Generating Currents
Consider the following circuit: a bar moves
on two rails that are connected at one end.
The whole setup has a magnetic field that
goes through it.
B

v


Generating Currents
If the bar is moving to the right, and the bar contains
electrons that are free to move (as in metal), then
the electrons are also moving to the right. There is
a magnetic force on the electrons pushing them
down with magnitude Fmagnetic = q v B.
B
Fmag on e  v 



Generating Currents
Thus, negative electrons should pile up at the
bottom of the moving bar, leaving a net
positive charge at the top of the bar. But
this should act just like a battery!
B
+

Fmag on e  v 

-

Generating Currents
To find the voltage of this “battery”, we note that as
the charges pile up at the ends of the rod, an
Electric field will be set up. The electrons will
continue to pile up until the Electric Force
(Fel =qE) balances the Magnetic Force.
B
+
Fmag on e  v 
Felec on e

-


Generating a Voltage
We now have, at equilibrium: SF = 0, or
Felec on e = Fmag on e , or qE = qvB , or
E = vB .
We know that the electric field is related to
voltage by E = -DV / Ds , (here Ds = L, the
length of the bar). Thus we have for the
voltage:
DV = v B L .
Generating Power
We have generated a voltage, and now to
generate electric power (P=IV) we need to
have that voltage drive a current. Since we
have completed the circuit by connecting the
ends of the rails, we will have a complete
circuit - and so we will get a current depending
on the resistance in the rails (V=IR).
Conservation of Energy
We have made an electric generator that can
generate electrical energy. But according to
the Law of Conservation of Energy, we can
only convert energy from one form into
another. In the case of the electric
generator, where does this energy come
from?
Generating Currents
Note that as electrons flow clockwise around the
circuit, this acts the same as a current of positive
charges going counterclockwise, as indicated on
the diagram below. Note that a current flows up
the bar.
B   +

 Fmag on e  v 
Felec on e

 

Generating Currents
Is there a magnetic force on this current due to its
flowing through a magnetic field? YES!
Note that the direction of the force on this current is
to the left. This will act to slow the bar down. In
effect, this apparatus converts the kinetic energy
of the bar into electric energy!
B   +

 Fmag on e  v 
Felec on e Fmag on I
 

Generalizing
We have from the previous apparatus:
DV = v B L .
We note that v = Dw/Dt where w is the width
of the circuit (distance from end of rails to
bar), so DV = (Dw/Dt) B L .
Can we take the D /Dt and apply it to all the
variables: DV = D(B L w) / Dt ?
From experiment, YES!
Faraday’s Law
We also note that wL = A (area of circuit).
We can also have N number of loops, so
we finally get: DV = D(N B A) / Dt . This
is called Faraday’s Law.
When we consider direction as well, we see
that the magnetic field, B, has to cut
through the area, A. If we assign a direction
to A that is perpendicular to the surface, we
get an even more general form:
DV = D[(N B A cos(qBA) ] / Dt .
Faraday’s Law
If the magnetic field is not constant over the
whole area, we need to break the area into
pieces and then sum up the pieces. This
leads to Faraday’s Law as:
DV = d[  B  dA ] / dt , where we have used
the dot product to represent the cos(qBA) .
Magnetic Flux
DV = d[  B  dA ] / dt
The quantity in brackets:  B  dA is called
the magnetic flux with units of Webers. It
is a measure of how much magnetic field
cuts through a certain area. Faraday’s Law
says there is a voltage produced whenever
this magnetic flux changes in time.
Webers = Volt-sec = T-m2 .
Magnetic Field
DV = d[  B  dA ] / dt
Because of its importance in magnetic flux,
the magnetic field, B, is sometimes called
the magnetic flux density.
Although we already have a unit for B, the
Tesla, because of B’s importance to
inductance and its relation to magnetic flux
density, it also has an equivalent unit called
the Weber/m2 = Tesla .
Lenz’s Law
DV = D[(N B A cos(qBA) ] / Dt or
DV = d/dt [  B  dA ]
The above formula is for determining the
amount of voltage generated. But what is
the direction of that voltage (what direction
will it try to drive a current)?
The answer is Lenz’s Law: the direction of
the induced voltage will tend to induce a
current to oppose the change in magnetic
field through the area.
Example #1
Using Lenz’s Law, which direction should the
induced current flow when the bar is moving to
the right as indicated below?
B

