Post on 14-Jan-2016
Inductance
Inductance
•Definition and Calculation of Self-Inductance
• To obtain an expression for the Energy Stored by an Inductor
•Definition and Calculation of Mutual-Inductance
Learning Objectives
The phenomenon of self-inductance was discovered by Joseph Henry in
1832 (Princeton University).
Joseph Henry 1797-1878
Let’s start with:
SELF-INDUCTANCE
When current in the circuit changes, the flux changes also, and a self-induced voltage appears in the circuit
Self Inductance
I constant, = 0
L is the self-inductance of the coil.
I increasing or decreasing
, = Vab>0ab
€
∝LdI
dt
(a) Definition used to find LSuppose a current I in a coil of N turns causes a flux B to thread each turn
€
NφB ∝ B∝ IThe self-inductance L is defined by the equation
€
NφB = LI
€
NφB = LI
€
L =NφB
I
(b) Definition that describes the behaviour of an inductor in a circuit
From Faraday’s Law of Induction
€
=−d
dtNΦB
€
=−d
dtNΦB = −
dLI
dt= −L
dI
dt
€
NφB = LI
dt
dIL−=
Two equivalent definitions of L
SI unit for inductanceV s A-1
This is called the henry (H)
dt
dIL−=
If a current changing by 1A/s is to generate 1V, the inductance is 1H.
Calculation of Self-Inductance
€
L =NΦB
I
The Self-Inductance of a Solenoid
n turns per unit length, radius Rand the length of the solenoid is l
Set up a current I, and we have a B field
nIB 0μ=Total number of turns is N=nl
Flux through each turn
€
φB = πR2B
€
NΦB = (nl)πR2B
= (nl)πR2(μ0nI) = πR2μ0n2Il
€
L =NΦB
I= μ0n
2πR2l
€
L =NΦB
I= μ0n
2πR2l
The inductance does not dependon current or voltage, it is a property
of the coil. (length, width, and number of turns per unit length)
Find the self-inductance of a solenoid of length 10 cm, area 5 cm2, and 100 turns.
n = 100/0.1 = 1000 turns/m
€
L = μ0n2πR2l
= 4π ×10-7( ) ×106 × 5 ×10−4
( ) × 0.1
= 6.27 ×10−5 HAt what rate must the current in the solenoid change to induce a voltage of 20 V?
Answer: 3.18 105 A/s dt
dIL−=
The Self-Inductance of a Toroid
ab
a
b
The Self-Inductance of a Toroida
b
Consider an elementary strip of area hdr
r
dr
h
€
B =μ0IN
2πr
( )∫=∫=b
aB hdrBAdB. ∫=
b
a r
drINh
π
μ
20
a
bINhln
20
πμ
=
€
dφB =μ0IN
2πrhdr
The Self-Inductance of a Toroid
⎟⎠
⎞⎜⎝
⎛=
=a
bhN
I
NL B ln
2
20
π
μ
Inductance – like capacitance – depends only on geometric factors
Unit of μ0 is H m-1
μ0 = 4π 10-7 H m-1
From the worked examples it can be seen that:
μ0 = 4π 10-7 wb/Am
€
L =NΦB
I= μ0n
2πR2l
The Energy Stored by an Inductor
I increasing
dt
dIL=
dt
dILIIP ==
The energy dU supplied to the inductor during an infinitesimal time interval dt is:
€
dU = Pdt = LI × dI
a b
(Faraday’s law in disguise)
The Energy Stored by an Inductor
The total energy U supplied while the current increases from zero to a final value I is
22
1
0
LIIdILUI
∫ ==
This energy is stored in the magnetic field
The energy stored in the magnetic field of an inductor is analogous to that in the electric field of a capacitor
22
1LIU =
C
QU
2
2
1=
Example: the energy stored in a solenoid
2
2
1LIU =
Energy per unit volume (magnetic energy density)
2202 2
1In
lR
UuB μ
π==
nIB 0μ=0
222
0 μμ B
In =
22202
1lIRn πμ=
MAGNETIC ENERGY DENSITY IN A VACUUM
0
2
2μB
uB =
The equation is true for all magnetic field configurations
Compare with the energy density in an electric field
202
1EuE =
The self-inductance L is defined by the equations
€
NφB = LI
dt
dIL−=
Review and Summary
Review and Summary• An inductor with inductance L carrying
current I has potential energy
22
1LIU =
•This potential energy is associated with the magnetic field of the inductor. In a vacuum, the magnetic energy per unit volume is
0
2
2μB
uB =
How would the self-inductance of a solenoid be changed if
(a)the same length of wire were wound onto a cylinder of the same diameter but twice the length?
(b)twice as much wire were wound onto the same cylinder?
(c)the same length of wire were wound onto a cylinder of the same length but twice the diameter?
(a) Since the diameter does not change, the number of turns and the area A remain constant. However, n2 is diminished by a factor of 4 and l is increased by a factor of 2. Thus L is reduced by a factor of 2.
(b) Using twice as much wire and making no other change, n2 and L are increased by a factor of 4.
(c) With twice the diameter, n2 is reduced by a factor of 4, but A is increased by the same factor; L is unchanged.
Mutual Inductance
A changing current in loop 1 causes a changing flux in loop 2 inducing a voltage
dt
dN B2
22
−=
122 iN B ∝
Mutual Inductance
(1) The mutual inductance M21 is defined by the equation:
12122 iMN B =(2) The mutual inductance M21 may also be
defined by the equation:
dt
diM 1
212 −=
Mutual InductanceIt can be proved that the same value is obtained for M if one considers the flux threading the first loop when a current flows through the second loop
2
11
1
22
i
N
i
NM BB
=
= (mutual inductance)
dt
diM
dt
diM 2
11
2 & −=−=
(mutually induced voltages)
A Metal DetectorSinusoidally varying current
Parallel to the magnetic field of Ct
Review and Summary•If two coils are near each other, a changing current in either coil can induce a voltage in the other. This mutual induction phenomenon is described by
dt
diM
dt
diM 2
11
2 & −=−=
where M (measured in henries) is the mutual inductance for the coil arrangement
Revision – Ampere’s Circuital Law
enclosed0 Ild.B∫ =μ
∫ ldB.
Where• is the line integral round a closed loopand• Ienclosed is the current enclosed by the loop
∫ ldB.
The B Field Due to a Long Straight Wire
( ) IrB 02 μπ =×
r2
IB 0
πμ
=
Magnetic Materials
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