Humans are clever!
With the magnetic force only pushing
perpendicular to the motion, it would seem
that we could not use this force to increase
the speed of particles and hence generate
energy.
However, humans being clever, consider the
following slides.
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 (Fmag= qvB).
B
+
Felec on e
Fmag on e 

-

v

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 , or DV = -E*Ds
(here Ds = L, the length of the bar and the minus
sign indicates the electric field goes from high
voltage to low voltage). 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   +
 Felec on e
v
Fmag 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   +

 Felec on e
v
Fmag on e  Fmag on I 
 

Generating Currents
Note that if no current flows, there will be no electric
power, and there will be no magnetic force on the
bar since there will be no current, and thus the bar
will not slow down and no kinetic energy will be
lost in the bar. Only if current flows and electric
power is generated will kinetic energy be lost.
B   +

 Felec on e
v
Fmag 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).
DV = (Dw/Dt) B L
Can we take the D /Dt and apply it to all
the variables: DV = D(w B L) / Dt ?
The answer, based on experiment, is 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 .
Lenz’s Law
DV = D[(N B A cos(qBA) ] / Dt
The above formula, Faraday’s Law, is for
determining the amount of voltage generated.
But what is the direction of that voltage (i.e.,
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.
Lenz’s Law
We’ll go over several cases in class. We’ll
also have a lab later in the semester to play
with this.
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 could also change the magnetic field
strength through the circuit to generate a
voltage - this is the basis of an inductor and a
transformer.
[We’ll consider both a little later after we look at AC voltages.]
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
(two complete rings
rather than one split ring)
N
S
crank
(turn at frequency, f)
Electric Generators
DV = D[(N B A cos(qBA) ] / Dt .
When we change the angle, qBA, with respect to
time (qBA=wt) , the calculus gives us the
following relation: DV = N B A wsin(qBA) , or
VAC = Vm sin(wt) where Vm=NBAw,is the
max voltage, 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.
AC Circuits
VAC = Vm sin(wt)
For this kind of AC voltage, we can determine
the amplitude (max) of the voltage (Vm=
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 (R=rL/A). Thus, I = V/R
= (NBAw/R) sin(wt) = Im sin(wt) , where
Im = NBAw/R = Vm/R . The alternating
voltage creates an alternating current!
From this we see that the electric power is:
P = I V = ImVm sin2(wt) .
AC Power
P = I V = ImVm 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)ImVm .
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 = Vm (1/2)1/2 and Irms = Im (1/2)1/2
so that we have
Pavg = Irms Vrms and Vrms = Irms R .
Note that the power formula and Ohm’s Law are
the same for DC and for AC-rms, but NOT for
instantaneous AC.
Transformers
Transformers work this way: An AC voltage:
VAC-1 = V1 sin(wt) , is used to generate a current: IAC-1
= I1sin(wt) in coil #1 which generates an oscillating
magnetic field in coil #1 since B is proportional to I:
BAC-1 = B1sin(wt); a second coil (#2) is inside coil
#1; this #2 coil then, by Faraday’s Law,
DV = D[(N B A cos(qBA) ] / Dt
has a voltage, VAC-2, induced in it. By adjusting the
number of loops in both coils, the induced (AC)
voltage in #2 can be different than that in coil #1. We
can adjust (or transform) the voltage up or down!
Transformers
For safety, we like to have rather low voltages
in the house.
For economy, since Plost = I2R, we like to have
low currents (which means high voltages) in
our transmission lines.
We can have both if we use transformers
located in our neighborhoods!
Inductors
A coil of wire can create a magnetic field if a
current is run through it. If that current
changes (as in the AC case), the magnetic field
created by the coil will change. Will this
changing magnetic field due to the changing
current through the coil cause a voltage to be
created across the coil? YES!
This is called self-inductance and is the basis
behind the circuit element called the inductor.
Inductors
Since the voltage created depends in this case
on the changing magnetic field,
DV = D[(N B A cos(qBA) ] / Dt .
and the field depends on the changing current,
B = (mo/4p) I DL sin(qIr) / r2 (DL=length)
we have: Vinductor = -L DI / Dt (L=inductance)
where the L (called the inductance) depends
on the shape and material
(just like capacitance and resistance).
Inductors
Vinductor = -L DI / Dt
Here the minus sign means that when the
current is increasing, the voltage across the
inductor will tend to oppose the increase,
and it also means when the current is
decreasing, the voltage across the inductor
will tend to oppose the decrease.
Units: Henry
From
Vinductor = -L dI /dt
L has units of Volt / [Amp/sec] which is
called a Henry:
1 Henry = Volt-sec / Amp .
For a solenoid: L = mpN2R2 / Length
(A Henry is a rather large amount of inductance.)
Energy Stored in an Inductor
We start from the definition of voltage: V =
PE/q (or PE = qV). But since the voltage
across an inductor is related to the current
change, we might express q in terms of I:
I = dq/dt, or dq = I dt. Therefore, we have:
Estored = S qi Vi =  V dq =  V I dt and now
we use VL = L dI/dt to get:
Estored =  (L dI/dt) I dt =  L I dI = (1/2)LI2.
Review of Energy in Circuits
There is energy stored in a capacitor (that has
Electric Field): Estored = (1/2)CV2 .
Recall that V is related to E (Electric Field).
There is energy stored in an inductor (that
has Magnetic Field): Estored = (1/2)LI2 .
Recall the I is related to B (Magnetic Field).
There is power dissipated (as heat) in a
resistor: Plost = RI2 .
Review of Circuit Elements
Resistor: VR = R I
where I = Dq/Dt
Capacitor: VC = (1/C)q
(from C = q/V)
Inductor: VL = -L DI/Dt
We can make an analogy with mechanics:
q is like x;
V is like F;
t is like t;
L is like m;
I = Dq/Dt is like v = Dx/Dt; C is like 1/k (spring);
DI/Dt is like a = Dv/Dt;
R is like air
resistance.
RL Circuit - DC
What happens when we have a resistor in series with an
inductor in a circuit with a battery?
From Conservation of Energy, the voltage changes
around the circuit must equal zero:
Vbattery – Vresistor – Vinductor = 0 , or
Vbattery = I*R + L*dI/dt .
If we replace I with v, and dI/dt with a, and R with b (air
resistance), and L with m, and identify Vbattery =
constant with the constant force of gravity, mg, we
have an equivalent equation:
mg = bv + ma, or mg – bv = ma (with down being the
positive direction for a falling object).
RL Circuit - DC
From the mechanical analogy, this should be
like having a mass with air resistance. If we
have a constant force (like gravity), the object
will accelerate up to a terminal speed (due to
force of air resistance increasing up to the point
where it balances the gravity).
SF=ma  -bv + mg = ma, or
m dv/dt + bv = mg
Mechanical Analogy: mass
falling with air resistance
15 0
10 0
ti m e in se c o n d s
6 4.0
5 6.0
4 8.0
4 0.0
3 2.0
2 4.0
0
1 6.0
50
8.0

