22.3 SELF INDUCTANCE L
22.3.1 Self Induction
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Figure 22.48 The figure shows a solenoid connected to a battery and a switch in an open-circuit condition. Since the switch is not closed, no electric current flows through the solenoid. As a result, no magnetic field is produced inside the solenoid, and the magnetic flux (Φ) as well as the flux linkage (NΦ) are zero. |
1. Consider a solenoid which is connected to a battery and a switch, forming an open circuit, as shown in Figure 22.48. The magnetic flux linkage through the solenoid is zero since there is no current in the solenoid, and so no magnetic field exists.
2. When the switch is closed, current begins to flow in the solenoid. The rising current produces a magnetic field, whose field lines pass through the solenoid and the magnitude increases with time. Hence, the magnetic flux linkage through the solenoid also increases with time.
3. Consequently, an emf has to be induced in the solenoid itself since the flux linkage changes with time. This process of producing an induced emf in the circuit due to a change of current flowing through the same circuit is known as self induction.
22.3.2 Effect of Self Induction on Current
1 A circuit in which self induction can occur
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| Figure 22.49 The figure shows a circuit consisting of a battery (emf E), a switch, an ohmic resistor of resistance R, and a solenoid with negligible resistance connected in series. The setup is used to study how current and back emf behave when the switch is opened or closed. |
Refer to the circuit shown in Figure 22.49. The ohmic resistor has resistance R and the solenoid has negligible resistance. The battery has emf E of negligible internal resistance. Consider the following two cases:
Case 1
Switch in contact with X
(a) Suppose the switch makes contact with point X. It is observed that at the moment the switch is closed, the electric filament lamp glows almost instantly at maximum brightness. This observation indicates that the current reaches maximum almost instantly.
(b) A current flows through the ohmic resistor in the closed circuit, increasing to a maximum value I0 given by
I0 = E / R
Case 2
Switch in contact with Y
(a) The switch makes contact with point Y. It is observed that now it takes a bit of time for the glow of the electric filament lamp to reach a similar maximum brightness, and not almost instantly, as in the previous case. This observation indicates that the current flowing through the solenoid takes a bit of time to reach the maximum value.
(b) The current flowing through the solenoid increases to a maximum value, also given by
I0 = E / R
since the external resistance and the emf in each circuit are the same.
The maximum current in each circuit is the same. However, there is a difference in the time taken by each current to rise to the maximum value. This difference is caused by the effect of self induction that occurs in the solenoid circuit but not in the resistor circuit.
2. How current rises with time
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| Figure 22.50 (a) Resistor circuit, When the switch is closed, the current quickly reaches its maximum value because there is no significant self-induction to oppose the change in current. (b) Solenoid circuit, When the switch is closed, the current increases gradually and takes time to reach its maximum value due to self-induction, which produces a back emf that opposes the increase in current. |
Figure 22.50 shows how the current in the resistor circuit and the current in the solenoid circuit vary with time upon closing the circuit. We notice the following:
(a) Resistor circuit (Figure 22.50(a)):
When the switch is closed, the current flowing through the resistor circuit almost instantly reaches the maximum value. No self induction takes place in the ohmic resistor, or its effect is so insignificant that it may be neglected.
(b) Solenoid circuit (Figure 22.50(b)):
When the switch is closed, the current flowing through the solenoid takes a bit of time to reach the maximum value. We can show that the current rises exponentially.
The slow rise in current is due to the fact that self induction takes place in the solenoid when the current changes. The effect of self induction opposes the increase of current, in accordance with Lenz’s law. Such opposition causes the current to increase with time at a slower rate. Hence, the effect prolongs the rise of current.
When the current has reached the maximum value it does not change any more. Since the current has become constant, the flux linkage through the solenoid also does not change with time, and so self induction does not occur.
3. How current falls with time
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| Figure 22.51 (a) Resistor circuit, When the switch is opened, the current drops almost instantly to zero because there is no significant self-induction to oppose the decrease in current. (b) Solenoid circuit: When the switch is opened, the current decreases gradually instead of instantly. This is due to self-induction in the solenoid, which produces a back emf that opposes the decrease in current, causing it to fall more slowly (exponentially). |
Figure 22.51 shows how the current in the resistor circuit and the current in the solenoid circuit vary with time when the switch is open. We notice the following:
(a) Resistor circuit
When the switch is switched off, the current almost instantly drops to zero (Figure 22.51(a)).
