Transcript Inductance

Inductance
We know already: changing magnetic flux creates an emf
changing current in a coil will induce a current in an adjacent coil
Coupling between coils is described by mutual inductance
𝑁2 Φ𝐵2 = 𝑀21 𝑖1
mutual inductance
Current i1 through coil 1 creates B-field and thus flux through coil 2.
If i1=i1(t) then dB/dt  0 and thus d/dt  0 inducing an emf in coil2.
From
𝑁2 Φ𝐵2 = 𝑀21 𝑖1
𝑑Φ𝐵2
𝑑𝑖1
= 𝑀21
𝑑𝑡
𝑑𝑡
For vacuum and material with constant susceptibility
M21 is a constant and given by
Using
ℰ2 = −𝑁2
𝑁2
𝑑Φ𝐵2
𝑑𝑡
ℰ2 = −𝑀21
𝑑𝑖1
𝑑𝑡
𝑀21
𝑁2 Φ𝐵2
=
𝑖1
Repeating the thought by driving a time dependent current, 𝑖2 , through coil 2
𝑑𝑖2
Induction of emf ℰ1 in coil 1 ℰ1 = −𝑀12
𝑑𝑡
𝑁1 Φ𝐵1
=
𝑖2
𝑑𝑖2
ℰ1 = −𝑀12
𝑑𝑡
with
𝑀12
𝑑𝑖1
𝑑𝑡
with
𝑀21 =
ℰ2 = −𝑀21
𝑁2 Φ𝐵2
𝑖1
For the vacuum case (but also in general) mutual inductance depends only on coil geometry
𝑀12 = 𝑀21 = 𝑀
with
𝑀=
𝑁1 Φ𝐵1 𝑁2 Φ𝐵2
=
𝑖2
𝑖1
Mutual inductance is measured in Henry where 1H=1Wb/A=1Vs/A=1Ωs=1J/A2
- Mutual inductance can give rise to cross-talk in electronic circuits 
- Mutual inductance has important applications, e.g., in transformers 
Many applications happen in AC circuits (see textbook).
Here a collection:
Transformers
Metal detectors
Here we have a brief look at the Tesla coil
Current i1 through coil 1 creates B-field
𝜇0 𝑖1 𝑁1
𝐵1 =
𝑙
and flux
Φ1 = 𝐵1 𝐴 = Φ2
solenoid 1 is long compared to solenoid 2
𝑀=
𝑁2 Φ𝐵2 𝜇0 𝑁1 𝑁2 𝐴
=
𝑖1
𝑙
M depends only on geometry and in particular
on product N1N2.
To see how a Tesla coil can create a vary large emf let’s have a look to
an example
Drive a current i2(t)through solenoid 2 (blue)
𝑀𝑖2
𝑁1
𝑑𝑖2
emf induced in solenoid 1 reads: ℰ1 = −𝑀
𝑑𝑡
Flux through solenoid 1 is given by Φ𝐵1 =
In an operating Tesla coil
high frequency alternating current
creates large amplitudes of di2/dt
and thus large amplitudes of alternating ℰ1
Primary and secondary resonant circuits tuned to
1
same frequency 𝑓 =
with f= 100 kHz to 1 MHz
2𝜋 𝐿𝐶
see slide 11 for derivation of resonance frequency
in LC-circuit
Hallmark of Tesla coil is the loose/critical coupling (large air gap) between the solenoids 1 and 2 to
prevent damage (insulation between tightly coupled solenoids would experience dielectric
breakdown).
From
http://tesladownunder.com/Tesla_coils_intro.htm
Note, I did not check
the scientific validity
of the information
provided on this
web-site
Self-Inductance and Inductors
The concept that a changing flux induces an emf can also be applied in the
case of a single solenoid
Self-inductance 𝐿 =
𝑑Φ
With ℰ = −𝑁
𝑑𝑡
𝑁Φ
𝑖
𝑑i
ℰ = −𝐿
𝑑𝑡
A device designed to have a particular inductance L is called an inductor
Circuit symbol
Examples from http://upload.wikimedia.org/wikipedia/commons/2/27/Aplikimi_i_feriteve.png