v


Example #1 (cont.)
Since the area through which the magnetic field is
going is increasing, the magnetic flux is
increasing. According to Lenz’s Law, the
induced current will flow in a direction so as to
create a magnetic field that will oppose the
increase. Here it means the current will create a
field going out of the loop and so using the RHR
the current must flow counterclockwise!
B 
I  Binduced 


v


Example #2
Consider the situation below in which the
North pole of a magnet is brought closer to
the center of a loop of wire.
Which way will the induced current be in the
wire?
FRONT VIEW
VIEW from LEFT
N B
S
v
B
Example #2 (cont.)
Since the magnetic field through the loop is
increasing, the flux is increasing. By
Lenz’s Law, the induced current will try to
create a magnetic field opposing the
increase, and the current direction is
determined by the RHR.
FRONT VIEW
VIEW from LEFT
 Iin
 Iin
S
N B
B
v   Bin
Bin


Example #2 (cont.)
If the North pole of the magnet is moved
away from the coil, then the external field
will decrease, and Lenz’s Law will say that
the induced current should create a
magnetic field to replace the decreasing
external magnetic field.
FRONT VIEW
VIEW from LEFT
 Iin
Iin 
S
N B
B
v
 Bin
Bin


Lenz’s Law
The Computer Homework assignment, Vol.
4 #3, deals with Lenz’s Law and will give
you practice with this as well.
Transformers and Inductors
DV = D[(N B A cos(qBA) ] / Dt .
We have already seen how changing the area
of a circuit in a magnetic field generates a
voltage.
We have also seen how changing the
magnetic field strength through the circuit
can generate a voltage - this is the basis of
an inductor and a transformer.
Transformer
We’ll consider an inductor in a little bit. But
now consider the situation in the figure
below. Note that circuit 1 is not electrically
connected to circuit 2. But it is connected
magnetically, since the magnetic field will
“flow” through the iron square ring.
I1 
B1
 B1
1
2
Transformer (cont.)
If I1 changes in circuit 1, then B1 changes.
But since the iron ring carries B1 through
circuit 2, a voltage and current will be
induced in circuit 2. The current in circuit
2 depends on how the current in circuit 1
changes (as well as the number of loops in
both circuit 1 and circuit 2).
I1 
B1
 B1
1
2
Transformer (cont.)
The voltage in circuit 2 depends on the change
in the current in circuit 1:
V2 = M12 dI1/dt
M12 is called the mutual inductance, and is
measured in Henries, where
henry = volt / [amp/sec] = volt-sec / amp.
I1 
B1
 B1
1
2
Mutual Inductance
Like capacitance and resistance, mutual
inductance depends NOT on voltage and/or
current, but on the geometry and materials.
In this case, the mutual inductance will
depend on the number of turns in both
circuits 1 and 2 as well as the geometries
(solenoids) used in circuits 1 and 2.
Electric Generators
Finally, we could change the direction of the
area in relation to the field - this is the basis
for the most common kind of generator.
This generator looks just like the electric
motor, except we put in rotational motion
and get current instead of putting in current
and getting rotational motion!
Electric Generator
If we use the crank to turn the area inside the
magnetic field, we are changing the magnetic
flux through the area. By Lens’ Law this
should generate a voltage and current!
crank
r
N
B
w
S
L
Electric Generators
DV = d/dt [(N B A cos(qBA) ] .
When we change the angle, qBA, with respect
to time: qBA=wt , we get the following
