CH 30 INDUCTANCE

Intended Learning Outcomes – after this lecture you will learn:

1. Self-induction and inductors
2. Building up of magnetic flux in an inductor, analogous to building up charge in a capacitor
3. Oscillation in an L-C circuit, analogous to a spring-mass system (simple harmonic oscillator)
4. R-L-C circuit, analogous to a damped harmonic oscillator
Textbook Reference: 30.2 – 30.6

Self-induction and Inductors

Typically: a solenoid with 𝑁𝑁 turns
Varying current → varying B field → induced emf
This is called self-induction
Assuming vacuum, or in a magnetic material with constant
𝑲𝑲𝒎𝒎 and ���⃗
𝑴𝑴 ∝ ��⃗
𝑩𝑩, then total flux through the coil
𝑁𝑁Φ𝐵𝐵 ∝ 𝑖𝑖
Define the inductance of the coil
𝑁𝑁Φ𝐵𝐵
𝐿𝐿 =
𝑖𝑖
𝐿𝐿 depends on the geometry of the coil

SI unit: T⋅m2 /A ≡ H (henry) H is a large unit, typically mH or µH

From Faraday’s law
𝑑𝑑Φ𝐵𝐵 𝑑𝑑𝑑𝑑
ℇ = −𝑁𝑁 = −𝐿𝐿 self-induced emf
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
𝑁𝑁 loops
emf induced in one loop

Inductors As Circuit Elements

Direction of ℰ determined by Lenz’s law

𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
𝑉𝑉𝑏𝑏𝑏𝑏 = −𝐿𝐿 <0 𝑉𝑉𝑏𝑏𝑏𝑏 = −𝐿𝐿 >0
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑

PHYS1114 Lecture 17 Inductance P. 1

Can be considered as a circuit element in, e.g., Kirchhoff’s law where 𝑖𝑖 from 𝑎𝑎 → 𝑏𝑏,
𝑑𝑑𝑖𝑖
𝑉𝑉𝑏𝑏𝑏𝑏 = 𝑉𝑉𝑏𝑏 − 𝑉𝑉𝑎𝑎 = −𝐿𝐿
𝑑𝑑𝑑𝑑
no need to worry whether 𝑖𝑖 is increasing/decreasing. The sign of 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑 will take care of the
direction of ℇ
�⃗ in the coil is non-electrostatic, and potential function is not defined. Why
Digression: Induced 𝑬𝑬
can it create a potential difference 𝑉𝑉𝑎𝑎𝑎𝑎 ?
If you worry about this, read P. 1020 of your textbook, “Inductors as Circuit Elements”

Magnetic Field Energy

As 𝑖𝑖 increases, induced emf resists it, therefore external source must supply energy. Can consider
the energy used in building up the magnetic field, or magnetic flux, c.f. charging of a capacitor
Assume inductor has no resistance, all energy stored in the magnetic field inside the inductor
External power delivered to inductor
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
𝑃𝑃 = = 𝑉𝑉𝑎𝑎𝑎𝑎 𝑖𝑖 = 𝐿𝐿𝐿𝐿
𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑
𝐼𝐼
1
⇒ 𝑈𝑈 = 𝐿𝐿 � 𝑖𝑖 𝑑𝑑𝑑𝑑 ⇒ 𝑈𝑈 = 𝐿𝐿𝐼𝐼 2
0 2
𝐼𝐼 is the final steady current
1
c.f energy stored in a capacitor is 2
𝐶𝐶𝑉𝑉 2
for a comparison between inductor and capacitor, see Summary I attached.

Magnetic Energy Density

Suppose the inductor is a thin toroidal solenoid whose
thickness ≪ 𝑟𝑟, cross sectional area 𝐴𝐴, and has 𝑁𝑁 turns
Magnetic field inside is uniform
𝜇𝜇0 𝑁𝑁𝑁𝑁
𝐵𝐵 =
2𝜋𝜋𝜋𝜋
Its inductance is
𝑁𝑁Φ𝐵𝐵 𝜇𝜇0 𝑁𝑁 2 𝐴𝐴
𝐿𝐿 = =
𝑖𝑖 2𝜋𝜋𝜋𝜋
Total energy stored in inductor after the current settles to a constant value 𝐼𝐼 is
1 2 1 𝜇𝜇0 𝑁𝑁 2 𝐴𝐴 2
𝑈𝑈 = 𝐿𝐿𝐼𝐼 = 𝐼𝐼
2 2 2𝜋𝜋𝜋𝜋
Magnetic energy density (total energy per unit volume) is
𝑈𝑈 1 𝑁𝑁𝑁𝑁 2 𝐵𝐵 2
𝑢𝑢 = = 𝜇𝜇 � � ⟹ 𝑢𝑢 =
2𝜋𝜋𝜋𝜋𝜋𝜋 2 0 2𝜋𝜋𝜋𝜋 2𝜇𝜇0

