Ch 34: Faraday’s Experiment
• Trying to induce a current using magnetic fields
• No induced current in “Y” loop with a DC circuit
• Saw a current when opening and closing the
switch (changing the magnetic field)
Electromagnetic Induction
Faraday’s Law - An induced emf is produced by
a changing magnetic field
– Can move magnet or loop
– Direction of motion controls direction of current
– No movement, no current
Predict the direction of the induced current
Motional EMF
• A current is caused by an electric field
• Current continues until FB = FE
FE = qE
qE = qvB
E = vB
FB = qvB
Which is the correct picture?
E = Blv (assumes B ┴ to v)
An airplane travels at 1000 km/hr in a region where
the earth’s magnetic field is 5 X 10-5T (vertical).
Calculate the potential difference between the
wing tips if they are 70 m apart.
1000 km/hr = 280 m/s
E = Blv
E = (5 X 10-5T )(70 m)(280 m/s) = 1.0 V
A metal bar of length l rotates in a magnetic
field B that is perpendicular to the plane of
rotation. It rotates at an angular speed of w.
a. Determine the potential difference between the
two ends of the bar (HINT: substitute for w, and
then integrate from 0 to l, the pivot to the end)
Motional EMF
• Moving a bar or wire produces charge separation
• If looped, produces a current
• Bar doesn’t want to move (Lenz’s law), must
exert a force
• Remember Fmag = IlB
Fpull = vl2B2
R
l = length
R = resistance
Example
Consider the following set-up. The bar is 10.0 cm
long.
a. Calculate the current needed for the bulb (P =
IV)
b. Calculate the resistance of the bulb
c. Calculate the speed needed to achieve this
current. (E = Blv)
d. Calculate the force required for the pull
EMF in a Moving Conductor
• Slide a conducting bar on the wire loop
• Increasing area
• What direction is the induced current? (right
hand rule)
Moving Conductor: Ex 2
Blood contains charged ions. A blood vessel is 2.0
mm in diameter, the magnetic field is 0.080 T,
and the blood meter registers a voltage of 0.10
mV. What is the flow velocity of the blood?
E = Blv
v = E /Bl
v = (1.0 X 10-4 V)
(0.080 T)(0.0020m)
v = 0.63 m/s
Magnetic Flux (flow)
FB = Magnetic Flux
FB = BAcosq
B = Magnetic Field (T)
A = area passes through (m2)
q = Angle ┴ to surface
If B ┴ to surface
– Cos 0o = 1
– Maximum flux
If B || to surface
– Cos 90o = 0
– No flux
Faraday’s Law of Induction
E = -NDFB
Dt
N = number of loops in a wire
DFB/Dt = change in magnetic flux over time
So why is it negative?
Lenz’s Law
An induced current’s magnetic field opposes the
original change in flux
• Always tries to keep magnetic field inside loop
constant.
• Use right-hand rule to predict direction of
current.
– Curve your fingers around the loop
– v is direction of the induced current
Lenz’s Law: Ex 1
Why is the direction of the current as indicated?
• Area is decreasing
• Flux is decreasing
• Induced current points into paper through ring
Lenz’s Law: Ex 2
What will happen to the current if you allow the
ring to relax to its original shape?
• Larger area
• Induced I will reverse direction
3 Ways to cause an emf
1. Change the magnetic field
2. Change area of loop
3. Rotate the loop (or magnet)
No flux
Maximum flux
Lenz’s Law: Ex 3a
Predict the direction of the induced current in the
following situations
• Counterclockwise current
• Magnet is going in (north in), need a current
pointing north out through the loop
• No current
• Magnetic flux is || to the loop
• Magnetic field decreasing
• Counterclockwise current to increase it
• Decreasing flux
• Clockwise current induced
B
• Initially no flux
• Flux increases to left
• Counterclockwise current
A long straight wire carries a current I as
shown.
a. Predict the direction of the magnetic field inside
the adjacent loop.
b. As the wire is pulled away from the loop,
predict the direction of the induced current.
Motional EMF
E = DFB
Dt
E = BDA
Dt
E = BlvDt
Dt
E = Blv (assumes B ┴ to v)
Lenz’s Law: Ex 4
A square coil of 100 loops is quickly pulled from
the magnetic field as shown in 0.10 s. Calculate
the change in flux.
FBfinal =0
FBinitial = BAcos0
FBinitial = (0.60 T)(0.050m)2(1)
FBinitial = 0.0015 Wb
DF = FBfinal – Fbinitial
DF = 0 – 0.0015 Wb = -0.0015 Wb
What Voltage and current are produced in the loop
(assume resistance = 100 W)
E = -NDFB
Dt
E = -(100)(-0.0015 Wb) = 1.5 V
0.10 s
V = IR
I = V/R = 1.5 V/100 W = 0.015 A (15 mA)
Faraday’s Law of Induction
E = -NDFB
Dt
E = NvlB
IR = NvlB
I = NvlB
R
(V = IR)
Faraday’s Law: Ex 1
A patient neglects to remove a 6.0 cm copper
bracelet (R = 0.010 W) before getting an MRI.
