6/1/2016
Ch 21: Induction
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
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
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What is the direction of the induced current?
Lenz’s Law: Ex 2
What will happen to the current if you allow the
ring to relax to its original shape?
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
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• No current
• Magnetic flux is || to the loop
• Magnetic field decreasing
• Counterclockwise current to increase it
• Decreasing flux
• Clockwise current induced
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• Initially no flux
• Flux increases to left
• Counterclockwise current
B
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.
If B ┴ to surface
– Cos 0o = 1
– Maximum flux
Magnetic Flux (flow)
FB = Magnetic Flux
FB = BAcosq
B = Magnetic Field (T)
A = area passes through (m2)
q = Angle ┴ to surface
Faraday’s Law of Induction
E = -N DFB = -NDBA
Dt
Dt
If B || to surface
– Cos 90o = 0
– No flux
N = number of loops in a wire
DFB/Dt = change in magnetic flux over time
B = Magnetic Field
A = Area
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A square coil of 100 loops is quickly pulled from
the magnetic field as shown in 0.10 s. Calculate
the change in flux.
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)
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
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 change in magnetic flux (DFB =
DBA)
b. Calculate the voltage through the bracelet based
on the change in flux.
c. Calculate the current through the bracelet
Motional EMF
• A current is caused by an electric field
• Current continues until FB = FE
FE = qE
FB = qvB
qE = qvB
E = vB
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Which is the correct picture?
• What direction is the induced current? (right
hand rule)
Motional EMF
DV = DFB
Dt
DV = BDA
Dt
DV = BlvDt
Dt
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
DV = Blv
DV = (5 X 10-5T )(70 m)(280 m/s) = 1.0 V
Force and 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
• (Shakelights)
Fpull = vl2B2
R
l = length
R = resistance
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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
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
Electric Generators (Dynamo)
• Generator is the reverse 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)
• Can consider angular rotation
E = NBAwsin wt
w = 2pf
(f is in Hertz)
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Generator: Ex 1
• 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
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 rad
(150 turns)
DC Generator
Alternator
• Split ring commutator
• Many windings smooth out the current
Counter EMF
• Engine turns the rotor
• Magnetic field produced
• Current induced in stationary stator coils
• 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
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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
Torque: Eddy Currents
Generators also have a counter torque
• Counter-torque
• Only produced when drawing current
• Larger the current, larger the counter torque
– Law of conservation of energy – more mechanical
input needed to produce larger currents
Eddy Currents
• Any conductor moving through magnetic field
will have eddy currents
• Electrons in atoms are moving as the metal
moves
• Magnetic field induced to oppose the change
•
•
•
•
•
Consider moving wheel below
Eddy currents Force opposes rotation
Braking a train car
Resistance for an exercise bike
Can produce a lot of heat
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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 = P S
VPIP = VSIs
IS = NP
IP
NS
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
Calculate the current in the primary
IS = NP
IP
NS
IP = ISNS = (0.400A)(30) = 0.030 A (30 mA)
NP
(400)
NP = NSVP = (30)(120V) = 400 turns
VS
9V
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Transformers: Ex 2
Calculate the power transformed
P = IV
P = (0.030 A)(120 V) = 3.6 W
(can use either primary or secondary)
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
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
• Transformers only work on ac
• dc only produces a secondary voltage when
switch is opened or closed
Ploss = I2R
Ploss = (5 A)2 (0.40 W) = 10 W
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
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Seismograph
• Magnet moves and creates current in coil
• Current translated into signal for earth’s
movement
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)
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 holds back the increase of current (a
back emf)
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
• Usually want to avoid inductance
– Resistors are wound in two directions to cancel the
inductance
• Inductors act 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: 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)
-6
L = 7.5 X 10 H or 7.5 mH
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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)
Energy in Inductors
(Energy density in 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)
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.
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LR Circuits
•
•
•
•
Electromagnets
Radio tuners
L is inductance
R is resistance of inductor and any other
resistance
• Inductor smoothes out the voltage drop/increase
Turning on current
I = Ioe-t/t
t = time constant (time to read 63% of max)
t=L
R
LR Circuits: Ex 1
– Current rises quickly, then levels off
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 = Io(1 - e-t/t)
t = L/R
t = 87.5 X 10-3 H/0.250 W = 0.350 s
Turning off current
– Opposite shape
– I = Imax e-t/t
Rank in order, from smallest to largest, the time
constants of the following circuits.
a
• Initially very low impedance
• Impedance rises with current
b
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)
c
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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
LR Circuits
•
•
•
•
DC
•Electrons flow constantly
•Electrons flow in only one
direction
•Batteries
Electromagnets
Radio tuners
L is inductance
R is resistance of inductor and any other
resistance
AC
•Electrons flow in short burst
•Electrons switch directions
(60 times a second)
•House current
• Initially very low impedance
• Impedance rises with current
I = V (1-e-t/t)
R
Turning on current
– Current rises quickly, then levels off
Turning off current
– Opposite shape
– I = Imax e-t/t
t = time constant (time to read 63% of max)
t=L
R
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LR Circuits: Ex 1
A solenoid has an inductance of 87.5 mH and a
resistance of 0.250 W. Find the time constant.
t = L/R
t = 87.5 X 10-3 H/0.250 W = 0.350 s
AC Current
• Current increases and decreases sinusoidally
I = Iocos2pft
Vrms = Vo/√2
Irms = Io/√2
2. Inductors and AC
•
•
•
•
Current lags the voltage by 90o
90o is ¼ of a cycle (360o)
No power is lost to heat
Energy is held in magnetic field, than returned to
source
How long will it take for the current to reach half
of its maximum value?