PHYS1114 Lecture 17 Inductance P. 2

 1
c.f electric energy density is 2 𝜖𝜖0 𝐸𝐸 2
For a magnetic material with constant permeability 𝜇𝜇 = 𝐾𝐾𝑚𝑚 𝜇𝜇0 , 𝑢𝑢 = 𝐵𝐵 2 /2𝜇𝜇
Just like the electric energy density, this result not only true for an ideal solenoid. It is true in
general for any magnetic field provided 𝜇𝜇 is constant

Example c.f. Example 30.5, P. 1026

A thin toroidal solenoid has 𝑁𝑁 = 200 turns, cross sectional area 𝐴𝐴 = 5.0 cm2 , and 𝑟𝑟 = 0.10 m
(4𝜋𝜋 × 10−7 Wb/Am)(200)2 (5.0 × 10−4 m2 )
𝐿𝐿 = = 40 𝜇𝜇H
2𝜋𝜋(0.10 m)
With a final current 𝐼𝐼 = 200 A, the total magnetic energy stored is
1
𝑈𝑈 = (40 × 10−6 H)(200 A)2 = 0.8 J
2
Not practical as an energy storage!

The R-L Circuit

Current Growth (building up flux inside inductor, c.f. charging in a R-C circuit)
Kirchhoff’s loop rule:
𝑑𝑑𝑑𝑑
ℰ − 𝑖𝑖𝑖𝑖 − 𝐿𝐿 =0
𝑑𝑑𝑑𝑑
For a summary of voltage change across a circuit
element, see Summary II attached
Qualitatively:
1. At 𝑡𝑡 = 0, 𝑖𝑖 = 0, 𝑑𝑑𝑑𝑑/𝑑𝑑𝑡𝑡 = ℰ/𝐿𝐿 a maximum
2. As 𝑖𝑖 grows, 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑 decreases, eventually to zero
3. Therefore 𝑖𝑖 increases from 0 to 𝐼𝐼 = ℰ/𝑅𝑅 (max when
no emf is induced in 𝐿𝐿, must be of the form
𝑖𝑖(𝑡𝑡) = 𝐼𝐼(1 − 𝑒𝑒 −𝑡𝑡⁄𝜏𝜏 )
Quantitatively:
𝑑𝑑𝑑𝑑 ℰ 𝑅𝑅
= − 𝑖𝑖
𝑑𝑑𝑑𝑑 𝐿𝐿 𝐿𝐿
With initial condition 𝑖𝑖(0) = 0
ℰ 𝑡𝑡
−
𝑖𝑖(𝑡𝑡) = �1 − 𝑒𝑒 𝑅𝑅 � 𝐿𝐿 ⁄
𝑅𝑅
Define time constant 𝜏𝜏 ≡ 𝐿𝐿/𝑅𝑅
ℰ
𝑖𝑖(𝑡𝑡) = �1 − 𝑒𝑒 −𝑡𝑡/𝜏𝜏 �
𝑅𝑅

PHYS1114 Lecture 17 Inductance P. 3

Power delivery and dissipation
Instantaneous power 𝑑𝑑𝑑𝑑
(multiply 𝑖𝑖 to Kirchhoff’s rule) 𝑖𝑖(𝑡𝑡)ℰ − 𝑖𝑖 2 (𝑡𝑡)𝑅𝑅 − 𝑖𝑖(𝑡𝑡) �𝐿𝐿 �=0
𝑑𝑑𝑑𝑑
Delivered by ℰ 𝑑𝑑 1
Stored in 𝐿𝐿, = 𝑑𝑑𝑡𝑡 �2 𝐿𝐿𝑖𝑖 2 �
Dissipated in 𝑅𝑅