The magnetic field changes from 1.00 T to
0.40 T in 1.2 s. Assume the field passes
perpendicular to the bracelet.
a. Calculate the magnetic flux for both T’s (FB =
Bacosq)
b. Calculate the voltage through the bracelet based
on the change in flux.
c. Calculate the current through the bracelet
Induced Electric Fields
Coulomb vs. non-Coulomb
1. Coulomb Electric field – created by positive and
negative charges
2. Non-Coulomb – created by a changing magnetic
field
Both exert forces on charges (F = qE)
Another version of Faraday’s Law
Electric Field outside a solenoid
A 4.0 cm diameter solenoid is wound with 2000
turns/meter. The current oscillates at 60 Hz
and has an amplitude (maximum) of 2.0 A.
Here are some equations to help you:
B = m0nI
I = I0sinwt
a. Determine the electric field inside the solenoid.
b. Determine the maximum electric field inside the
solenoid.
Electric Generators (Dynamo)
• Generator is the inverse of
a motor
• AC Generator shown
• Rotation through magnetic
field induces I
• Current flows first one way,
then the other
• Segments ab and cd are moving conductor
• (Side segments have force in wrong direction)
E = Blv┴
v┴ = vsinq
E = 2NBlvsinq
• Can consider angular rotation
q = wt
v = wr = w(h/2)
E = 2NBlvsinq
E = 2NBlw(h/2) sin wt
E = wNBAsin wt
h = length of ad or bc
lh = Area
Remember
w = 2pf
f = frequency (Hertz)
w (radians/s)
• Over 99% of electricity in US produced by
generators
–
–
–
–
Coal/oil/gas plants
Wind power
Nuclear
Water
• 60 Hz in US and Canada
• 50 Hz in some others
Generator: Ex 1
A 60-Hz generator rotates in a 0.15 T magnetic
field. If the area of the coil is 0.020 m2, how
many loops must it contain for a peak output of
170 V?
E = NBAwsin wt
assume wt = 90
E = NBAw
 = E /BAw
w = 2pf = 2p(60Hz) = 377 s-1
N = 150 loops
Generator: Ex 2
A 60-Hz generator rotates in a 0.010 T magnetic
field. If the area of the coil is 2.0 m2, how many
loops must it contain for a peak output of 160 V?
(21 turns)
DC Generator
• Split ring commutator
• Many windings smooth out the current
Alternator
• Engine turns the rotor
• Magnetic field produced
• Current induce in stationary stator coils
Microphones
• Coil moves in and out of magnetic field with
sound
• emf induced in the coil
• Current is then sent to speakers, recorders, etc..
Tape Heads
Recording
• Changing current in coil
creates magnetic field
• Magnetizes the metal on the
tape
Playback
• Changing magnetic field from
tape induces current in coil
• Digital tape only has 1’s and
0’s
Seismograph
• Magnet moves and creates current in coil
• Current translated into signal for eart’s
movement
Transformers
• Increase or decrease AC
voltage
• TV – increase voltage for
picture tube
• Power packs – decrease
voltage
• Utility poles – decrease
voltage for house
YEAH!!! MY
FAVORITE
TOPIC!!!!
• Two coils linked by soft iron core
• Can be intertwined
• Flux from primary induces a current in the
secondary (99% efficient)
• Vary number of loops to control voltage
VS = NS
VP NP
• Step-up Transformer – Increases voltage
• Step-down Transformer – Decreases voltage
• POWER can’t increase (can’t get something for
nothing)
P = VI
PP = PS
V PI P = V SI s
IS = N P
IP N S
Transformers: Ex 1
A transformer for a radio reduces the voltage from
120 V to 9.0 V. The secondary has 30 turns and
the radio draws 400 mA. Calculate the turns in
the primary.
VS = NS
VP NP
NP = NSVP = (30)(120V) = 400 turns
VS
9V
Calculate the current in the primary
IS = N P
IP N S
IP = ISNS = (0.400A)(30) = 0.030 A (30 mA)
NP
(400)
Calculate the power transformed
P = IV
P = (0.030 A)(120 V) = 3.6 W
(can use either primary or secondary)
Transformers: Ex 2
An average of 120 kW of power is sent to a small
town 10 km from the power plant. The
transmission lines for a total resistance of 0.40 W.
Calculate the power lost to resistance if the
power is transmitted at 240 V vs. 24,000 V.