Imax = V/R
t = tln2
I = ½ V/R
t = (0.350 s)(0.693)
-t/t
1 V = V (1-e )
t = 0,243 s
2 R R
½ = 1-e-t/t
e-t/t = ½
et/t = 2
ln(et/t) = ln2
1. Resistors and AC
• Current changes with voltage
• I and V are in phase
• Energy is lost as heat
P = IV = Irms2R = Vrms2/R
V = IXL
XL = inductive reactance
XL = 2pfL
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Inductors and AC: Ex 1
A coil has a resistance of 1.00 W and an inductance
of 0.300 H. What current is in the coil at 120 V
dc?
In dc frequency = 0 so XL = 0
V = IR
I = V/R = 120 V/ 1.00 W = 120 A
3. Capacitors and AC
• DC - no current flows once plates are charged
• AC – Current flows constantly (plates charge and
discharge)
• Current leads voltage by 90o
• No power loss to heat
What current is in the coil in the voltage is 120 V
(rms) at 60.0 Hz?
XL = 2pfL
XL = 2p(60.0 Hz)(0.300 H) = 113 W
(ignore the 1 W of resistance, not in phase)
V = IXL
I= V/XL
I = 120 V/113 W = 1.06 A (much lower)
V = IXC
XC = 1
2pfC
(the larger the capacitance, the more charge it can
hold: less retarding of current flow)
Capacitors act as filters
– Prevent spikes in current flow
– Keep current steady
Capacitors and AC: Ex 1
What are the peak and rms currents of a circuit if C
= 1.0 mF and Vrms = 120 V. The frequency is
60.0 Hz.
XC =
1 =
2pfC
XC = 2700 W
1
2p(60 Hz)(1.0 X 10-6F)
Vrms = Vo/√2
Vo = Vrms√2
Vo = (120 V)(√2) = 170 V
Io = Vo/XC
Io = 170 V/2700 W
Io = 63 mA
Irms = Vrms/XC
Irms = 120 V/2700 W = 44 mA
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Resistor Current and
Voltage in phase
V = IR
Inductor Current lags the
voltage by 90o
XL = 2pfL
Capacitor Current leads the
voltage by 90o
XC = 1
2pfC
Phasor Diagrams
• Can analyze voltages on a graph
• Can resolve the vectors and create a triangle
f = Phase Angle (angle at which voltage is out of
phase from current)
• Resolving the vectors gives the instantaneous
voltage
LRC Circuits
•
•
•
•
Most circuits have L, R, and C’s in them
Voltages across R, L and C are not in phase
Vo
VRo + VLo + Vco
Currents are in phase
• As time goes on, phasor diagram rotates
• Resolved voltage changes
Impedance (Z)
• Impedance = total resistance to the flow of
current (from L, R, and C)
• Electrical devices are often impedance matched
(tuner connected to an amplifier)
Z = √R2 + (XL – XC)2
Z = √R2 + (2pfL – 1/ 2pfC)2
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Vrms = IrmsZ
Vo = IoZ
Power dissipated (lost) to impedance
tan f = XL – XC
R
Pave = I2rmsZcosf
Pave = IrmsVrmscosf
cosf = Power factor
cosf = 1 for pure resistor
cosf = 0 for pure inductor or capacitor
cos f = R
Z
Impedance: Ex 1
A circuit has R=25.0 W, L = 30.0 mH, and C = 12.0
mF. Calculate the impedance of the circuit if
they are connected to a 90.0 V ac(rms), 500 Hz
source. Also calculate the phase angle.
XL = 2pfL
XL = 2p(500 Hz)(0.030 H) = 94.2 W
XC = 1/2pfC
XC = 1/2p(500 Hz)(12 X 10-6F) = 26.5 W
Calculate the rms current
V = IZ
Irms = Vrms/Z = 90.0 V/ 72.2 W
Irms = 1.25 A
Z = √R2 + (XL – XC)2
Z = √(25.0W)2 + (94.2 W – 26.5 W)2
Z = 72.2 W
tan f = XL – XC
R
tan f = 94.2 W – 26.5 W
25.0 W
f = 69.7o
Calculate the power loss in the circuit
Pave = IrmsVrmscosf
Pave = (1.25 A)(90.0 V)cos(69.7o)
Pave = 39.0 W
Calculate the voltage drop across each element
(VR)rms = IrmsR = (1.25 A)(25.0 W) = 31.2 V
(VL)rms = IrmsXL = (1.25 A)(94.2 W) = 118 V
(VC)rms = IrmsXC = (1.25 A)(26.5 W) = 33.1 V
Voltages do not add to 90.0 V (out of phase)
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Resonance in AC: Oscillators
• LC circuits have a frequency for maximum
current
• Used for tuning radio stations
• Resonant frequency
fo = 1
2p √LC
• Tuners vary either L or C to tune in a station
• Can also be used to broadcast (radio, cellphone)
• When R is small, pure LC circuit
• When switch closed
– Capacitor discharges
– Current creates magnetic field (energy stored in
inductor’s magnetic field)
– Current then flows back to plates (stored in
capacitor’s electric field)
– Electromagnetic oscillation
Resonant Frequency: Ex 1
A radio tunes in a station at 980 kHz at a
capacitance of 3 mF. What is the inductance of
the circuit?
fo = 1
2p √LC
ANS: 8.8 X 10 -9 H or 8.8 pH
Energy Stored in a Magnetic Field
For a solenoid:
U = ½ B2 Al
mo
Note that Al is the volume of the cylinder (could be
the volume of other shapes)
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