Transcript Chapter 20

Chapter 20
Induced Voltages and
Inductance
Faraday’s Experiment
• A primary coil is connected to a battery and a
secondary coil is connected to an ammeter
• The purpose of the secondary circuit is to detect
current that might be produced by a (changing)
magnetic field
• When there is a steady current in the primary circuit, the
ammeter reads zero
Faraday’s Experiment
• When the switch is opened, the ammeter reads a
current and then returns to zero
• When the switch is closed, the ammeter reads a current
in the opposite direction and then returns to zero
• An induced emf is produced in the secondary circuit by
the changing magnetic field
Magnetic Flux
• The emf is actually induced by a change in the quantity
called the magnetic flux rather than simply by a
change in the magnetic field
• Magnetic flux (defined similar to that of electrical flux)
is proportional to both the strength of the magnetic
field passing through the plane of a loop of wire and
the area of the loop
• For a loop of wire with an area A in a
uniform magnetic field, the flux is (θ
is the angle between B and the
normal to the plane):
ΦB = BA = B A cos θ
Magnetic Flux
• When the field is perpendicular to the plane of the
loop, θ = 0 and ΦB = ΦB, max = BA
• When the field is parallel to the plane of the loop, θ =
90° and ΦB = 0
• The flux can be negative, for example if θ = 180°
• SI unit of flux: Weber
• Wb = T. m²
Wilhelm Eduard Weber
1804 – 1891
Magnetic Flux
• The value of the magnetic flux is proportional to the
total number of magnetic field lines passing through
the loop
• When the area is perpendicular to the lines, the
maximum number of lines pass through the area and
the flux is a maximum
• When the area is parallel to
the lines, no lines pass
through the area and the
flux is 0
Electromagnetic Induction
• When a magnet moves toward a
loop of wire, the ammeter shows the
presence of a current
• When the magnet moves away from
the loop, the ammeter shows a
current in the opposite direction
• When the magnet is held stationary,
there is no current
• If the loop is moved instead of the
magnet, a current is also detected
Electromagnetic Induction
• A current is set up in the circuit as
long as there is relative motion
between the magnet and the loop
• The current is called an induced
current because is it produced by
an induced emf
Faraday’s Law and Electromagnetic
Induction
• The instantaneous emf induced in a circuit equals
the time rate of change of magnetic flux through the
circuit
• If a circuit contains N tightly wound loops and the
flux changes by ΔΦB during a time interval Δt, the
average emf induced is given by Faraday’s Law:
 B
  N
t
Faraday’s Law and Lenz’ Law
• Since ΦB = B A cos θ, the change in the flux, ΔΦB, can
be produced by a change in B, A or θ
• The negative sign in Faraday’s Law is included to
indicate the polarity of the induced emf, which is found
by Lenz’ Law:
• The current caused by the induced emf travels in the
direction that creates a magnetic field with flux
opposing the change in the original flux through the
circuit
 B
  N
t
Heinrich Friedrich
Emil Lenz
1804 – 1865
Faraday’s Law and Lenz’ Law
• Example
• The magnetic field, B, becomes smaller with time and
this reduces the flux
• The induced current will produce an induced field,
Bind, in the same direction as the original field
 B
  N
t
Chapter 20
Problem 18
A square, single-turn wire loop 1.00 cm on a side is placed inside a
solenoid that has a circular cross section of radius 3.00 cm, as shown
in the figure. The solenoid is 20.0 cm long and wound with 100 turns
of wire. (a) If the current in the solenoid is 3.00 A, find the flux through
the loop. (b) If the current in the solenoid is reduced to zero in 3.00 s,
find the magnitude of the average induced emf in the loop.
Motional emf
• A straight conductor of length ℓ
moves perpendicularly with constant
velocity through a uniform field
• The electrons in the conductor
experience a magnetic force
F=qvB
• The electrons tend to move to the
lower end of the conductor
• As the negative charges accumulate
at the base, a net positive charge
exists at the upper end of the
conductor
Motional emf
• 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
q E = q v B;
E = v B;
• There is a potential difference
between the upper and lower ends of
the conductor
Motional emf
• The potential difference between the
ends of the conductor (the upper end
is at a higher potential than the lower
end):
ΔV = E ℓ = B ℓ v
• 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
Motional emf in a Circuit
• As the bar (with zero resistance) is
pulled to the right with a constant
velocity under the influence of an
applied force, 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
• The changing magnetic flux through
the loop and the corresponding
induced emf in the bar result from the
change in area of the loop
Motional emf in a Circuit
• The induced, motional emf, acts like a
battery in the circuit
B v
  B v and I 
R
• 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 be in a
direction such that it opposes the
change in the external magnetic flux
(Lenz’ Law)
Motional emf in a Circuit
• The flux due to the external field is
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
• If the bar is moving toward the left, the
magnetic flux through the loop is