(b) Solenoid circuit
When the circuit is switched off, the current takes a bit of time to decrease to zero (Figure 22.51(b)).
The slow fall of current is again due to the fact that self induction takes place in the solenoid when the current changes. The effect of self induction opposes the decrease of current, in accordance with Lenz’s law. Such opposition causes the current to decrease with time at a slower rate. The effect prolongs the fall of current. We can show that the current falls exponentially.
22.3.3 Back emf
1 Meaning of back emf
Back emf is an induced emf produced in a circuit due to the change of current that flows through the same circuit. This induced emf produces an effect which tends to oppose the change experienced by the current.
2 Back emf in circuit
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| Figure 22.52 The diagram shows an electrical circuit consisting of a power supply E, an ohmic resistor with resistance |
(a) Consider the open circuit shown in Figure 22.52. The solenoid has negligible resistance and the ohmic resistor has resistance R. Suppose that the switch is closed and the current that flows through the solenoid increases with time. Then the flux linkage through the solenoid also increases with time, thus producing an induced emf Eb across the solenoid, in accordance with Faraday’s law of electromagnetic induction.
(b) This induced emf is the back emf. It is produced as long as the current keeps on increasing. Figure 22.53 show the time interval within which a back emf is induced.
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| Figure 22.53 The figure shows that back emf ($E_b$) is produced only during the time when the current is increasing. As the current changes, an induced emf is generated, but once the current becomes constant, the back emf disappears. |
If the current stops changing, like when it has reached the maximum constant value I0, the back emf will no longer appear across the solenoid. When the current becomes constant, the flux linkage through the solenoid does not change anymore. Hence, self induction does not occur.
(a) The back emf opposes the emf E applied to the circuit, in accordance with Lenz’s law, as the current increases.
(b) Hence, the voltage V across the resistor, before the current becomes constant, is
V = E − Eb
(c) The current I at a particular instant is
I = V / R
= (E − Eb) / R
Notice that the p.d. V across the resistor is ‘weakened’ due to the opposition of the back emf. Hence, the current driven by the p.d. V is smaller than that without back emf produced in the circuit. This explains why it takes the current longer time to reach the maximum value.
4 Instantaneous value of back emf
Consider the circuit shown in Figure 22.52. Self induction occurs in the solenoid when the current I flowing through the solenoid increases with time. A back emf Eb is developed across the solenoid. According to Faraday’s law of electromagnetic induction,
Eb ∝ dΦ / dt
where dΦ/dt is the rate of change of magnetic flux linkage through the solenoid. But magnetic flux linkage Φ is determined by the current I. We have
I ∝ Φ
Hence,
Eb ∝ dI/dt\
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| Figure 22.54 The figure shows that the back emf (Eb) decreases as the rate of change of current (dI/dt) decreases. This means Eb is proportional to dI/dt, given by the equation Eb = −L dI/dt, where the negative sign follows Lenz’s law. |
Figure 22.54 shows how Eb varies with dI/dt. Notice that as the gradient dI/dt decreases with time, the value of Eb decreases accordingly, due to the fact that Eb is proportional to dI/dt.
Hence, the back emf is given by
Eb = −L dI/dt
where L is a constant known as the self inductance of the solenoid (or, in general, the circuit component in which self induction occurs). The negative sign is for taking into account Lenz’s law.
5 An approximation of back emf within short time interval
For a change of current which has taken place over a very short time interval, Δt, we may make the following approximation:
dI/dt ≅ ΔI/Δt
where ΔI is the change of current. Using this approximation, the average back emf within this time interval is given by
Eb = −L (ΔI/Δt)
22.3.4 Inductor
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Figure 22.55 The figure shows the circuit symbol used to represent an inductor. An inductor is a component that has self inductance, such as a coil, solenoid, or toroid, and ideally has negligible resistance. |
A circuit element which possesses mainly self inductance is known as an inductor. An ideal inductor is one which does not possess resistance but only self inductance, or the resistance is so small that it may be neglected. Examples of inductors are coils, solenoids and toroids. The symbol used to represent an inductor is shown in Figure 22.55.
Note: Not only coils and solenoids have self inductance, but also wires and cables. The self inductance of wires and cables is insignificant, especially if they are short, and current flowing through them fluctuates at low frequency, like 60 Hz. However, at very high frequencies, the inductance may become large and no longer be negligible.