At 240 V
P = IV
I = P/V = 120,000 W/240 V = 500 A
Ploss = I2R
Ploss = (500 A)2 (0.40 W) = 100 kW
At 24,000 V
P = IV
I = P/V = 120,000 W/24,000 V = 5 A
Ploss = I2R
Ploss = (5 A)2 (0.40 W) = 10 W
• Transformers only work on ac
• dc only produces a secondary voltage when
switch is opened or closed
Counter EMF
• Counter (back) emf – as a motor turns, an emf is
induced that opposes the motion (Lenz’s law)
• Counter emf is less than the external voltage
when under a load
• The slower a motor rotates, the less counter emf
Counter EMF: Ex 1
The windings of a DC motor have a resistance of
5.0 W. When the motor reaches full speed, the
counter emf is 108 V. What is the current when
the motor is just starting up, and when it reaches
full speed (voltage = 120 V)
Just starting up (almost no counter emf)
V = IR
I = V/R = 120 V/5.0 W = 24 A
At full speed (V = 120 V – 108 V = 12 V)
I = V/R = 12 V/5.0 W = 2.4 A
Current is VERY high at start
– Lights may dim when refrigerator starts
– Lights dim if on when starting a car
Counter EMF: Ex 2
If a blender or drill jams (motor can’t turn), the
device may burn out. Why?
– No counter emf
– Current can be very high
– Wires may heat up
Self-Inductance: Solenoids
• Solenoid (inductor) – coil of wire (choke coil)
• L = inductance of the coil (Henry’s)
• As current increases in an inductor, an induced
emf is created
• Induced emf retards the increase of current (like
a back emf)
• Usually want to avoid inductance
– Resistors are wound in two directions to cancel the
inductance
• Acts as a resistor for alternating current
(impedance)
• Ex
– dc current can burn out a transformer
– ac has self-inductance (impedance) that limits the
current
Inductance of a solenoid
L = m0N2A
l
m0 = 4p X 10-7 T m/A
N = number of turns
A = cross-sectional area
l = length of solenoid
Inductance: Ex 1
Calculate the inductance of a solenoid with 100
turns, a length of 5.0 cm, and a cross sectional
area of 0.30 cm2.
L = m0N2A
l
L = (4p X 10-7 T m/A)(100)2(3 X 10-5m2)
(0.05 m)
L = 7.5 X 10-6 H or 7.5 mH
Inductance: Ex 2
The same solenoid is now filled with an iron cores
(m = 4000 m0). Calculate the inductance
L = (4000)(7.5 X 10-6H)
L = 0.030 H or 30 mH
Inductance: Ex 3
An inductor is made by tightly wrapping 0.300 mm
diameter wire around a 4.00 mm diameter
cylinder. Calculate the length of cylinder needed
to produce an inductance of 10 mH. (5.7 cm)
Inductors in Circuits
Used to store charge at high voltage (spark plug
within a car)
A 1.0 A current passes through a 10 mH inductor
coil. Calculate the potential difference if the
current drops to zero in 5.0 ms.
Energy in Inductors
(Energy software in a solenoid)
A 10 mH inductor is 5.7 cm long and 4.0 mm in
diameter. It carries a 100 mA current.
a. Calculate the energy stored in the inductor.
(5.0 X 10-8 J)
b. Calculate the solenoid volume. (7.16 X 10-7
m3)
c. Calculate the magnetic field density (U/volume
= 0.070 J/m3)
d. Calculate the magnetic field strength. (4.2 X
10-4 T)
LC Circuits
•
•
•
•
Oscillating circuit
Responds at natural frequency (resonance)
Used in cell phones to pick up a signal
Dials change in resonance frequency
w= 1
LC
LC: Ex 1
A 1.00 mH inductor is to be used for an AM radio.
Calculate the capacitance needed to pick up a
frequency of 902 kHz to listen to Rush
Limbaugh.
LR Circuits
•
•
•
•
Electromagnets
Radio tuners
L is inductance
R is resistance of inductor and any other
resistance
• Inductor smoothes out the voltage drop/increase
• Initially very low impedance
• Impedance rises with current
I = Ioe-t/t
t = time constant (time to read 63% of max)
t=L
R
Turning on current
– Current rises quickly, then levels off
Turning off current
– Opposite shape
– I = Imax e-t/t
LR Circuits: Ex 1
A solenoid has an inductance of 87.5 mH and a
resistance of 0.250 W. Find the time constant.
a. Calculate the time constant
b. Calculate the time needed to reach ½ of the
maximum current. (I = Ioe-t/t)
t = L/R
t = 87.5 X 10-3 H/0.250 W = 0.350 s
Rank in order, from smallest to largest, the time
constants of the following circuits.
a
b
c
The switch in the following circuit has been in
position a for a long time (VL = 0).
a. Calculate the current initially (100 mA)
b. Calculate the current at 5.0 ms after the switch is
thrown. (61 mA)
c. At what time has the current decayed to 1% of
its initial value? (46 ms)
DC vs AC
• DC = Direct current
– Electrons flow constantly
– Electrons only flow in one direction (negative to
positive)
– Batteries provide DC current
DC vs AC
• AC = Alternating current
–
–
–
–
Electrons switch directions
“Pulsed current”
Home electricity
More efficient for power transmission over large
distances
– USA uses 60 Hertz (60 cycles per second), many
other countries use 50 Hz
DC vs AC
DC
•Electrons flow constantly
•Electrons flow in only one
direction
•Batteries
AC
•Electrons flow in short burst
•Electrons switch directions
(60 times a second)
•House current