Faraday’s law can be used to describe the process of induction. In modern vector differential form
Faraday’s law is given by
∇ × E = −
∂ B
∂t
.
(1)
Electromagnetic forces are described by the Lorentz force law
F = Q ( E + v × B ) , (2) where v is the velocity of charge Q . For typical circuits, the charges move in wires. Faraday’s law and the Lorentz force law always give the correct classical physics of induction. This includes the behavior of transformers and generators.
While these equations are correct, it can often be simpler to work with integral equations.
When Faraday was experimentally developing his law he was essentially working with an integral equation. Faraday’s experimental law can be written as
V = − dΦ d t
, (3) where V is the “Electromotive Force” ( EMF ), and the flux Φ is a surface integral of the magnetic flux density B Z
Φ = B · d s .
(4)
S
The EMF is what we measure in the laboratory. Historically, a “galvanometer” would be used to measure the EMF . A galvanometer is a very sensitive mechanical meter. Today we would typically use electronic meters (such as an oscilloscope), not mechanical galvanometers. The electromotive force is defined as the path integral of the force per charge around a closed circuit, so integrating
(2) gives I
F
V = · d l =
I
( E + v × B ) · d l .
(5)
C
Q
C
Equating this EMF (5) with the time derivative of the flux (4) gives
I d
( E + v × B ) · d l = −
Z
B · d s .
C d t
S
(6)
This is Faraday’s law (3) written in terms of the fields E and B . This integral form of Faraday’s law works correctly for both stationary circuits like transformers and moving circuits like generators.
In practical use, we will be measuring the EMF , so we need to calculate the change of flux term on the right hand side (RHS) of (6) to predict the V we will measure. If the circuit is stationary, then the time derivative can be moved inside the integral, but what if the circuit is moving like in a generator? Fortunately for engineers, there is an integral theorem that gives the answer. Reynolds
Revision: April 23, 2012
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developed the theorem for fluid mechanics, but we can apply the mathematical theorem to the electromagnetics case. The “Reynolds Transport Theorem” for an arbitrary vector field A is d
Z
A · d s =
Z
∂ A
· d s +
Z
( ∇ · A ) v · d s −
I v × A · d l , d t
S S
∂t
S C
(7) where the surface S and the corresponding boundary C move with velocity v . Apply this transport theorem to the RHS of (6), noting that the divergence of B is always zero d
Z
V = − B · d s = −
Z
∂ B
· d s +
I v × B · d l .
d t
S S
∂t
C
(8)
The first term on the right is the “transformer EMF ”, and the second term on the right is the
“motional EMF .” Equating the integral expression for the EMF , (RHS of (5)) to the RHS of (8) produces I Z
∂ B
( E + v × B ) · d l = − · d s +
I v × B · d l .
(9)
C S
∂t
C
Canceling the common terms gives
I Z
∂ B
E · d l = − · d s .
(10)
C S
∂t
This is just the equation obtained by surface integrating and applying Stokes’ theorem to (1). So we have shown that given the Lorentz force law (2), Faraday’s experimental law (3), and the definition of EMF (5) one can derive the integral Faraday’s law (6). With a bit more work the differential form of Faraday’s law (1) could also be obtained. Kirchhoff’s voltage law (KVL) is the statics form of (10), but with time dependant B fields the RHS need not be zero. Note that the velocities in
(2), (5) and (6) (velocity of charge) are not the same as the velocities in (7) and (8) (velocity of circuit). Yet the v × B · d l terms in (5), (6) and (8) are equal. Thought question: why are the v × B · d l terms equal? See Question 1 below.
The potential V is the line integral of E between two points. Written in terms of the readout and test leads of a Digital Volt Meter (DVM), we get
Z
V
DVM DISPLAY
= −
RED
E · d l .
BLACK
(11)
Actually the DVM measures EMF as given by (5) or (6) , where the path integral is closed through the meter. For this experiment, and usually with transformers, we do not have any motional EMF , so the v × B · d l term is zero. If we then assume the DVM has infinite impedance and “perfect electrical conductor” (PEC) wires, then (5) reduces to (11). With zero motional EMF around a closed loop V equals the sum of the voltages around the loop. Extending (11) to a circuit loop, the line integral of E around a closed loop is the sum of the voltages around the loop, i.e., a KVL sum
X
V i
= − E · d l .
(12)
LOOP
C
Then Faraday’s law (10) can be written with a KVL sum as
X
Z
V i
=
∂ B
· d s .
∂t
LOOP
S
(13)
Note the KVL sum is not necessarily zero, because of the induction term. The RHS of (13) describes the transformer effect.