relation: DV = N B A wsin(qBA) , or
VAC = Vo sin(wt) where Vo= NBAw,
and where w=dqBA/dt =2pf .
This kind of voltage is an alternating voltage
(AC voltage) since the sine function
alternates between positive and negative.
Electric Generators
You should be able to DESIGN your own
generator based on the voltage (V) and
frequency (w) you want out of the
generator.
AC Circuits
VAC = Vo sin(wt)
For this kind of AC voltage, we can determine
the amplitude of the voltage (Vo= NBAw) .
But since the average of sine is zero, how
do we treat the average?
What is usually important is the power
delivered by the electric circuit. From
P=IV we see that both the current and the
voltage are important.
AC Circuits
From Ohm’s Law, we have V = IR, where R
is a constant that depends on the material
and geometry of the materials used to
conduct the current. Thus, I = V/R =
(NBAw/R) sin(wt) = Io sin(wt) , where Io =
NBAw/R = Vo/R .
From this we see that the electric power is:
P(t) = I(t)*V(t) = IoVo sin2(wt) .
AC Power
P(t) = I(t)*V(t) = IoVo sin2(wt)
Note that the Power involves the square of the
sine function, and so the Power oscillates
but is always positive.
But what we are usually interested in is the
average power. From the calculus, we find
that the average of sin2(q) = 1/2. Thus:
Pavg = (1/2)IoVo .
Average of sin2(q)
Avg of sin2(q) = S sin2(qi) / S(1)
This is approximate, so we could extend the
summation to an integral:
Avg of sin2(q) =  sin2(q) dq /  (1) dq , where
the limits of integration go from 0 to 2p for
both the numerator and the denominator.
Another way
A tricky way to get this without integrating is:
Note now that sin2(q) + cos2(q) = 1 for all q’s,
so the average of [sin2(q) + cos2(q)] = 1 .
But the average of [sin2(q) + cos2(q)] = 1 =
average of sin2(q) + average of cos2(q) .
Sine and cosine differ only by the starting
point, so the average of sin2(q) = average of
cos2(q) over a complete oscillation.
Therefore, average of sin2(q) = 1/2 !
RMS Voltage and Current
In order to work with AC circuits just as we
did with DC circuits, we create a voltage
and current called rms (root mean square).
Vrms = Vo (1/2)1/2 and
Irms = Io (1/2)1/2 so that we have
Pavg = Irms Vrms = (1/2) IoVo , and
Vrms = Irms R .
Eddy Currents
Another consequence of Faraday’s Law is the
existence of eddy currents. We’ll look at a
demonstration in class to see these eddy
currents at work.
Extension of Ampere’s Law
Recall from Part 3: Ampere’s Law:
closed loop B  dL = moIencircled
Consider a circuit with a capacitor that is
being charged. There is a current flowing to
the capacitor, and so there will be a
magnetic field that encircles the wire.
However, there is no current flowing across
the capacitor. Will there be a magnetic field
encircling the charging capacitor?
Ampere’s Law (extended)
Gauss’ Law for Electric Fields:
 E  dA = Qenclosed/eo
Recall also that I = dq/dt. These two can be
combined to give for a charging capacitor:
Ieffective = eo d/dt[  E  dA] .
With this, Ampere’s Law can be extended:
closed loop B  dL = moIencircled + moeo d/dt [  E  dA]
Maxwell’s Equations
We now have the four main laws for electric
and magnetic fields. Together they are
called Maxwell’s Equations. The four laws
are:
Maxwell’s Equations
Gauss’ Law for Electric fields
closed area E  dA = Qenclosed / eo
Gauss’ Law for Magnetic fields
closed area B  dA = 0
Ampere’s Law
closed loop B  dL = moIencircled + moeo d/dt [  E  dA]
Faraday’s Law
closed loop E  dL = -DV = -d/dt [  B  dA ] .