Lecture 11
Faraday’s Law and Electromagnetic Induction
and
Electromagnetic Energy and Power Flow
In this lecture you will learn:
• More about Faraday’s Law and Electromagnetic Induction
• The Non-uniqueness of Voltages in Magnetoquasistatics
• Electromagnetic Energy and Power Flow
• Electromagnetic Energy Stored in Capacitors and Inductors
• Appendix (some proofs)
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Faraday’s Law Revisited
r
r
r
r
r
r
∂
∂
∫ E . ds = − ∫∫ B . da = − ∫∫ µo H . da
∂t
∂t
A closed contour
Faraday’s Law: The line integral of Efield over a closed contour is equal to
–ve of the time rate of change of the
magnetic flux that goes through any
arbitrary surface that is bounded by
the closed contour
B
Important Note: In electroquasistatics the line integral of E-field over a closed
contour was always zero
r r
r
∫ E . ds = ∫ (− ∇φ ) . ds = 0
In magnetoquasistatics this is NOT the case
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
1
Electromagnetic Induction and Kirchoff’s Voltage Law
Kirchoff’s voltage law comes from the electroquasistatic
r
r
approximation:
R1
∫ E . ds = 0
-
r
r
⇒ ∫ E . ds = I R1 + I R2 − V = 0
⇒
+
+ V
-
I
V
I=
(R1 + R2 )
-
+
Now consider a circuit through which the magnetic
flux is changing with time (Kirchoff’s voltage law is
violated)
⇒
r
r
r
r
dλ
∂
∫ E . ds = − ∫∫ B . da = −
dt
∂t
R1
-
r
r
dλ
⇒ ∫ E . ds = I R1 + I R2 − V = −
dt
⇒
R2
I=
V
−
1
I
+
λ (t )
+ V
-
dλ
(R1 + R2 ) (R1 + R2 ) dt
-
+
R2
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Lenz Law
R1
Suppose an induced current I is flowing through the wire:
⇒
I=−
+
-
dλ
1
(R1 + R2 ) dt
λ (t )
I
The induced current in the wire produces its own
magnetic field
-
+
R2
Lenz Law is just an easy way to remember in which direction the induced current flows
The law states that the induced current will flow in a direction such that its own
magnetic field opposes the time variation of the magnetic field that produced it
Example: Suppose the magnetic flux through the wire loop shown above was
increasing with time (so that dλ/dt > 0).
Lenz would tell us that the induced current would flow in the clockwise direction so
that its own magnetic field would oppose the increasing magnetic flux through the loop
In the equation above this fact comes out from the negative sign on the right hand side
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
2
Non-Uniqueness of Voltages in Magnetoquasistatics - I
R1
We have:
+
-
r
r
dλ
∫ E . ds = I (R1 + R2 ) = −
dt
λ (t )
V1
I
V2
-
+
R2
Question: What is the voltages difference V2-V1 ?
One may be tempted to write …….
V1 − I R2 = V2
⇒
I=0
V2 − I R1 = V1
⇒
V2 − V1 = 0
….. which cannot be correct since we know that: I = −
dλ
1
(R1 + R2 ) dt
What went wrong? Our usual concepts of circuit theory and potentials which are
based on conservative E-fields are not valid when time varying magnetic fields
are present
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Non-Uniqueness of Voltages in Magnetoquasistatics - II
Lets do some real experimental measurements and see what we get for V2-V1
C2
voltmeter
V
+
Faraday’s Law for contour C1:
r r
r
r
∂ ∫∫ B . da
∫ E . ds = −
∂t
dλ
⇒ I R1 + I R2 = −
dt
V2
I
C1
+ R2 -
⇒
V = V2 − V1 = −
Faraday’s Law for contour C2:
⇒
+
λ (t )
V1
-
r r
r
r
∂ ∫∫ B . da
∫ E . ds = −
∂t
⇒ V − I R1 = 0
R1
R1 ⎛ dλ ⎞
⎜
⎟
R1 + R2 ⎝ dt ⎠
This value for V2-V1 would actually be measured
experimentally if the measurement is done as
shown above
V = V2 − V1 = IR1
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
3
Non-Uniqueness of Voltages in Magnetoquasistatics - III
Lets do another real experimental measurement
and see what we get for V2-V1
C3
V1
V
+
I
R1
-
C1
+
λ (t )
V2
+ R2 voltmeter
Faraday’s Law for contour C1:
r r
r
r
∂ ∫∫ B . da
∫ E . ds = −
∂t
dλ
⇒ I R1 + I R2 = −
dt
⇒
Faraday’s Law for contour C3:
r r
r
r
∂ ∫∫ B . da
∫ E . ds = −
∂t
⇒ − V − I R2 = 0
⇒
V = V2 − V1 =
R2 ⎛ dλ ⎞
⎜
⎟
R1 + R2 ⎝ dt ⎠
This value for V2-V1 would actually be measured
experimentally if the measurement is done as
shown above
Lesson: The result of a voltage measurement
depends on how exactly you do the
measurement when time varying magnetic
fields are present
V = V2 − V1 = − IR2
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Inductors and Faraday’s Law - I
Consider an inductor with a time varying current:
Inductance = L
The magnetic flux is given by:
λ (t ) = L I (t )
I (t )
+
V (t ) _
Consider Faraday’s law over the closed contour C:
λ
C
r
r
r
r
r
r
∂
∂
∫ E . ds = − ∫∫ B . da = − ∫∫ µo H . da
∂t
∂t
d λ (t )
d I (t )
⇒ V (t ) =
=L
dt
dt
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
4
Inductors and Faraday’s Law - II
Inductance = L
Question: is there electric field
anywhere else besides at the
terminals??