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As an example suppose we have a simple induction circuit
+
B(t)
+
+
B=0 B=0
where M
1 and M
2 are high impedance voltmeters. For simplicity, assume no current flows in the high impedance meters, and ignore the inductance of the resistor loop. Further, the magnetic field
B ( t ) is assumed to be uniform inside the resistor loop of area A , and B is zero elsewhere. What do the voltmeters show? First use Faraday’s law, (10) or (13), to find the current in the resistor circuit loop. Apply the left side of Faraday’s law to the resistor loop (a KVL sum), and equate to the induction term, then solve for the current
Faraday law apply to resistor loop find flux integral solve for current
X
LOOP
V
IR
IR
I i
=
=
=
=
Z
∂ B
· d s
Z
S
∂t
∂ B
· d s
A
∂B
∂t
A ∂B
A
.
R ∂t d s = A
∂B
∂t
(14)
(15)
(16)
(17)
Now find M
1 reading using known current
Faraday law apply to loop containing so
M
1
& R
X
Z
∂ B
M
1
V i
= · d s
LOOP Z
S
∂t
− IR =
M
1
0 · d s = 0
S
= IR.
(18)
(19)
(20)
Lastly find M
2 reading using known current
Faraday law apply to loop containing so
M
2
& R
X
Z
∂ B
− M
2
V i
=
LOOP
S
∂t
+ IR = A
∂B
∂t
M
2
= 0 .
· d s (21)
(22)
(23)
Note that M
1
= M
2
, even though both voltmeters are connected to the same points. While this may seem strange, it is this effect that allows simple transformer circuits to make interesting things like directional couplers, etc. Unfortunately it can also cause serious measurement errors for the unwary.
Hint for real world work: Do not include stray flux in your measurement circuits.
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1. Why is the v × B · d l term in (5) zero for stationary circuits? Remember that v is the velocity of the charge. Hints: Cross product is zero for parallel vectors and vector identity
A · ( B × C ) = ( A × B ) · C = ( C × A ) · B .
2. Solve for I in terms of B , and use the resulting I to find M
1 and M
2 in the following circuit.
+
B=0
+
B(t)
+
B=0
+
Oscilloscope, signal generator, clamp-on AC ammeter probe (magnetic core with winding), single and dual resistor loops, BNC T connector, and cables.
1. Put the BNC T on the generator main output. Use a BNC to BNC cable to connect the sync output of the generator to the external trigger input of the oscilloscope. Connect the clamp-
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on probe output cable to the signal generator main output. The output of the clamp-on is driven by the generator.
2. Now build the circuit of Figure 1, with the core clamped around the loop and probes as shown below. Open the clamp and put the single-resistor loop on the core so that the core passes through the resistor loop. Using two BNC to grabber (or alligator) cables connect the two oscilloscope channels to the resistor loop. One pair of clips should connect to the loop from the right, the other pair should connect from the left. Both black leads should connect to the same point (outside the core), and both red leads should connect to the other terminal of the resistor at the center of the core.
Core
Red Red
M2 R M1
Black Black
3. Set the generator to about 100Hz sine-wave and adjust the scope to obtain stable triggering.
4. Set the oscilloscope inputs to the maximum vertical sensitivity on both channels. Adjust the generator output to show 80-100% of full scale trace(s) and set the time base set to show one or two cycles. Sketch the waveforms and record the scale factors.
5. Remove the probe that connects to the loop side of the resistor ( M
2 in Figure 1) and use a BNC to BNC cable to connect this oscilloscope channel to the generator output at the T connector. Adjust the scope so that the generator signal is on screen, and use the generator voltage and the resistor voltage to calculate the number of turns in the clamp-on probe.
Assume an ideal transformer with N
P
/N
S
= V
P
/V
S
.
N
=
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6. Built the circuit of Figure 2. Open the clamp, remove the single resistor loop, and put the dual resistor loop on the core so that the core passes through the resistor loop. Using two
BNC to grabber (or alligator) cables connect the two oscilloscope channels to the resistor loop. One pair of clips should connect to the loop from the right, the other should connect from the left. Both black leads should connect to the same point (outside the core), and both red leads should connect to other terminal of the resistors at the center of the core.
7. Adjust the generator output to show 80-100% of full scale traces and set the time base to show one or two cycles. Sketch the waveforms and record the scale factors.
8. (Trick Question) Which trace is the correct voltage across the resistors in the above sketch?
9. Keeping in mind that the black leads must be connected to the same point outside the core
(our oscilloscopes do not have differential inputs), how do you rearrange the probes to get both channels to show identical voltages? Test your concept by making it work, and draw a schematic similar to Figure 2.
10. Show your setup to the laboratory instructor.
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