Current Decay (draining flux inside inductor, c.f. discharging in a R-C circuit)
After the current reaches the constant value 𝐼𝐼0 in the previus case, then remove ℰ
Qualitatively:
𝑖𝑖 must decrease from initial value 𝐼𝐼0 to zero (when all
energy stored in 𝐿𝐿 is drained), therefore must be of the
form
𝑖𝑖(𝑡𝑡) = 𝐼𝐼0 𝑒𝑒 −𝑡𝑡/𝜏𝜏
Quantitatively:
Put ℰ = 0 in loop rule
𝑑𝑑𝑑𝑑
−𝑖𝑖𝑖𝑖 − 𝐿𝐿 = 0
𝑑𝑑𝑑𝑑
⇒ 𝑖𝑖(𝑡𝑡) = 𝐼𝐼0 𝑒𝑒 −𝑡𝑡/𝜏𝜏
Instantaneous power
𝑑𝑑𝑑𝑑
𝑖𝑖 2 (𝑡𝑡)𝑅𝑅 + 𝑖𝑖(𝑡𝑡) �𝐿𝐿 � = 0
𝑑𝑑𝑑𝑑

Dissipated in 𝑅𝑅 < 0, energy stored in 𝐿𝐿 decreasing

Question
In the circuit, before the current settles to a constant value
a) when 𝑆𝑆1 is closed and 𝑆𝑆2 is open,
𝑉𝑉𝑎𝑎𝑎𝑎 ( > / < ) 0 and 𝑉𝑉𝑏𝑏𝑏𝑏 ( > / < ) 0
b) when 𝑆𝑆1 is open and 𝑆𝑆2 is closed, and current is
flowing in the direction shown,
𝑉𝑉𝑎𝑎𝑎𝑎 ( > / < ) 0 and 𝑉𝑉𝑏𝑏𝑏𝑏 ( > / < ) 0

Answer: see inverted text on P. 1030

PHYS1114 Lecture 17 Inductance P. 4

The L-C Circuit – analogy of a harmonic oscillator
An electrical oscillation, energy transfer between electric and magnetic energy
c.f. a mechanical oscillation (spring and mass), energy transfer between PE and KE

Quantitatively: from loop rule

𝑑𝑑𝑑𝑑 𝑞𝑞
−𝐿𝐿 − = 0
𝑑𝑑𝑑𝑑 𝐶𝐶
Since 𝑖𝑖 = 𝑑𝑑𝑑𝑑/𝑑𝑑𝑡𝑡
𝑑𝑑 2 𝑞𝑞 1
⇒ + 𝑞𝑞 = 0
𝑑𝑑𝑡𝑡 2 𝐿𝐿𝐿𝐿
c.f. in a mass and spring system with
𝑞𝑞 = 𝑄𝑄 cos(𝜔𝜔𝜔𝜔 + 𝜙𝜙) spring constant 𝑘𝑘
𝑖𝑖 = −𝜔𝜔𝜔𝜔 sin(𝜔𝜔𝜔𝜔 + 𝜙𝜙) 𝑑𝑑2 𝑥𝑥 𝑘𝑘
+ 𝑥𝑥 = 0
𝑑𝑑𝑡𝑡 2 𝑚𝑚
1
𝜔𝜔 = � solution is 𝑥𝑥 = 𝐴𝐴 cos(𝜔𝜔𝜔𝜔 + 𝜙𝜙), 𝜔𝜔 =
𝐿𝐿𝐿𝐿
�𝑘𝑘/𝑚𝑚 = 2𝜋𝜋𝜋𝜋
Amplitude 𝑄𝑄 and phase 𝜙𝜙 determined by initial
conditions, e.g,
If 𝑞𝑞(0) = 𝑄𝑄 and 𝑖𝑖(0) = 0, then 𝜙𝜙 = 0
If 𝑞𝑞(0) = 0, then 𝜙𝜙 = ±𝜋𝜋/2

See simulation at http://www.walter-fendt.de/html5/phen/oscillatingcircuit_en.htm