+
I (t )
V (t ) _
λ
C’
Consider Faraday’s law over the new closed contour C’:
r
r
r
r
r
r
∂
∂
∫ E . ds = − ∫∫ B . da = − ∫∫ µo H . da
∂t
∂t
r
r
1 d λ (t )
− V (t ) + ∫vertical E . ds = −
2 dt
r
r V (t )
⇒ ∫vertical E . ds =
2
There must be E-field inside the loop !!
This E-field is not very easy to determine analytically.
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
The Current-Charge Continuity
Equation in Electromagnetism - I
r
r r ∂ε E
Start from:
∇×H = J +
∂t
Take divergence on both sides:
⇒
(
)
( )
r
r ∂∇. ε E
0 = ∇ .J +
∂t
Use Gauss’ law:
To get:
( )
0
r
r
r ∂∇. ε E
∇ . ∇ × H = ∇ .J +
∂t
r
∇. ε E = ρ
r ∂ρ
− ∇ .J =
∂t
Current-charge continuity equation
What does this equation mean??
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
5
The Current-Charge Continuity Equation in Electromagnetism - II
Suppose the current density is spatially varying as shown below
∆x
x
Jx (x )
x + ∆x
J x ( x + ∆x )
The difference in current density at x and x+∆x must result in piling up of
charges in the infinitesimal strip …….
∂ ρ ( x , t ) ∆x
∂t
J x ( x + ∆x , t ) − J x ( x , t ) ∂ ρ ( x , t )∆x
⇒−
=
∂t
∆x
∂J x ( x , t ) ∂ ρ ( x , t )
⇒−
=
∂x
∂t
J x ( x , t ) − J x ( x + ∆x , t ) =
Now generalize to 3D and get the desired result:
r
r r
∂ ρ (r , t )
− ∇ . J (r , t ) =
∂t
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
The Current-Charge Continuity Equation in Electromagnetism - III
• A continuity equation of the form:
r
r r
∂ ρ (r , t )
− ∇ . J (r , t ) =
∂t
establishes a relation between the flow rate of a quantity (in the present case
the flow rate of charge density) and the time rate of change of the quantity
• A continuity equation is in fact a “conservation law”
For example, the current-charge continuity equation expresses charge conservation.
