long ideal air-core solenoid is connected in series to a battery of emf 12 V and a resistor of 20 ohm, as shown in Figure 22.67. A coil is tightly wound over the central portion of the solenoid.
(a) Derive an expression for the mutual inductance of the coil-solenoid combination.
(b) The solenoid has a diameter of 5.0 cm and 60 turns per cm of length. The coil has 100 turns. Determine
(i) the mutual inductance of the combination
(ii) the magnetic flux linkage through the coil
(iii) the average emf induced in the coil when the circuit is switched off and the current drops to zero in 20 ms.
Answer
(a) Suppose current I flows through the solenoid. The current produces a magnetic field. Hence there is magnetic flux Φcoil passing through the coil of Ncoil turns. The mutual inductance M of the coil-solenoid combination is given by
M = Ncoil Φcoil / I
The magnetic flux through the coil Φcoil is equal to the magnetic flux through the central portion of the solenoid and is given by
Φcoil = BA cos 0°
where B is the magnetic flux density in the central portion of the solenoid. We have
B = μ0 n I
Hence,
M = Ncoil Φcoil / I
= Ncoil BA / I
= (Ncoil(μ0 nI)A) / I
= μ0 nNcoilA
M = μ0 nNcoilA
(b) (i)
where
n = 60 / (1.0 × 10−2) = 6 × 103 m−1
and
A = 1/4 πd²
= 1/4 π × (5.0 × 10−2)² = 1.96 × 10−3 m²
Hence,
M = (4π × 10−7)(6 × 103)(100)(1.96 × 10−3)
= 1.48 × 10−3 H
(ii) magnetic flux linkage through coil = NcoilBA
where
B = μ0 nI
and
I = E / (r + R)
= 12 / (0 + 20) = 0.60 A
Hence,
B = (4π × 10−7)(6 × 103)(0.6)
= 4.52 × 10−3 T
flux linkage = NcoilBA
= (100)(4.52 × 10−3)(1.96 × 10−3)
= 8.86 × 10−4 Wb
(iii) Average induced emf
Ecoil = − (change of flux linkage / time taken)
= − (0 − 8.86 × 10−4) / (20 × 10−3) = 0.044 V
EXAMPLE 22.39
 |
| A smaller circular coil with turns and radius $R_S$ is placed inside a larger circular coil with $N_L$ turns and radius $R_L$. The centres of both coils coincide and their planes are parallel, allowing magnetic interaction between the coils for mutual induction. |
A circular coil with NS turns and radius RS is placed inside a larger circular coil with NL turns and radius RL. The centres of the coils coincide and their planes are parallel to each other, as shown in Figure 22.68.
(a) Derive an expression for the mutual inductance of the pair of coils. Explain why the value of the mutual inductance calculated using this expression is only an approximation.
(b) Given the following data:
RS = 0.50 cm; RL = 10 cm
NS = 10; NL = 30
A current of 3.5 A flows through the large coil. Estimate
(i) the mutual inductance of the coils
(ii) the average emf induced in the small coil if the current in the large coil drops to zero in 10 ms.
Answer
(a) The mutual inductance M of the coils is given by
M = (NS ΦS) / I
where ΦS is the magnetic flux through the small coil and I is the current in the large coil.
The magnetic flux density B at the centre of the coils produced by the current I is given by
B = (μ0NLI) / (2RL)
Assume that the magnetic field passing normally through the whole cross section of the small coil is uniform. Then the magnetic flux ΦS is
ΦS = BAS cos 0°
M = (NSBAS) / I
= (NS / I) ( (μ0NLI) / (2RL) ) AS
= (μ0NSNLAS) / (2RL)
In the derivation of the expression above, we make one assumption: The magnetic field is uniform throughout the whole plane of the small coil, and its strength at any point on this plane is equal to μ0NLI / 2RL. If the radius RS of the small coil is large so that the area of the cross section is not small, then this assumption is not true. The magnetic field will not be uniform anymore on the plane.
(b) (i)
AS = πRS2
= π(0.50 × 10−2)2 = 7.85 × 10−5 m²
M = (μ0NSNLAS) / (2RL)
= (4π × 10−7)(10)(30)(7.85 × 10−5) / 2(0.10)
= 1.48 × 10−7 H ≅ 0.15 μH
(ii) The emf ES induced in the small coil is given by
ES = −M (ΔI / Δt)
= −(1.48 × 10−7) ( (0 − 3.5) / (10 × 10−3) ) = 5.2 × 10−5 V
Application of Mutual Induction and Mutual Inductance
Mutual induction is widely used in modern electrical and electronic systems to transfer energy efficiently without direct contact.
⚡ Transformer
Uses two coils (primary and secondary) to transfer electrical energy. A changing current in the primary coil induces emf in the secondary coil, allowing voltage transformation.
📱 Wireless Charging
Charging pads generate a changing magnetic field that induces current in the device, enabling charging without cables.
🍳 Induction Cooker
Produces heat directly in cookware by inducing currents in the metal, making cooking faster and more efficient.
⚙️ Electric Generators
Changing magnetic fields induce emf in coils, enabling conversion of mechanical energy into electrical energy.
🔒 Electrical Isolation
Allows energy transfer between circuits without direct connection, improving safety and reducing electrical noise.
📡 Sensors & Devices
Used in inductive sensors and communication systems to detect objects or signals without physical contact.
🔍 Conclusion
Mutual induction is a fundamental principle in electromagnetism that enables energy transfer between circuits without direct electrical contact. It plays a crucial role in many modern technologies, including transformers, wireless charging systems, and electronic devices.
By understanding how a changing current in one coil can induce an emf in another, we can design efficient systems for energy transfer, improve electrical safety, and enhance the performance of various electrical and electronic applications.
A changing current → produces a changing magnetic field → induces emf in another coil.
Post a Comment for "Mutual Induction and Mutual Inductance: Definition, Formula, and Explanation"