PHYS1114 Lecture 17 Inductance P. 5

Analogy between the mass-spring and inductor-capacitor system

total energy in a L-C circuit is conserved, just like a harmonic oscillator

The L-R-C Circuit – analogy of a damped harmonic oscillator

𝑑𝑑𝑑𝑑 𝑞𝑞
−𝑖𝑖𝑖𝑖 − 𝐿𝐿 − =0
𝑑𝑑𝑑𝑑 𝐶𝐶
𝑑𝑑 2 𝑞𝑞 𝑅𝑅 𝑑𝑑𝑑𝑑 1
⇒ 2
+ + 𝑞𝑞 = 0
𝑑𝑑𝑡𝑡 𝐿𝐿 𝑑𝑑𝑑𝑑 𝐿𝐿𝐿𝐿
1 𝑅𝑅 2
𝑞𝑞(𝑡𝑡) = 𝐴𝐴𝑒𝑒 −(𝑅𝑅 ⁄2𝐿𝐿)𝑡𝑡
cos �� − 𝑡𝑡 + 𝜙𝜙�
𝐿𝐿𝐿𝐿 4𝐿𝐿2

Oscillation frequency is no longer 𝜔𝜔 = 1/√𝐿𝐿𝐿𝐿, but

1 𝑅𝑅 2 1 𝑅𝑅 2
− 2>0 𝜔𝜔′ = � − 2
𝐿𝐿𝐿𝐿 4𝐿𝐿 𝐿𝐿𝐿𝐿 4𝐿𝐿
⇒ 𝑅𝑅 < 2�𝐿𝐿⁄𝐶𝐶 𝑅𝑅 > 2�𝐿𝐿⁄𝐶𝐶 , slow
exponential decay

Oscillate with angular

frequency 𝜔𝜔′

𝑅𝑅 = 2�𝐿𝐿⁄𝐶𝐶 , 𝜔𝜔′ = 0, 𝑞𝑞 ∝ 𝑒𝑒 −(𝑅𝑅⁄2𝐿𝐿)𝑡𝑡 ,

no oscillation, quickly damped to zero

For an analogy of forced harmonic oscillator, need an AC emf

PHYS1114 Lecture 17 Inductance P. 6

Summary I Comparison and Inductors and Capacitors

Inductor Capacitor

Ideal system: thin toroidal solenoid large parallel plates

Store magnetic flux 𝑁𝑁Φ𝐵𝐵 Charge q

Inductance 𝐿𝐿 = 𝑁𝑁Φ𝐵𝐵 /𝑖𝑖 Capacitance C = q / V

In a LR circuit, i through inductor grows In a RC circuit, V (or q) across capacitor

(building up flux) or decays (draining flux) grows (building up charge) or decays
with time constant 𝜏𝜏 = 𝐿𝐿/𝑅𝑅 (draining charge) with time constant 𝜏𝜏 =
𝑅𝑅𝑅𝑅

1 1
Energy stored in field 𝑈𝑈 = 2 𝐿𝐿𝑖𝑖 2 𝑈𝑈 = 𝑞𝑞 2 ⁄2𝐶𝐶 = 2 𝐶𝐶𝑉𝑉 2

Energy density 𝑢𝑢 = 𝐵𝐵 2 /2𝜇𝜇0 1

𝑢𝑢 = 2 𝜖𝜖0 𝐸𝐸 2

Summary II Voltage drop traversing a circuit element along direction of i

i i i

a b a b
a b
Potential drop Potential drop
Potential drop 𝑑𝑑𝑑𝑑
𝑉𝑉𝑏𝑏𝑏𝑏 = −𝑖𝑖𝑖𝑖 𝑉𝑉𝑏𝑏𝑏𝑏 = −𝐿𝐿 𝑑𝑑𝑑𝑑
𝑉𝑉𝑏𝑏𝑏𝑏 = −𝑞𝑞/𝐶𝐶

In case of C and L, always assume building up charge/flux. If they are draining, the
respective derivatives 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑 and 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑 will take care of the sign automatically.
If travel direction is opposite to 𝑖𝑖, reverse the sign, 𝑉𝑉𝑎𝑎𝑎𝑎 = −𝑉𝑉𝑏𝑏𝑏𝑏
Potential rise/drop across an external emf is determined by its polarity and is independent of
the direction of 𝑖𝑖
Kirchhoff’s loop rule can be stated as: the algebraic sum of potential rises and drops in a
loop is zero

PHYS1114 Lecture 17 Inductance P. 7

Clicker Questions

PHYS1114 Lecture 17 Inductance P. 8

Ans: Q30.2) A, Q30.6) C, Q-RT30.1) DABC

PHYS1114 Lecture 17 Inductance P. 9

See https://en.wikipedia.org/wiki/Joseph_Henry

PHYS1114 Lecture 17 Inductance P. 10