decreasing with time – the induced
current must be clockwise to produce
its own flux into the page
Chapter 20
Problem 59
A conducting rod of length ℓ moves on two horizontal frictionless rails. A
constant force of magnitude 1.00 N moves the bar at a uniform speed of 2.00
m/s through a magnetic field that is directed into the page. (a) What is the
current in an 8.00-Ω resistor R? (b) What is the rate of energy dissipation in
the resistor? (c) What is the mechanical power delivered by the constant
force?
Lenz’ Law – Moving Magnet Example
• As the bar magnet is moved to the right toward a
stationary loop of wire, the magnetic flux increases
with time
• The induced current produces a flux to the left, so the
current is in the direction shown
• When applying Lenz’ Law, there are two magnetic
fields to consider: changing external and induced
AC Generators
• Alternating Current (AC) generators convert
mechanical energy to electrical energy
• Consist of a wire loop rotated by some external means
(falling water, heat by burning coal to produce steam,
etc.)
• As the loop rotates, the magnetic flux through it
changes with time inducing an emf and a current in the
external circuit
AC Generators
• The ends of the loop are connected to slip rings that
rotate with the loop; connections to the external circuit
are made by stationary brushes in contact with the slip
rings
• The emf generated by the rotating loop can be found
by ε = 2 B ℓ v= 2 B ℓ v sin θ
• If the loop rotates with a constant angular speed, ω,
and N turns ε = N B A ω sin ω t
AC Generators
• The magnetic force on the charges in the wires AB and
CD is perpendicular to the length of the wires
• An emf is generated in wires BC and AD
• The emf produced in each of these wires is ε = B ℓ v=
= B ℓ v sin θ
DC Generators
• Components are essentially the same as that of an ac
generator
• The major difference is the contacts to the rotating
loop are made by a split ring, or commutator
• The output voltage always has the same polarity
• The current is a pulsing current
DC Generators
• To produce a steady current, many loops and
commutators around the axis of rotation are used
• The multiple outputs are superimposed and the output
is almost free of fluctuations
Motors
• Motors are devices that convert electrical energy into
mechanical energy (generators run in reverse)
• A motor can perform useful mechanical work when a
shaft connected to its rotating coil is attached to some
external device
• As the coil begins to rotate, the induced back emf
opposes the applied voltage and the current in the coil
is reduced
Self-inductance
• Self-inductance occurs when the changing flux
through a circuit arises from the circuit itself
• As the current increases, the magnetic flux through a
loop due to this current also increases inducing an emf
that opposes the change in magnetic flux
• As the magnitude of the current increases, the rate of
increase lessens and the induced emf decreases
• This opposing emf results in a gradual increase of the
current
Self-inductance
• The self-induced emf is proportional to the time rate of
change of the current
I
  L
t
• L: inductance of a coil (depends on geometric factors)
• The negative sign indicates that a changing current
induces an emf in opposition to that change
• The SI unit of self-inductance: Henry
• 1 H = 1 (V · s) / A
 B
 B I
  N
 N
t
I t
 B
LN
I
Joseph Henry
1797 – 1878
Chapter 20
Problem 42
An emf of 24.0 mV is induced in a 500-turn coil when the
current is changing at a rate of 10.0 A/s. What is the
magnetic flux through each turn of the coil at an instant
when the current is 4.00 A?
Inductor in a Circuit
• Inductance can be interpreted as a measure of
opposition to the rate of change in the current (while
resistance is a measure of opposition to the current)
• As a circuit is completed, the current begins to
increase, but the inductor produces an emf that
opposes the increasing current
• As a result, the current doesn’t change from 0 to its
maximum instantaneously
RL Circuit
• When the current reaches its maximum, the rate of
change and the back emf are zero
• The time constant, , for an RL circuit is the time
required for the current in the circuit to reach 63.2% of
its final value:
L
 
R
• The current can be found by:

I  1  e t /  
R
Chapter 20
Problem 48
Consider the circuit shown in the figure. Take ε =
6.00 V, L = 8.00 mH, and R = 4.00 Ω. (a) What is the
inductive time constant of the circuit? (b) Calculate
the current in the circuit 250 μs after the switch is
closed. (c) What is the value of the final steady-state
current? (d) How long does it take the current to
reach 80.0% of its maximum value?
Energy Stored in a Magnetic Field
• The emf induced by an inductor prevents a battery
from establishing an instantaneous current in a
circuit
• The battery has to do work to produce a current
• This work can be thought of as energy stored by the
inductor in its magnetic field
PEL = ½ L I2
Answers to Even Numbered Problems
Chapter 20:
Problem 10
34 mV
Answers to Even Numbered Problems
Chapter 20:
Problem 16
(a) left to right
(b) no induced current
(c) right to left
Answers to Even Numbered Problems
Chapter 20:
Problem 30
1.00 m/s
Answers to Even Numbered Problems
Chapter 20:
Problem 32
13 mV
Answers to Even Numbered Problems
Chapter 20:
Problem 44
(a) 1.00 kΩ
(b) 3.00 ms