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Handout 10: Inductance
Self-Inductance and inductors
In Fig. 1, electric current is present in an isolate
circuit, setting up magnetic field that causes a magnetic
flux through the circuit itself. This flux changes when the
current changes. Thus, any circuit with varying current
will have an emf induced in it by variation of its own
magnetic field. Such emf is called a “self-induced emf”.
By Lenz’s law, the self-induced emf always opposes the
change in the current that caused the emf.
From Fig. 1, the current 𝐼 in the circuit causes the
total magnetic flux 𝛷 through the coil. We define “selfinductance” L of the circuit as
𝐿 =
Φ
.
𝐼
The unit of the inductance is henry (H). The selfinductance is also called simply the “inductance”. From
Faraday’s law, we have an expression for the selfinduced emf
ℰ = −
Figure 1: Current in the coil causes
magnetic field and hence magnetic flux
through the coil itself
𝑑Φ
𝑑 𝐿𝐼
𝑑𝐼
= −
= −𝐿 .
𝑑𝑡
𝑑𝑡
𝑑𝑡
A device that is designed to have a particular
inductance is called an “inductor”. The usual symbol for
an inductor is
.
..
Solenoid Consider a solenoid (Fig. 2) of lengths
ℓ containing N turns. The cross-section area of the
solenoid is 𝐴. When the current 𝐼 flows in it, from
Ampere’s law, the magnetic field inside the solenoid is
given by 𝐵 = 𝜇0 𝑁𝐼 ℓ. The total magnetic flux is
Φ = 𝐵 𝑁𝐴 = 𝜇0 𝑁 2 𝐼𝐴 ℓ.
Since inductance 𝐿 = Φ I, we have an expression of
inductance of the solenoid
𝜇0 𝑁 2 𝐴
𝐿 =
.
ℓ
Note that the inductance is a constant depending on the
dimensions and the number of turns of the solenoid.
Figure 2: A solenoid
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Example 1 A cylindrical solenoid of length 0.50 m
contains 150 turns. The radius of the solenoid is 2.0 cm.
Calculate the inductance.
Example 2 An inductor has inductance 0.26 H and
carries a current what is increasing at uniform rate
0.015 As-1. Determine the self-induced emf.
Example 3 An air-core toroid with cross-section area A
and mean radius 𝑅 is closely wound with N turns.
a) Show that the inductance is given by
𝐿 =
𝜇0 𝑁 2 𝐴
.
2𝜋𝑅
b) Suppose 𝑁 = 200 turns, 𝐴 = 5.0 cm2 and 𝑅 = 0.1
m, calculate 𝐿.
Energy stored in inductor
Because of the self-induced emf ℰ, the voltage
across the inductor is 𝑉 = 𝐿 𝑑𝐼 𝑑𝑡 . The energy is
transferred to the inductor at rate 𝑃 = 𝐼𝑉 = 𝐿𝐼 𝑑𝐼 𝑑𝑡 .
The energy supplied, 𝑈, is equal to the integral 𝑃𝑑𝑡.
While the current increases from zero to any 𝐼, the
energy store is given by
𝐼
1 2
𝐿𝐼 .
2
0
It must be noted that resistors and inductors
behave differently. When the current (either increasing
𝑈 =
𝐿𝐼 𝑑𝐼 =
3
or decreasing) passes through a resistor, energy is
dissipated as heat. By contrast, when the current in the
inductor is increasing, the energy is stored in the
inductor. When the current is decreasing, the energy is
released from the inductor. When the current is
constant, there is no energy flow in or out.
Solenoid Self-inductance of a solenoid is given
by 𝐿 = 𝜇0 𝑁 2 𝐴 ℓ. Therefore, the energy stored
1 2
1 𝜇0 𝑁 2 𝐴 2
1 𝜇0 𝑁𝐼
𝑈 = 𝐿𝐼 =
𝐼 =
2
2
ℓ
2𝜇0
ℓ
2
𝐴ℓ .
