Lecture 22. Inductance. Magnetic Field Energy. Inductance Magnetic field...Lecture 22. Inductance....
Transcript of Lecture 22. Inductance. Magnetic Field Energy. Inductance Magnetic field...Lecture 22. Inductance....
Lecture 22. Inductance. Magnetic Field Energy.
Outline:
Self-induction and self-inductance.
Inductance of a solenoid.
The energy of a magnetic field.
Alternative definition of inductance.
Mutual Inductance.
1
Energy Transformations in EM Waves and Circuits
2
The energy of EM waves traveling in vacuum is stored in both 𝐸 and 𝐵 fields (in equal amounts).
Circuit elements that store the 𝐸 field energy - capacitors.
Circuit elements that store the 𝐵 field energy – inductors.
By combining capacitors and inductors, we can build the EM oscillators (e.g., generators of EM waves).
Maxwell’s Equations
� 𝐸 ∙ 𝑑𝑙𝑙𝑜𝑜𝑜
= −𝑑𝑑𝑑� 𝐵 ∙ 𝑑𝐴
𝑠𝑢𝑢𝑢
Faraday’s Law of electromagnetic induction; a time-dependent ΦB generates E.
� 𝐵 ∙ 𝑑𝑙𝑙𝑜𝑜𝑜
= 𝜇0 � 𝐽 + 𝜖0𝜕𝐸𝜕𝑑 ∙ 𝑑𝐴
𝑠𝑢𝑢𝑢
Generalized Ampere’s Law; B is produced by both currents and time-dependent ΦE.
�⃗� = 𝑞 𝐸 + �⃗� × 𝐵
The force on a (moving) charge.
3
Gauss’s Law: charges are sources of electric field (and a non-zero net electric field flux through a closed surface); field lines begin and end on charges.
No magnetic monopoles; magnetic field lines form closed loops.
� 𝐸 ∙ 𝑑𝐴𝑠𝑢𝑢𝑢
=𝑄𝑒𝑒𝑒𝑙𝜀0
� 𝐵 ∙ 𝑑𝐴𝑠𝑢𝑢𝑢
= 0
ℰ = −𝑑Φ𝐵
𝑑𝑑
(Self) Inductance
The flux depends on the current, so a wire loop with a changing current induces an “additional” e.m.f. in itself that opposes the changes in the current and, thus, in the
magnetic flux (sometimes called the back e.m.f.).
Inductance: (or self-inductance)
This definition applies when there is a single current path (see below for an inductance of a system of distributed currents 𝐽 𝐹 ).
𝑳 ≡𝜱𝑰
9
𝑩 𝒆
𝐼 𝑑
Variable battery
𝐿 – the coefficient of proportionality between Φ and 𝐼, purely geometrical quantity. 𝓔 = −
𝒅𝜱𝑩
𝒅𝒆= −𝑳
𝒅𝑰𝒅𝒆
Units: T⋅m2/A = Henry (H)
The back e.m.f. :
Inductance as a circuit element:
𝐹𝐹𝐹𝐹𝑑𝐹𝐹: ℰ = −𝑑Φ𝐵
𝑑𝑑
Inductance of a Long Solenoid
The magnetic flux generated by current 𝐼:
𝐿 =Φ𝐵
𝐼=𝜇0𝑁2
l 𝜋𝐹2 = 𝜇0𝑛2l 𝜋𝐹2
l
(n – the number of turns per unit length)
10
Long solenoid of radius r and length l with the total number of turns 𝑁:
𝐿 ≡Φ𝐵
𝐼
Φ𝐵 = 𝐵𝜋𝐹2𝑁 = 𝐵 = 𝜇0𝑁l𝐼 =
𝜇0𝑁2𝐼l
𝜋𝐹2 = 𝜇0𝑛2𝐼 l 𝜋𝐹2
- the inductance scales with the solenoid volume l 𝜋𝐹2; - the inductance scales as N2: B is proportional to 𝑁l and the net flux ∝ N.
Let’s plug some numbers: l =0.1m, N=100, r=0.01m
𝐿 = 𝜇0𝑛2l 𝜋𝐹2 = 4𝜋 ∙ 10−7 ∙1000.1
2
∙ 0.1 ∙ 𝜋 0.01 2 ≈ 40𝜇𝜇
How significant is the induced e.m.f.? For 𝑑𝑑𝑑𝑑
~104𝐴/𝑠 (let’s say, ∆𝐼~ 10 mA in 1 µs) ℰ = 𝐿
𝑑𝐼𝑑𝑑 =
4 ∙ 10−5 𝜇 × 1 ∙ 104𝐴𝑠 = 0.4𝑉
Note that
Inductance vs. Resistance
13
Inductor affects the current flow only if there is a change in current (di/dt ≠ 0). The sign of di/dt determines the sign of the induced e.m.f. The higher the frequency of current variations, the larger the induced e.m.f.