The effect of an inductor in a circuit
Let’s compare resistor and inductor
Current flowing through resistor R
gives rise to potential drop 𝑉𝑎𝑏 = 𝑖𝑅
Emf ℰ
= −𝐿
𝑑i
𝑑𝑡
opposes current change
Potential drop 𝑉𝑎𝑏 = 𝐿
Note: Sign opposite to emf
𝑑𝑖
𝑑𝑡
Example: Inductance of an air core toroidal solenoid
1) Determine B from Ampere’s law
r
𝐵 𝑑𝑟 = 𝜇0 𝑁 𝑖
2) Determine flux through one loop Φ = 𝐵𝐴 = 𝜇0
3) Determine emf of solenoid and compare with
𝑁 2𝐴
𝑑Φ
𝑁 2 𝐴 𝑑𝑖
𝐿 = 𝜇0
ℰ = −𝑁
= −𝜇0
2𝜋𝑟
𝑑𝑡
2𝜋𝑟 𝑑𝑡
𝑁𝐴𝑖
2𝜋𝑟
𝑖
𝐵 = 𝜇0 𝑁
2𝜋𝑟
ℰ = −𝐿
𝑑i
𝑑𝑡
Magnetic field Energy
We will see that similar to the electric field there is energy stored in a
magnetic field
Energy stored in an Inductor
Let’s calculate the energy input U needed to establish a current
I in an ideal (zero resistance) inductor with inductance L
𝑑𝑖
With 𝑉𝑎𝑏 = 𝐿
we obtain for the power, P, delivered to the inductor
𝑑𝑡
𝑑𝑖
𝑃 = 𝑉𝑎𝑏 𝑖 = 𝐿𝑖
𝑑𝑡
𝑡
𝑑𝑖
For the energy delivered after time, t, we obtain 𝑈 = 𝐿
𝑖 ′ 𝑑𝑡′
0 𝑑𝑡
Changing integration variable from t to i we obtain
𝐼
𝑈=𝐿
𝑖 𝑑𝑖 =
0
1 2
𝐿𝐼
2
Energy stored in an inductor when permanent current
I is flowing
We now want to use 𝑈 =
1 2
𝐿𝐼
2
to see that the energy is stored in the field
(very much in analogy to the transition from energy in a capacitor to energy stored in the electric field)
𝑁 2𝐴
Let’s recall the inductance L of a toroidal inductor 𝐿 = 𝜇0
2𝜋𝑟
Volume, V, which is filled with a magnetic field of magnitude 𝐵 = 𝜇0 𝑁
𝑖
2𝜋𝑟
V= 2 𝜋𝑟 𝐴 where A is the area of the cross-section
𝑖
𝑁 2𝐴
𝐵
=
𝜇
𝑁
for i=I
𝐿 = 𝜇0
0
2𝜋𝑟
2𝜋𝑟
1 2
1 𝑁 2 𝐴 2𝜋𝑟𝐵 2
1 𝑁 2𝐴 2
The energy 𝑈 = 𝐿𝐼 can be expressed as 𝑈 = 𝜇0
(
)
𝐼 = 𝜇0
2
2
2𝜋𝑟
𝜇
𝑁
2 2𝜋𝑟
0
1 𝐴2𝜋𝑟𝐵2
1 𝑉𝐵2
𝑈=
=
2 𝜇0
2 𝜇0
which yields the energy density u=U/V
1 𝐵2
𝑢=
2 𝜇0
1 𝐵2
for vacuum or 𝑢 =
in a magnetic
2 𝜇 material
R-L circuits
𝑑i
ℰ − 𝑅𝑖 − 𝐿 = 0
𝑑𝑡
t=0 is time when switch is closed all the
voltage drops across L and thus i(t=0)=0
Kirchhoff’s loop rule
For𝑡 → ∞ I becomes stationary and is limited only by R
ℰ
𝐼=
𝑅
Solving the differential equation:
𝑖
𝐿
0
𝑑𝑖′
=
ℰ − 𝑅𝑖′
𝑡
𝑑𝑡′
0
Substitution x = ℰ − 𝑅𝑖 yields dx = −𝑅𝑑𝑖
ℰ −𝑅𝑖
−𝐿
ℰ
𝑑𝑥
=
𝑥
𝑡
𝑑𝑡′
0
ℰ
ℰ −Rt
𝑖 = (1 − e L )
𝑅
𝑅
−𝐿 ℰ − 𝑅𝑖
ln
=t
𝑅
ℰ
ℰ − 𝑅𝑖 =
R
−Lt
ℰe
i
𝐼 = ℰ/R
𝑖 𝑡 = 𝜏 = ℰ/R(1−1/e) =0.63ℰ/R
𝜏 = 𝐿/𝑅
t
What happens if we release energy stored in the solenoid
Kirchhoff’s loop rule:
𝑑i
𝑅𝑖 + 𝐿 = 0
𝑑𝑡
𝐿 𝑖 𝑡
− ln
=𝑡
𝑅
𝐼0
𝐿
−
𝑅
𝑖=
ℰ − 𝑅𝑖 − 𝐿
𝑖
𝑑𝑖′
=
𝑖′
𝐼0
R
−Lt
𝐼0 e
i
𝐼0
I0/e
𝜏 = 𝐿/𝑅
t
𝑑i
=0
𝑑𝑡
𝑡
𝑑𝑡′
0
L-C circuits
L-C circuit shows qualitative new behavior
Because there is no power dissipation, energy once
stored in C or L will periodically redistribute between
energy in E-field and B-field
From http://en.wikipedia.org/wiki/LC_circuit
Kirchhoff’s loop rule
𝑞
𝑑i
− −𝐿 =0
𝐶
𝑑𝑡
𝑞
𝑑2q
− −𝐿 2 =0
𝐶
𝑑𝑡
𝑑2q 1
+
𝑞=0
𝑑𝑡 2 𝐿𝐶
Compare with harmonic oscillator
𝜔=
1
𝐿𝐶