Since 𝜇0 𝑁𝐼 ℓ = 𝐵 inside the solenoid and 𝐴ℓ is the
volume of the solenoid. The energy density (energy per
unit volume), is given by
𝑢 =
𝑈
𝐵2
=
.
𝐴ℓ
2𝜇0
This is a general expression for energy density stored in
magnetic field. It should be compared with expression
for energy density stored in electric field 𝑢 = 1 2 𝜀0 𝐸 2
(see handout 4).
Example 4 A electric-power industry wants to store
energy in a large inductor. What inductance would be
needed to store 1.00 kW h-1 of energy in a coil carrying a
200-A current?
Example 5 On average, electromagnetic waves contain
equal amount of magnetic energy density and electric
energy density. Determine the ratio of electric field 𝐸 to
to magnetic field 𝐵. Comment on this result.
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RL circuit
Circuit in Fig. 3 consists of a battery with emf 𝓔, a
resistor with resistance 𝑅 and an inductor with
inductance 𝐿. Initially, the switch 𝑆1 and 𝑆2 are open.
When 𝑆1 is closed, the current 𝐼 flows from the
battery and it is increasing from zero. At any instant, the
voltage across the resistor is 𝑉𝑅 = 𝐼𝑅 and the selfinduced emf of the inductor is ℰ𝐿 = −𝐿 𝑑𝐼 𝑑𝑡 . From
Kirchhoff’s law, 𝐸 =
𝑉,
ℰ−𝐿
𝑑𝐼
= 𝐼𝑅
𝑑𝑡
⟹
−
Figure 3: RL circuit
𝐿 𝑑𝐼
ℰ
= 𝐼− .
𝑅 𝑑𝑡
𝑅
We can solve the above equation as follows:
𝐼
0
𝑡
𝑑𝐼
= −
𝐼− ℰ 𝑅
𝐼
ℰ
ln 𝐼 −
𝑅
0
𝐼− ℰ 𝑅
ln
−ℰ 𝑅
= −
0
𝑅
𝑑𝑡
𝐿
𝑅
𝑡
𝐿
Figure 4: The maximum current ℰ 𝑅 .
After one time constant 𝜏, the current is
1 − 1 𝑒 ≃ 63% of the final value
𝑅
= − 𝑡.
𝐿
The above equation can be rearranged, yielding current
𝐼 as a function of time 𝑡:
𝐼 =
ℰ
1 − 𝑒− 𝑅
𝑅
𝐿 𝑡
.
The current increases with time and it reaches the
maximum value ℰ 𝑅 as 𝑡 ⟶ ∞. This can be seen from
the graph in Fig. 4.
The time constant for an RL circuit is defined as
𝜏 = 𝑅 𝐿. When 𝑡 = 𝜏, the current is 1 − 1 𝑒 ≃ 63% of
the maximum value.
When the switch 𝑆1 in Fig. 3 has been closed for a
while and that the current reaches the maximum value
𝐼 = ℰ 𝑅 . The switch 𝑆1 is open and 𝑆2 is closed at the
same time. From Kirchhoff’s law,
−𝐿
𝑑𝐼
= 𝐼𝑅.
𝑑𝑡
The initial current is 𝐼 = ℰ 𝑅 . One can solve the above
equation to obtain
Figure 5: The decay of current in an
RL circuit.
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𝐼 =
ℰ −𝑅
𝑒
𝑅
𝐿 𝑡
.
This equation is plotted in Fig. 5. When 𝑡 = 𝜏, the
current is 1 𝑒 ≃ 37% of the initial value.
LC circuit
An LC circuit, consisting of an inductor and a
capacitor, is an example of oscillatory circuit. The
charge can current oscillates in time, in contrast to the
RL circuit where the current and charge vary
exponentially with time.