Energy of Magnetic Field To increase a current in a solenoid, we have to do some work: we work against the back e.m.f.
𝑈 = �𝑃 𝑑 𝑑𝑑𝑢
𝑖
= �𝐿𝐿 ∙ 𝑑𝐿𝑑
0
=12 𝐿𝐼
2
V Let’s ramp up the current from 0 to the final value 𝐼
(𝐿 𝑑 is the instantaneous current value):
𝑉 𝑑 = 𝐿𝑑𝐿 𝑑𝑑𝑑 𝑃 𝑑 = 𝑉 𝑑 𝐿 𝑑 = 𝐿
𝑑𝐿 𝑑𝑑𝑑 𝐿 𝑑
P(power) - the rate at which energy is being delivered to the inductor from external sources.
The net work to ramp up the current from 0 to 𝐼 :
𝐿 𝑑
𝐿 𝑑 𝑑𝑈 = 𝑃 𝑑 𝑑𝑑
This energy is stored in the magnetic field created by the current.
𝐿 = 𝜇0𝑛2l 𝜋𝐹2 = 𝜇0𝑛2 ∙ 𝑣𝑣𝑙𝑣𝑣𝑣
𝑈𝐵 =12 𝜇0𝑛
2 𝑣𝑣𝑙𝑣𝑣𝑣 ∙ 𝐼2= 𝐵 = 𝜇0𝑛𝐼 =𝐵2
2𝜇0∙ 𝑣𝑣𝑙𝑣𝑣𝑣 = 𝑣𝐵 ∙ 𝑣𝑣𝑙𝑣𝑣𝑣
The energy density in a magnetic field:
14
𝒖𝑩 =𝑩𝟐
𝟐𝝁𝟎 (compare with 𝑣𝐸 = 𝜀0𝐸2
2 )
This is a general result (not solenoid-specific).
𝑼 =𝟏𝟐𝑳𝑰𝟐
Quench of a Superconducting Solenoid
MRI scanner: the magnetic field up to 3T within a volume ~ 1 m3. The magnetic field energy:
Magnet quench: http://www.youtube.com/watch?v=tKj39eWFs10&feature=related
𝑈 = 𝑣𝑣𝑙𝑣𝑣𝑣 ∙ 𝑣𝐵 = 1𝑣3 3𝑇 2
2 ∙ 4𝜋 ∙ 10−7 𝑊𝑊𝐴 ∙ 𝑣
≈ 4𝑀𝐽
The heat of vaporization of liquid helium: 3 kJ/liter. Thus, ~ 1000 liters will be evaporated during the quench.
The capacity of the LHe dewar: ~2,000 liters.
15
Another (More General) Definition of Inductance
16
𝑈𝐵 =12𝐿𝐼2 = �
𝐵2
2𝜇0𝑑𝜏
𝑎𝑙𝑙 𝑠𝑜𝑎𝑒𝑒
Advantage: it works for a distributed current flow (when it’s unclear which loop we need to consider for the flux calculation).
𝑳 ≡𝟐𝑼𝑩𝑰𝟐
Alternative definition of inductance:
Example: inductance per unit length for a coaxial cable
17
Uniform current density in the central conductor of radius a:
𝑈𝐵 =1
2𝜇0� 𝐵 𝐹 2𝑑𝜏𝑏
0
𝐵 𝐹 =
𝜇0𝐼2𝜋𝐹
, 𝐹 < 𝐹 < 𝑊
𝜇0𝐼𝐹2𝜋𝐹2
, 𝐹 < 𝐹
Straight Wires: Do They Have Inductance?
𝐼
𝐼
𝐹 𝑊
𝐿 =2𝑈𝐵𝐼2
Using 𝑳 = 𝜱/𝑰: 𝐿 𝜇 ≈
𝜇02𝜋 l ∙ 𝑙𝑛
l𝐹 ≈ 2 ∙ 10−7l 𝑣 ∙ 𝑙𝑛
l𝐹
l = 1𝑐𝑣 𝐿 ≈ 10−9𝜇
𝑈𝐵 =1
2𝜇0�
𝜇0𝐼2𝜋𝐹
2𝑑𝜏
𝑏
𝑎
=1
2𝜇0�
𝜇0𝐼2𝜋𝐹
2∙ l ∙ 2𝜋𝐹𝑑𝐹
𝑏
𝑎
≈ (𝑊~l ) ≈𝜇0𝐼2
4𝜋∙ l ∙ 𝑙𝑛
l𝐹
l
~l 𝐵 ≈𝜇0𝐼2𝜋𝐹 Φ ≈ l�
𝜇0𝐼2𝜋𝐹 𝑑𝐹
l
𝑎
=𝝁𝟎𝑰𝟐𝝅 ∙ l ∙ 𝒆𝒆
l𝒆
𝑳 𝑯 ≈𝝁𝟎𝟐𝝅
l ∙ 𝒆𝒆l𝒆
at home
Using 𝑳 = 𝟐𝑼𝑩/𝑰𝟐:
Inductance vs. Capacitance
𝑈𝐵 =𝐵2
2𝜇0∙ 𝑣𝑣𝑙𝑣𝑣𝑣 = 𝑣𝐵 ∙ 𝑣𝑣𝑙𝑣𝑣𝑣
𝐿 =Φ𝐵𝐼 =
2𝑈𝐵𝐼2
𝑈𝐵 =12𝐿𝐼2 =
12Φ𝐵
2
𝐿
𝐶 =𝑄𝑉
=2𝑈𝐸𝑉2
𝑈𝐸 =12𝐶𝑉2 =
12𝑄2
𝐶
𝑈𝐸 =𝜀0𝐸2
2 ∙ 𝑣𝑣𝑙𝑣𝑣𝑣 = 𝑣𝐸 ∙ 𝑣𝑣𝑙𝑣𝑣𝑣
18
Mutual Inductance Let’s consider two wire loops at rest. A time-dependent current in loop 1 produces a time-dependent magnetic field B1. The magnetic flux is linked to loop 1 as well as loop 2. Faraday’s law: the time dependent flux of B1 induces e.m.f. in both loops.