22.3.5 Self Inductance L
1 Definition
Self inductance is the property of a circuit or circuit component which can induce an emf in the circuit or component itself.
2 Unit
The unit of self inductance is the henry (H).
3 Meaning of 1 H
At a particular instant, let the current changes at the rate of
dI/dt = 1 A s−1
At that same instant, let the back emf produced be
Eb = 1 V
Then we have
1 V = L(1 A s−1)
L = 1 H
When the current flowing through an inductor changes at the rate of 1 A s−1 and produces a back emf of 1 V across the inductor, then the self inductance of the inductor is 1 H.
4 L in terms of N, Φ and I
Suppose current I flows through a coil with N turns. At a particular instant the current changes at the rate of dI/dt. Then at that instant, the back emf Eb produced in the coil is
Eb = −L dI/dt
Back emf is produced because self induction occurs within the coil. Hence, Faraday’s law of electromagnetic induction has to be obeyed. We have
Eb = −N dΦ/dt
where dΦ/dt is the rate of change of magnetic flux through the coil.
L dI/dt = N dΦ/dt
LI = NΦ
since L and N are constants. Hence, the self inductance of the coil is given by
L = NΦ / I
where NΦ is the magnetic flux linkage through the coil produced by current I.
Notice that the magnitude of the magnetic flux is determined by the magnitude of the current.
Φ ∝ I
= kI
where k is a constant which is determined by the geometrical properties of the coil. Hence, the expression
L = NkI / I
= Nk
indicates that the self inductance of the coil has to be determined by the geometrical properties of the coil.
22.3.6 Self Inductance of a Solenoid
1
Suppose current I flows through a long air-core solenoid with length l and N turns. The self inductance L of the solenoid is given by
LI = NΦ
2
The magnetic flux Φ is given by
Φ = BA cos 0°
where B and A are the magnetic flux density and the across sectional area of the solenoid respectively. The magnitude of B inside the central portion of the solenoid is given by
B = μ0 (N/l) I
3
Assume that the magnitude of B is uniform throughout the whole length of the solenoid. Then the magnetic flux linkage NΦ through the solenoid is given by
NΦ = N(BA)
= Nμ0 (N/l) IA
LI = μ0 (N2/l) IA
LI = μ0 (N2/l) IA
4
Hence, the self inductance of a long air-core solenoid is determined by the geometrical properties of the solenoid, like
(a) the number of turns (N)
(b) the length of the solenoid (l)
(c) the diameter of the cross section of the solenoid (A)
Note: (a) We may rewrite the expression given above as
L = μ0 (N2/l2) lA
= μ0 n2(lA)
where lA is the cylindrical volume of the air space within the solenoid.
(b) The equation for L is derived based on the assumption that the magnetic flux density B is uniform throughout the whole length of the solenoid. This is not the case because B at each open end of the solenoid is only one half of that at the central region of the solenoid. Hence, the true value of L is less than that obtained using the equation. However, if the solenoid is very long, then the magnetic field inside the solenoid is uniform over a very long length of the solenoid. In this case, the reduction of the magnetic field strength at both open ends may be neglected. We may assume that the uniformity of the field extends to both ends.
EXAMPLE 22.24
Show that the unit henry is equivalent to the base units of kg m2 s−2 A−2.
Answer
We have
L = E / (dI/dt)
Hence, the henry is composed of
⇒ kg m2 s−2 A−2 volt / (current/time) ⇒ J C−1 / A s−1 ⇒ (kg m2 s−2)(A s)−1 / A s−1
EXAMPLE 22.25
Back emf of 5.0 V is developed across coil X when the current flowing through it changes at the rate of 25 A s−1. Determine the self inductance of X. At a particular instant, a back emf of 5.0 V is also developed across another coil, Y, which has self inductance of 350 mH. Determine the rate at which the current flowing through the coil must change at that instant.
Answer
Coil X:
|Eb| = L (dI/dt)
5.0 = L(25)
L = 0.20 H
Coil Y:
5.0 = 0.350 (dI/dt)
dI/dt = 14.3 A s−1
EXAMPLE 22.26
A 10 Ω resistor, a 300 mH ideal solenoid, and a battery of emf 12 V and negligible internal resistance, are connected in series, as shown in Figure 22.52. Initially the switch is open. It is then closed. Determine
(a) the rate of change of current in the circuit at the moment when the switch is closed
(b) the rate of change of current at the instant when
(i) the p.d. across the resistor is 3.0 V
(ii) the p.d. across the resistor is 80 % of the back emf
(iii) the current is 0.50 A
(c) the maximum current that can flow through the solenoid
Answer
(a) At the moment when the switch is closed, current has not yet flow in the circuit although current is about to start flowing at certain rate. Hence, the p.d. V across the resistor is zero.
