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Motional electromotive force
1. Motional EMF
x x x x x x
x x x x x x
x x x x x x
x x x x x x
x x x x x x
v
B -
- -
-
-
-
-
E
-
FE = -eE
FB = -evB
eE = evB
E = vB
Vind = LE = LvB
2. x x x x x x
x x x x x x
x x x x x x
x x x x x x
x x x x x x
v
L
d V dt
Fi = LdB Ff = L(d+vdt)B
|Vind | = dF/dt
= LvdtB/dt
= LvB
3. Motional emf
As the negative charges accumulate at the
base, a net positive charge exists at the
upper end of the conductor
As a result of this charge separation, an
electric field is produced in the conductor
Charges build up at the ends of the
conductor until the downward magnetic
force is balanced by the upward electric
force
There is a potential difference between the
upper and lower ends of the conductor
4. Motional emf, cont
The potential difference between the ends
of the conductor can be found by
• V = B v L
• The upper end is at a higher potential than the
lower end
A potential difference is maintained across
the conductor as long as there is motion
through the field
• If the motion is reversed, the polarity of the
potential difference is also reversed
5. Motional emf in a Circuit
Assume the moving bar
has zero resistance
As the bar is pulled to
the right with velocity v
under the influence of
an applied force, F, the
free charges experience
a magnetic force along
the length of the bar
This force sets up an
induced current because
the charges are free to
move in the closed path
6. Motional emf in a Circuit, cont
The changing
magnetic flux through
the loop and the
corresponding induced
emf in the bar result
from the change in
area of the loop
The induced, motional
emf, acts like a
battery in the circuit
R
v
B
I
and
v
B
7. Lenz’ Law Revisited – Moving
Bar Example
As the bar moves to the
right, the magnetic flux
through the circuit
increases with time
because the area of the
loop increases
The induced current
must in a direction such
that it opposes the
change in the external
magnetic flux
8. Lenz’ Law, Bar Example, cont
The flux due to the external field in
increasing into the page
The flux due to the induced current must
be out of the page
Therefore the current must be
counterclockwise when the bar moves to
the right
9. Lenz’ Law, Bar Example, final
The bar is moving
toward the left
The magnetic flux
through the loop is
decreasing with
time
The induced
current must be
clockwise to to
produce its own
flux into the page
10. Lenz’ Law, Moving Magnet
Example
A bar magnet is moved to the right toward a
stationary loop of wire (a)
• As the magnet moves, the magnetic flux increases with
time
The induced current produces a flux to the left,
so the current is in the direction shown (b)
11. Lenz’ Law, Final Note
When applying Lenz’ Law, there are
two magnetic fields to consider
• The external changing magnetic field
that induces the current in the loop
• The magnetic field produced by the
current in the loop
12. Generators
Alternating Current (AC) generator
• Converts mechanical energy to electrical
energy
• Consists of a wire loop rotated by some
external means
• There are a variety of sources that can
supply the energy to rotate the loop
These may include falling water, heat by
burning coal to produce steam
13. AC Generators, cont
Basic operation of the
generator
• As the loop rotates, the
magnetic flux through it
changes with time
• This induces an emf and a
current in the external
circuit
• The ends of the loop are
connected to slip rings
that rotate with the loop
• Connections to the
external circuit are made
by stationary brushed in
contact with the slip rings
14. AC Generators, final
The emf generated by the
rotating loop can be found
by
V =2 B ℓ v=2 B ℓ v sin θ
If the loop rotates with a
constant angular speed, ω,
and N turns
V = N B A ω cos ω t
V = Vmax when loop is
parallel to the field
V = 0 when when the loop
is perpendicular to the field
16. Angular velocity, w:
Amount of angle (radian) turned in a second
wt = total angle of turn during time t
2pf = w period T = 1/f = 2p/w
t = 0: F = 0
wt1
t = t1: F = A B sin(wt1)
Vind = - dF/dt
= - ABsin(wt1)/t1
Vind = - dF/dt
= - ABwcos(wt)
AC generation
18. 170 V
-170 V
When we say 120 V, it means rms value!!
P v2 or i2
vrms = 0.707va
19. Power Transmission
City of Gainesville has 120,000 population. On average
approximately 200 W/person of electric power is required.
Let’s assume that GRU transmit power with 120 V. How much
current Should be carried in power line?
Pt = 120,000 x 200 W = 24,000,000 W = 24 MW
P = IV, I = P/V = 24,000,000/120 = 200,000 A
However, if we deliver power with 500,000 V,
I = 24,000,000/500,000 = 48 A
Now Joule heating due to wire resistance (R) is reduced
By (48/200,000)2 = 5.8 x 10-8