ℰ2 = −𝑑Φ1→2𝑑𝑑 Φ1→2 = 𝑀1→2𝐼1
𝑩𝟏
loop 1 𝐼1
loop 2
Primitive “transformer”
The e.m.f. in loop 2 due to the time-dependent I1 in loop 1:
The flux of B1 in loop 2 is proportional to the current I1 in loop 1. The coefficient of proportionality 𝑀1→2 (the so-called mutual inductance) is, similar to L, a purely geometrical quantity; its calculation requires, in general, complicated integration. Also, it’s possible to show that
𝑀1→2 = 𝑀2→1 = 𝑀
so there is just one mutual inductance 𝑀.
ℰ2 = −𝑀𝜕𝐼1𝜕𝑑 𝑀 =
Φ1→2𝐼1
𝑀 can be either positive or negative, depending on the choices made for the senses of transversal about loops 1 and 2. Units of the (mutual) inductance – Henry (H).
19
Calculation of Mutual Inductance
20
Calculate M for a pair of coaxial solenoids (𝐹1 = 𝐹2 = 𝐹, l 1 = l 2 = l, 𝑛1 and 𝑛2 - the densities of turns).
𝐵1 = 𝜇0𝑛1𝐼1 Φ = 𝐵1𝜋𝐹2 = 𝜇0𝑛1𝐼1𝜋𝐹2
𝑀 =𝑁2Φ𝐼1
= 𝑁2𝜇0𝑛1𝜋𝐹2 =𝜇0𝑁1𝑁2𝜋𝐹2
l
the flux per one turn
l
𝑀1→2 = 𝑀2→1 = 𝑀
𝐿 = 𝜇0𝑛2l 𝜋𝐹2 =𝜇0𝑁2 𝜋𝐹2
l 𝑀 = 𝐿1𝐿2
Experiment with two coaxial coils.
The induced e.m.f. in the secondary coil is much greater if we use a ferromagnetic core: for a given current in the first coil the magnetic flux (and, thus the mutual inductance) will be ~𝐾𝑚 times greater.
ℰ2 = 𝑀∗ 𝑑𝐼1𝑑𝑑 = 𝐾𝑚𝑀
𝑑𝐼1𝑑𝑑
no-core mutual inductance
Appendix I: Inductance of a Circular Wire Loop
Note that we cannot assume that the wire radius is zero:
- diverges! Φ~ �𝜇0𝐼2𝜋𝐹 𝑑𝐹
𝑏
"0"
~𝑙𝑛𝑊
"0" What to do: introduce some
reasonably small non-zero radius.
Example: estimate the inductance of a circular wire loop with the radius of the loop, b, being much greater than the wire radius a. 𝑊
2𝐹
The field dies off rapidly (~1/r) with distance from the wire, so it won’t be a big mistake to approximate the loop as a straight wire of length 2πb, and integrate the flux through a strip of the width b:
2πb
b 𝐵 ≈𝜇0𝐼2𝜋𝐹
Φ ≈ 2𝜋𝑊 �𝜇0𝐼2𝜋𝐹
𝑑𝐹𝑏
𝑎
= 𝜇0𝑊𝐼 ∙ 𝑙𝑛𝑊𝐹
𝐿 =Φ𝐼≈ 𝜇0𝑊 ∙ 𝑙𝑛
𝑊𝐹
𝐿 𝑊 = 1𝑣, 𝐹 = 10−3𝑣 ≈ 4𝜋 ∙ 10−7𝑊𝑊𝐴 ∙ 𝑣 1𝑣 ∙ 𝑙𝑛103 ≈ 10𝜇𝜇
22