It says that charge cannot just disappear. In the integral form it becomes:
r r
r ∂
r
dQ (t )
− ∫∫ J (r , t ) . da =
∫∫∫ ρ (r , t ) dV =
∂t
dt
If there is a net inflow of charge into a closed volume
then that must result in an increase in the total charge
Q inside that closed volume
closed
volume
J
• In the next few slides we will try to find a power-energy
continuity equation for the electromagnetic field
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
6
The Electromagnetic Power-Energy Continuity Equation - I
• We know that electric and magnetic fields have energy. Let W be the energy
density (i.e. energy per unit volume) of the electromagnetic field
r
W (r , t ) = Energy density (units: Joules/m3)
• Suppose we had a vector that expressed the energy flow rate (or energy flux) in
Joules/(m2-sec) for the electromagnetic field
r r
S (r , t ) = Energy flow rate (units: Joules/(m2-sec))
• Then the conservation of electromagnetic energy must be expressed in a
continuity equation of the form:
r
r r
∂ W (r , t )
− ∇ . S (r , t ) =
∂t
Compare with the current-charge continuity equation:
r
r r
∂ ρ (r , t )
− ∇ . J (r , t ) =
∂t
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
The Electromagnetic Power-Energy Continuity Equation - II
Start from the two Maxwell’s equations:
r
r
∂ µo H
∇×E = −
∂t
r
r r ∂ε E
∇×H = J +
∂t
r
r
Take the dot-product of the first equation with H and of the second equation with E
to get:
r r
(∇ × E ). H = − 21 ∂ µo∂Ht .H
r
r
r r
r r r r 1 ∂ ε E .E
∇ × H .E = J .E +
2 ∂t
(
)
Subtract the above two to get:
r r
r r
(∇ × E ). H − (∇ × H ). E = − 21 ∂ µo∂Ht .H − J . E − 21 ∂ ε ∂Et . E
r
r
r
r
r r
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
7
The Electromagnetic Power-Energy Continuity Equation - II
r r
r r
(∇ × E ). H − (∇ × H ). E = − 21 ∂ µo∂Ht .H − J . E − 21 ∂ ε ∂Et . E
r
r
r
r
Use the vector identity on the left hand side:
To get:
Define:
r r
(
) (
)
(
)
r r
r r
r r
∇ . A× B = ∇ × A .B − ∇ × B . A
r r
r r
r v
∂ ⎛ µ H .H ε E . E ⎞ r r
⎟ + J .E
- ∇ . E × H = ⎜⎜ o
+
∂t ⎝
2
2 ⎟⎠
(
)
r r
r r
r r
S (r , t ) = E (r , t ) × H (r , t )
r r
r r
r r
r r
⎛ µo H (r , t ) .H (r , t ) ε E (r , t ) . E (r , t ) ⎞
⎟⎟
⎜
W (r , t ) = ⎜
+
2
2
⎠
⎝
And get:
r
r r
r r
∂ W (r , t ) r r
+ J (r , t ) . E (r , t )
− ∇ . S (r , t ) =
∂t
This is what we
wanted (other than
the last term on the
right hand side)
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
The Electromagnetic Power-Energy Continuity Equation - III
r r
r r
r r
Poynting Vector: S (r , t ) = E (r , t ) × H (r , t )
r r
• S (r , t ) is called the Poynting vector and it is the energy flow per second per
unit area
r r
• S (r , t ) points in the direction of power flow
r r
• S (r , t ) has the units of Joules/(m2-sec) or Watt/m2
Energy Density:
r r
r r
r r
r r
⎛ µo H (r , t ) .H (r , t ) ε E (r , t ) . E (r , t ) ⎞
⎟⎟
W (r , t ) = ⎜⎜
+
2
2
⎝
⎠
r
• W (r , t ) is the energy density of electromagnetic field
r
• W (r , t ) has the units of Joules/m3
r
• W (r , t ) has two parts:
r r
r r
µo H (r , t ) .H (r , t )
r r
2
r r
ε E (r , t ) . E (r , t )
2
Energy density of the
magnetic field
Energy density of the
electric field
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
8
The Electromagnetic Power-Energy Continuity Equation - IV
Power dissipation:
r
r r
r r
∂ W (r , t ) r r
+ J (r , t ) . E (r , t )
− ∇ . S (r , t ) =
∂t
• The last term in the continuity equation J.E indicates power dissipation
• Its units are Joules/(m3-sec)
• This term represents the energy that is lost from the electromagnetic field
Essentially what is happening is that if the medium is conductive then in the
presence of electric fields currents will be produced and the last term represents
the usual I 2R losses in conductors or resistors
r r
r
Integral Form: − ∫∫ S (r , t ) . da =
Net power flow
into a closed
volume
r r
r r
r
∂
∫∫∫ W (r , t ) dV + ∫∫∫ J (r , t ) . E (r , t ) dV
∂t
closed
volume
Rate of increase of the total
electromagnetic energy
inside the closed volume
=
S
Rate of the total energy
loss through dissipation
+
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Example: Parallel Plate Capacitor
area = A
φ=0
φ=V
+
+
+
+
+
+
+
+
Recall that:
x
-
ε
0
⎛1 x⎞
⎟
⎝2 d ⎠
φ (x ) = V ⎜ −
Ex (x ) = −
d
+
∂φ ( x ) V
=
∂x
d
V
Electric field energy density = ε E x ( x ) . E x ( x )
2
Total electric field energy =
ε Ex (x ). Ex (x )
2
1
= CV 2
2
=
ε ⎛V ⎞
2
⎜ ⎟
2⎝d ⎠
2
ε V
( A d ) = ⎛⎜ ⎞⎟ ( A d )
2⎝d ⎠
C=
εA
d
For all (linear) capacitors the total stored electric field energy equals CV 2/2
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
9
Example: Parallel Plate Inductor
y
I
W
µ H d l µo dl
⇒L= o x
=
I
W
Hx = K =
Hx
d
x
(L is the total inductance)
Magnetic field energy density =
Total magnetic field energy =
K zˆ
W
The structure has length ℓ in the z-direction
Recall that:
K=
µo H x . H x
2
µo H x . H x
2
1 2
= LI
2
=
I
W
− K zˆ
µo ⎛ I ⎞2
⎜ ⎟
2 ⎝W ⎠
2
(W d l ) = µo ⎛⎜ I ⎞⎟ (Wd l )
2 ⎝W ⎠
For all (linear) inductors the total stored magnetic field energy equals LI 2/2
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Power Flow in Electroquasistatics and Circuit Theory - I
Consider a complex electrical circuit shown below:
_
V1
I3
I1
+
+
_
V3
A complex electrical circuit
_
V2
+
I4
I2
_
V4
+
If you wanted to calculate the total electrical power P going into the circuit you
would do the following sum:
P = ∑ Vn I n = V1 I1 + V2 I2 + V3 I3 − V4 I 4
n
Where does this formula comes from?