20 0
0
sp e e d in m / s
M a s s fa l lin g w i th a ir re s i s ta n c e
RL Circuit - DC (cont.)
If we connect the resistor and the inductor to a
battery and then turn the switch on, from the
mechanical analogy we would expect the
current (which is like velocity) to begin to
increase until it reaches a constant amount.
From conservation of energy:
Vbattery = Vresistor + Vinductor
where
Vbattery = constant; VR = I*R, and VL = L*dI/dt .
LR Circuit - qualitative look
From the circuit point of view, initially we have
zero current so there is no VR =I*R (voltage
drop across the resistor). Thus the full voltage of
the battery is trying to change the current,
hence VL = Vbattery, and so dI/dt = Vbattery /L.
However, as the current increases, there is more
voltage drop across the resistor, VR = I*R,
which reduces the voltage across the inductor
(VL = Vbattery - VR), and hence reduces the rate
of change of the current, dI/dt = VL/L !
RL Circuit - DC: I(t) versus t
t in milliseconds
46
41
36
31
26
21
16
11
6
1.2
1
0.8
0.6
0.4
0.2
0
1
I in amps
I(t) vs t for RL Circuit
Review of Circuit Elements
Resistor: VR = R I
where I = Dq/Dt
Capacitor: VC = (1/C)q
(from C = q/V)
Inductor: VL = -L DI/Dt
We can make an analogy with mechanics:
q is like x;
V is like F;
t is like t;
L is like m;
I = Dq/Dt is like v = Dx/Dt; C is like 1/k (spring);
DI/Dt is like a = Dv/Dt;
R is like air
resistance.
Inductive Reactance - AC
Since VL = -L DI/Dt, for an AC current we will have a
voltage induced that will oppose the changing current.
This opposition will tend to limit the current in the
circuit and behave in some sense like a resistance. We
call this action Reactance.
For an inductor, since VL = -L DI/Dt, VL will be big if L
is big and/or if w is big (causing DI/Dt to be big). A more
detailed calculation for reactance (or resistance to an AC
circuit) gives: XL = wL and is used in the Ohm’s Lawlike relation: VL = XLI
(the bigger XL, the smaller the I or the bigger the VL)
Capacitive Reactance - AC
There is a similar effect for a capacitor in an AC
circuit. For a capacitor, since VC = q/C, VC
will be small if C is big and/or if w is big
(causing little q to accumulate in the short time).
A more detailed calculation for reactance gives
XC = 1/(wC), and it is used in the Ohm’s Lawlike relation: VC = XCI
(the bigger XC, the smaller the I or the bigger the VC)
AC Circuits
A resistor obviously limits the current in a circuit. But,
as we just saw, a capacitor and an inductor also limit
the current in an AC circuit. However, the reactances
do not just add together. Using the fundamental
relations and the calculus, we come up with the
concept of impedance, Z: V = IZ where Z takes
into account all three reactances: XR=R, XL=wL and
XC= 1/wC:
Z = [R2 + (wL - 1/wC)2]1/2.
Power, however, is still: Pavg = I2R (not P=I2Z).
Oscillations
Newton’s Second Law: SF = ma can be written
as: SF - ma = 0 .
This is like Conservation of Energy: SV = 0 .
If we put an inductor with a capacitor with an AC
voltage, we have the analogy with a mass connected to
a spring that has an oscillating applied force.
In each of these cases, we get resonance.
We’ll demonstrate this in class with a mass and spring.
This is the basis of tuning a radio!
AC Circuits
V = IZ where Z = [R2 + (wL - 1/wC)2]1/2 .
Note that when (wL - 1/wC) = 0, Z is smallest
and so I is biggest! This is the condition for
resonance. Thus when w=[1/LC]1/2, we have
resonance. This is equivalent to the resonance of a
spring when w=[k/m]1/2 .
Computer Homework Vol. 4, #4, gives practice
with problems involving inductance (L) and
impedance (Z).