V = E − Eb = 0
Eb = E = 12 V
The magnitude of the back emf Eb is given by
Eb = L (dI/dt)
12 = (0.300) (dI/dt)
dI/dt = 12 / 0.300
= 40.0 A s−1
(b) (i) P.d. across resistor
V = E − Eb
3.0 = 12 − Eb
Eb = 12 − 3.0
= 9.0 V
The magnitude of back emf is given by
Eb = L (dI/dt)
9.0 = (0.300) (dI/dt)
dI/dt = 9.0 / 0.300 = 30.0 A s−1
(ii)
V = E − Eb
0.80Eb = E − Eb
Eb = E / 1.80 = 12 / 1.80 = 6.67 V
6.67 = (0.300) (dI/dt)
dI/dt = 6.67 / 0.300 = 22.2 A s−1
(iii) When the current becomes 0.50 A, the p.d. V across the resistor is
V = IR
= (0.50)(10) = 5.0 V
But
V = E − Eb
5.0 = 12 − Eb
Eb = 12 − 5.0 = 7.0 V
dI/dt = Eb / L = 7.0 / 0.300 = 23.3 A s−1
(c) Maximum current
I0 = E / R = 12 / 10 = 1.2 A
EXAMPLE 22.27
Refer to Example 22.26 above. At a particular instant the current flowing through the solenoid is 0.20 A. Estimate the current 1.0 ms later.
Answer
When the current is 0.20 A, the p.d. V across the resistor is
V = IR
= (0.20)(10) = 2.0 V
V = E − Eb
2.0 = 12 − Eb
Eb = 12 − 2.0 = 10 V
Eb = L (ΔI/Δt)
10 = (0.300) (ΔI / 1.0 ms)
ΔI = (10 / 0.300)(1.0 × 10−3) = 3.33 × 10−2 A
ΔI = If − Ii
3.33 × 10−2 = If − 0.20
If = 0.20 + 0.03 = 0.23 A
EXAMPLE 22.28
The current flowing through a solenoid changes from 5.00 A to 5.50 A in 2.0 ms. The back emf produced across the solenoid due to the change of current is 5.5 V. Estimate the self inductance of the solenoid.
Answer
ΔI = If − Ii = 5.50 − 5.00 = 0.50 A
Δt = 2.0 ms
ΔI / Δt = 0.50 / (2.0 × 10−3) = 250 A s−1
Eb = L (ΔI / Δt)
L = Eb / (ΔI / Δt)
= 5.5 / 250 = 0.022 H
EXAMPLE 22.29
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| Figure 22.56 The figure shows how current varies with time in a series resistor–inductor circuit, used to determine the voltage across each component. |
An ohmic resistor of 10 Ω is connected in series to an ideal inductor of self inductance 500 mH. The circuit is connected to a power supply so that a current flows through the two elements. The current varies with time in the manner as shown in Figure 22.56. Determine the voltage across each component
(a) 0.20 s
(b) 0.60 s
after connecting the power supply to the circuit.
Answer
(a) Inductor: At time t = 0.20 s, the rate of change of current is
dI/dt = (2.0 − 0) / (0.50 − 0) = 4.0 A s−1
voltage across inductor = back emf = L (dI/dt)
= (0.500)(4.0) = 2.0 V
Resistor: Let the current at t = 0.20 s be I.
I / 0.20 s = 2.0 A / 0.50 s
I = 0.20 (2.0 − 0) / (0.50 − 0) = 0.80 A
voltage across resistor = IR
= (0.80)(10) = 8.0 V
(b) Inductor: At time t = 0.60 s, the current is constant, meaning dI/dt = 0.
voltage across inductor = 0
Resistor: At t = 0.60 s, the current I = 2.0 A.
voltage across resistor = (2.0)(10) = 20 V
EXAMPLE 22.30
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| Figure 22.57 The figure shows how the back emf changes with time when the current in an ideal inductor decreases. The back emf is produced only during the time when the current is changing. |
Answer
1. We have back emf Eb ∝ − dI/dt, where dI/dt is the gradient.
2. The back emf initially increases linearly. Hence, the negative current gradient, dI/dt, must increase accordingly, meaning the gradient becomes more and more negative.