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
10
Power Flow in Electroquasistatics and Circuit Theory - II
closed volume
_
V1
I3
I1
+
_
+
V3
A complex electrical circuit
_
V2
I4
+
_
I2
r r
r r
+
V4
r r
Start from the Poynting Vector: S (r , t ) = E (r , t ) × H (r , t )
r
(
r
r
r
)
r
Total power flow into the closed volume = P = − ∫∫ S . da = − ∫∫ E × H . da
r
In electroquasistatics: E = −∇ φ
Therefore:
(
)
r
r
P = ∫∫ ∇φ × H . da
( )
r
r
r
r
P = ∫∫ ∇ × (ϕ H ). da − ∫∫ φ (∇ × H ). da
r
r
(
r
Use the vector identity: ∇ × ϕ F = (∇ϕ ) × F + ϕ ∇ × F
To get:
r
r
In magnetoquasistatics: ∇ × H = J
)
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Power Flow in Electroquasistatics and Circuit Theory - III
closed volume
_
V1
I3
I1
+
+
_
V3
A complex electrical circuit
_
V2
I4
+
I2
0
( )
(
)
r
r
r
r
P = ∫∫ ∇ × ϕ H . da − ∫∫ φ ∇ × H . da
(
_
V4
+
curl of a vector integrated over a
closed surface is always zero
)
⇒
r
r
P = − ∫∫ φ ∇ × H . da
⇒
r r
P = − ∫∫ φ J . da = ∑ Vn I n
n
⇒
P = ∑ Vn I n
n
Just what we wanted to prove
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
11
Appendix: Energy Stored in Capacitors is QV/2
U=
r r
ε E .E
dV
2
r
r
r
1
1
−
∫∫∫ ε E . ∇φ dV = −
∫∫∫ ∇ . φ ε E − φ ∇ . ε E dV
2 S + S1 + S2
2 S + S1 + S2
r r 1
0
1
−
∫∫ φ ε E . da +
∫∫∫ φ ρ dV
2 S + S1 + S2
2 S + S1 + S2
r0 r 1
r r 1
r r
1
− ∫∫ φ ε E . da − ∫∫ φ ε E . da − ∫∫ φ ε E . da
2S
2 S1
2 S2
r r 1
r r
1
− ∫∫ φ ε E . da − ∫∫ φ ε E . da
V1=V
2 S1
2 S2
∫∫∫
S + S1 + S2
=
=
=
=
(
1
1
Q1 V1 + Q2 V2
2
2
1
= QV
2
=
)
( )
S
S1
V
S2
V2=0
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
Appendix: Energy Stored in Inductors is Iλ/2
r r
µo H . H
U = ∫∫∫
dV
2
S
r
r r
r
r
1 r
1
= ∫∫∫ H . ∇ × A dV = ∫∫∫ ∇ . H × A + A . ∇ × H dV
2S
2S
r
1 r r 0 r 1 r
= ∫∫ H × A . da + ∫∫∫ A . ∇ × H dV
2S
2S
r
1 r
= ∫∫∫ A . ∇ × H dV
2S
1 r r
I
C
= ∫∫∫ A . J dV
2S
r r
1
= I ∫C A . ds
2
1
= Iλ
2
(
(
)
)
S
ECE 303 – Fall 2007 – Farhan Rana – Cornell University
12