Consider an LC circuit in Fig. 6. Initially, the
capacitor has charge 𝑞0 and the switch is open. When
the switch is closed, the charge flows from the capacitor
to the inductor. The potential difference across the
capacitor is 𝑞 𝐶 . The increasing current in the inductor
causes a self-induced emf ℰ𝐿 = −𝐿 𝑑𝐼 𝑑𝑡 . The current
can be written as 𝐼 = 𝑑𝑞 𝑑𝑡. From Kirchhoff’s law,
𝐿
𝑑𝐼 𝑞
+ = 0
𝑑𝑡 𝑐
⇒
Figure 6: LC circuit
𝑑2 𝑞
𝑞
+
= 0.
𝑑𝑡 2 𝐿𝐶
The above equation looks like a SHM equation for
electric charge 𝑞 with angular frequency 𝜔 = 1 𝐿𝐶 .
The solution, subject to the initial condition
𝑞 0 = 𝑞0 , is given by
Figure 7: Oscillating charge of LC
circuit with period 𝑇 = 2𝜋 𝜔 = 2𝜋 𝐿𝐶
𝑞 = 𝑞0 cos 𝜔𝑡.
The amplitude of charge oscillation is 𝑞0 . The variation
of charge is illustrated in Fig. 7. The current varies as
𝐼 =
𝑑𝑞
= −𝜔𝑞0 sin 𝜔𝑡.
𝑑𝑡
The amplitude of the current is 𝜔𝑞0 and the phase of the
current differs from that of charge by 𝜋 2.
Figure 8 shows the graphs of energy against time.
The energy in the capacitor is given by
𝑈𝐶 =
𝑞2
𝑞02
=
cos 2 𝜔𝑡,
2𝐶
2𝐶
and the energy in the inductor by
Figure 8: Variation of energies stored in
capacitor, inductor and total energy over
time.
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𝑈𝐿 =
1 2
1
𝑞02
𝐿𝐼 = 𝐿𝜔2 𝑞02 sin2 𝜔𝑡 =
sin2 𝜔𝑡.
2
2
2𝐶
The total energy 𝑈𝐶 + 𝑈𝐿 = 𝑞02 2𝐶 is constant over time.
Therefore, there is no energy lost.
RLC circuit in series
In the LC circuit, the total energy is constant and
the charge in the capacitor oscillates with constant
amplitude. Introducing a resistor into the circuit will
cause energy lost and, hence, the amplitude of charge
oscillation decreases with time. This is analogous to
damped harmonic oscillator in mechanics.
Consider an RLC circuit shown in Fig. 9. Intially,
the charge 𝑞0 is stored in the capacitor. The capacitor
discharges. The voltage across the capacitor is 𝑞 𝐶 . The
induced emf across the inductor is ℰ𝐿 = −𝐿 𝑑𝐼 𝑑𝑡 . The
voltage across the resistor is 𝐼𝑅. From Kirchhoff’s law,
−𝐿
Figure 9: RLC circuit
𝑑𝐼 𝑞
− = 𝐼𝑅.
𝑑𝑡 𝑐
By writing 𝐼 = 𝑑𝑞 𝑑𝑡, Kirchhoff’s law can be expressed
in the form
𝑑 2 𝑞 𝑅 𝑑𝑞
1
+
+
𝑞 = 0.
2
𝑑𝑡
𝐿 𝑑𝑡 𝐿𝐶
This second-order differential equation is analogous to
damped harmonic motion where the mass is replaced by
the charge 𝑞.
The solution is different for the underdamped
(small 𝑅) and overdamped (large 𝑅) cases. Here we are
interested in a specific case when 𝑅 2 < 4𝐿 𝐶 . The
solution takes the form
𝑞 = 𝑞0 𝑒 −(𝑅
2𝐿)𝑡
cos 𝜔𝑡 ,
𝜔 =
1
𝑅2
− 2.
𝐿𝐶 4𝐿
The plot of charge against time is shown in Fig. 10. The
amplitude of the oscillation decreases exponentially.
Figure 10: Graph of charge against time for
an underdamped RLC circuit