3. The negative gradient increases to a maximum value at the instant when the back emf reaches the maximum value.
4. From that instant onwards, the negative current gradient starts to decrease.
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| Figure 22.58 The figure shows how current changes with time as the negative gradient gradually decreases. When the back emf becomes zero, the current stops changing and remains constant. |
5. When the back emf finally becomes zero, the current gradient is zero. After that, the current is constant. Figure 22.58 shows how the current varies with time.
EXAMPLE 22.31
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| Figure 22.59 The graph shows the variation of back emf across an ideal inductor with time. The back emf is negative when the current increases, zero when the current is constant, and positive when the current decreases, in accordance with the relation ε = −L (dI/dt). |
The graph in Figure 22.59 shows how the current flowing through an ideal inductor of self inductance 200 mH varies with time. Sketch a graph to show how the back emf across the inductor varies with time.
Answer
(a) At time t = 0, the current gradient, dI/dt, is zero. Since back emf E ∝ − dI/dt, the initial back emf is zero.
(b) For time interval between 0 and about 1.8 ms, the negative current gradient increases. Hence, during this time interval E must increase.
(c) At about t = 1.8 ms, the current gradient is a maximum, hence, producing a maximum back emf E1.
E1 = −L (dI/dt)
= −(0.200 H) (−1.0 A / 0.4 ms)
= +0.50 kV
(d) After that the current gradient decreases and finally becomes zero at about t = 3.0 ms. E = 0 at that instant.
(e) For the time interval between t = 3.0 ms and t = 4.2 ms, the positive current gradient increases. Hence, the back emf increases.
(f) At about t = 4.2 ms, the gradient is a maximum, hence producing a maximum back emf E2.
E2 = −(0.200 H) (+1.0 A / 0.25 ms)
= −0.80 kV
(g) At about t = 5.8 ms, the current gradient becomes zero. At that instant, the back emf is zero.
(h) After that, the negative current gradient increases until it reaches a maximum value at about t = 6.9 ms, producing a maximum back emf E3.
E3 = −(0.200 H) (−0.55 A / 0.4 ms)
= +0.27 kV
(i) The current gradient is reduced to zero at about t = 8.2 ms. At that instant, the back emf is zero.
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Figure 22.60 The graph shows how back emf varies with time. It is zero when the current gradient is zero, negative when the current increases, and positive when the current decreases. Maximum values occur at about t = 4.2 ms (−0.80 kV) and t = 6.9 ms (+0.27 kV). |
(j) The changes of the back emf E with time are shown in Figure 22.60.
Applications of Self Inductance and Back EMF
⚡ Electrical Transformers
Self inductance plays an important role in transformers. It helps transfer energy between coils through changing magnetic fields, allowing efficient voltage step-up or step-down in power distribution.
🔌 Inductors in Circuits
Inductors are widely used in electronic circuits to control current changes. They are essential in filters, oscillators, and power supplies to stabilize current flow.
💡 Energy Storage
Inductors store energy in their magnetic fields when current flows through them. This stored energy can be released when the current decreases, useful in switching circuits and power systems.
⚙️ Electric Motors
Back emf is important in electric motors. It regulates the current automatically, preventing excessive current and protecting the motor from damage.
📡 Communication Systems
Inductors are used in tuning circuits such as radios and communication devices. They help select specific frequencies by working with capacitors in resonance circuits.
🔋 Switching Power Supplies
In modern electronics, inductors are used in switching regulators to efficiently convert voltage levels while minimizing energy loss.
Conclusion
Self inductance is a fundamental property of electrical circuits where a changing current induces an emf within the same circuit. This phenomenon, known as self induction, plays a crucial role in controlling how current increases and decreases over time.
The presence of back emf, as explained by Lenz’s law, opposes any change in current, resulting in a gradual rise and fall rather than an instantaneous change. This behavior is especially significant in components such as solenoids and inductors.
The self inductance (L) depends on the physical and geometrical properties of the circuit, particularly in coils and solenoids. It determines how strongly a circuit resists changes in current and is measured in henry (H).
Understanding self inductance and back emf is essential in analyzing real-world electrical systems, from simple circuits to advanced technologies such as transformers, motors, and communication devices.
In summary, self inductance not only explains the behavior of current in circuits but also forms the foundation for many practical electrical and electronic applications.
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