# induction by 14bz5m6

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```									Ch – 33 Electromagnetic Induction
Reading Quiz – Ch. 33
1. Currents circulate in a piece of metal that is
pulled through a magnetic field. What is the
correct name for these currents?
a. Eddy currents        c.Flux currents.
b. Induction currents d.Faraday currents.

2.    The magnetic flux is a measure of the
magnetic field:
a.   while electromagnetic induction occurs
b.   parallel to a closed loop
c.   passing through a closed loop
d.   when it is changing
Learning Objectives – Ch 33
• Understand how a changing magnetic field
will induce a potential difference – this is the
phenomenon of electromagnetic induction.
• To observe the experimental evidence for
electromagnetic induction.
• To understand and use Lenz’s law for
induced currents.
• To learn of Faraday’s law as a new law of
nature.
• To understand basic applications of
electromagnetic induction to technology.
• To gain a qualitative understanding of
electromagnetic waves.
• To analyze circuits with inductors.
Induced Current
• A current creates a
magnetic field
• Is the opposite true?
• Can a magnetic field
create, or induce a
current?
• Faraday was hoping
that the magnetic field
generated by the
current on the left
circuit would induce a
current in the wire on
the right.
• But no luck, as the
meter shows.
• So he shut down for
lunch.
• But when he opened the
switch to cut off the
current in the left circuit,
the meter suddenly
moved, showing a
momentary current in the
wire on the right.
• It quickly went back to
zero.
• Baffled, he closed
the switch again.
• And noticed that the
meter jumped again
momentarily, and this
time the meter needle
went the other way.
• I bet he actually did it
a bunch of times
before he noticed that
it went the other way..
• Faraday found that he could induce a
current in a closed wire, but only if the
magnetic field through the coil is changing.
• This is an informal statement of Faraday’s
Law.
Motional emf
An induced current in a
circuit can be          Changing            Stationary circuit
created 2 ways:         magnetic field
(since current is
with meter
changing)
1. Change the strength
of the magnetic field
through a stationary
circuit. That’s what
opening and closing
the switch.
Motional emf
Circuit
changes
Constant B
size due to
The other way to induce      field into
the page
a current:
2. Change the size or
orientation of the
circuit in a stationary
magnetic field.
We shall look at this
second method first.
This is motional emf.
Motional emf
• The external B field
causes a magnetic
force on positive and
negative charges
moving to the right.
• The electron holes
move up and the
electrons move down.
Motional emf
• Since the rest of the
loop isn’t moving,
there is no magnetic
force on it, so the
electron holes will
flow along the wire to
get back to the more
negative side.
• Wow, electric current
with no power source
except a moving wire!
Why aren’t we all fat
and happy?
Motional emf
• We’ll find out soon.
• Another result of the
charge separation is
an E field.
• This E field also
causes a force, which
is in the opposite
direction as the force
due to the B field.
Motional emf
• Electron holes continue to
move up, but only until
FE down equals FB up, at
which charge separation
ceases.
• As long as the wire keeps
moving, there will be a
charge separation.
• The magnetic force is
doing work to maintain
that charge separation.
Workbook exercises
• 1 b,d,e
Workbook exercises - answers
• b: ccw
• d: ccw
• e: 0
Motional emf
Recall emf (motional or otherwise) is the
work done per unit charge, i.e. a potential
difference.

∆V = - Eds

When FE equals FB (assume α = 900):
qE = qvB or E = vB
Motional emf
For a wire of length L,
moving in a direction
perpendicular to an
external B field:
l
∆V = -
  vBds
0
emf = v LB
Motional emf
• The emf due to the
charge separation exists
whether or not the loop is
closed (battery analogy).
• If there is a closed loop:
I = vLB/R
• The induced current is
due to magnetic forces on
moving charges.
Why aren’t we fat and happy?
• Once there is a
current flowing
through the wire,
charges move in 2
directions.
– They all move right
with the moving wire.
– Current moves up.
• Yet another magnetic
force generated, this
time to the left.
Why aren’t we fat and happy?
• This magnetic force
opposes the original
velocity of the moving
wire.
• The moving wire will slow
down and stop.
• Need a constant Fpull to
the right to make the
contraption work.
• That’s why we’re not fat
and happy.
Why aren’t we fat and happy?
• To keep the wire at
constant speed, and
continue the emf and
current Fpull = Fmag
F = ILB (Ch. 32)
Fpull = Fmag = ILB
I = vLB/R
F = vL2B2/R
Why aren’t we fat and happy?
Fpull = Fmag
Wpull = Wmag
The rate at which work
is done on the circuit
equals the power
dissipated by the
circuit:
P = I2R = v2L2B2/R
Numerical Example
• #3 = end of chapter
Magnetic Flux

m  A  B

Φm = AB cos θ

Units: 1 weber = 1Wb = 1Tm2
Magnetic Flux in a non-uniform
field
• Divide the loop into
many small pieces

d m  B  dA
• The flux is the sum of
all these:

 m   B  dA
Lenz’s Law
• There is an induced
current in a closed
conducting loop only
if the magnetic flux is
changing (either B, A
or θ). The direction of
the induced current is
such that the induced
magnetic field
opposes the change.
Using Lenz Law
1. Determine the direction of the external magnetic
field.
2. Determine how the flux is changing. Is it increasing,
decreasing, or staying the same?
3. Determine the direction of an induced magnetic field
that will oppose the change in the flux.
– Increasing: induced magnetic field points opposite the external
magnetic field.
– Decreasing: induced magnetic field points in the same direction
as the external magnetic field.
– Constant: no induced magnetic field.
4. Determine the direction of the induced current. Use
the right-hand rule.
Recall that a current in
a circuit can be
created 2 ways:
– Change the size or
orientation of the
circuit in a stationary
magnetic field.
– Change the strength
of the magnetic field
through a stationary
circuit.
Both of these create a
changing magnetic
flux.
The current exists
because the
changing magnetic
flux has induced an
emf. In a closed
circuit with a
resistance, R:
I = ε/R
The current is a
consequence of the
emf.
The emf is a
consequence of Φm
An emf is induced in a conducting coil of N turns if
the magnetic flux through the coil changes. The
magnitude of the emf is equal to the rate of
change of the magnetic flux:

d m
N
dt

The direction is given by Lenz Law.
Recall the expression
for emf of a wire of
length L, moving in an
external B field:
l
∆V = -
  vBds
0
emf = v LB
give us the same?
ε = dΦ/dt = d(xLB)/dt
L, B are constant so
ε = (LB)dx/dt
ε = v LB
For the case shown,
the induced B field
will be in which
direction?
Workbook exercise #14 (p.33-8)
Draw a graph of the current, given the graph
of the magnetic field
I0-1s: negative constant
I1-2s: no current
I2-3s: positive constant, smaller magnitude
than I0-1s
Numerical Problem
The resistance of the
loop is 0.10 Ω.
A. Is the magnetic field
strength increasing
or decreasing?
B. What is the rate of
change of the the
magnetic field?
Numerical Problem
A. Induced current is
shown moving ccw.
RH rule indicates a
magnetic field out of
the page, opposing
external field.
Therefore, external
magnetic field must
have been
increasing.
B. Rate of change is
2.34 T/s
Inductors
• An inductor is a device that produces a uniform
magnetic field when a current passes through it.
A solenoid is an inductor.
• The magnetic flux of an inductor is proportional
to the current.
• For each coil (turn) of the solenoid:
Φper coil = A•B
Φsol = N(A•B) = NAB = NA(u0NI/ℓ) = (Au0N2/ℓ)Isol
• This is actually a self-inductance
Inductors
• The proportionality constant is defined as L, the
inductance:
Lsol = Φsol /I = Au0N2/ℓ
• Note that the inductance, L depends only on
the geometry of the inductor, not on the current.
• The unit of inductance is the henry
1 H = 1 Wb/Ampere
The circuit symbol for an inductor:
Potential difference across an
inductor
• For the ideal inductor,
R = 0, therefore
potential difference
across the inductor
also equals zero, as
long as the current is
constant.
• What happens if we
increase the current?
Potential difference across an inductor
• Increasing the current
increases the flux.
• An induced magnetic field
will oppose the increase
by pointing to the right.        Induced          Induced
current
• The induced current is                            field

opposite the solenoid
current.
• The induced current
carries positive charge to
the left and establishes a
potential difference
across the inductor.                 Potential
difference
Potential difference across an inductor
The potential difference
across the inductor
can be found using
Induced
d m
Induced
current

N
field

dt
Where Φm = Φper coil
Φsol = N Φper coil
We defined Φ = LI
dΦsol/dt = L |dI/dt|
Potential
difference
Potential difference across an inductor
• If the inductor current is
decreased, the induced
magnetic field, the
induced current and the
potential difference all
change direction.
• Note that whether you
increase or decrease the
current, the inductor
always “resists” the
change with an induced
current.
The sign of potential difference across an
inductor
∆VL = -L dI/dt
• ∆VL decreases in the
direction of current flow if
current is increasing.
• ∆VL increases in the
direction of current flow if
current is decreasing.
• ∆VL is measured in the
direction of current in the
circuit
Conceptual Question - Inductors
Which of the following statements could be true?
a. I is from a to b and constant.
b. I is from a to b and increasing.
c. I is from a to b and decreasing.
d. I is from b to a and steady.
e. I is from b to a and increasing.
f. I is from b to a and decreasing.
Faraday’s Law Problem (#34, p.1078)

The top figure shows a 5-
turn 1.0-cm diameter coil
with R=0.10 Ω inside a                    COIL

2.0-cm solenoid. The
solenoid is 8.0 cm long,
Solenoid cross section at t = 0.02s
has 120 turns and carries
the current shown in the
graph. A positive current
is cw when seen from the
left. Determine the
direction and magnitude
of current in the coil.
Setting up the problem
The mathematical
Law:                                              COIL

d m
N                                Solenoid cross section at t = 0.02s

dt

I = ε/R
Icoil = 1/R (N dΦ/dt)
For each of the following
variables, decide whether the
quantity refers to the solenoid
or the coil: R, N, Φ
Setting up the problem
Icoil = 1/R (N dΦ/dt)
For each of the following
COIL
variables, decide whether
the quantity refers to the
solenoid or the coil:            Solenoid cross section at t = 0.02s

R, N, Φ
Answers: R – coil, N – coil
Φ – coil
Icoil = 1/Rcoil (Ncoil dΦcoil/dt)
Setting up the problem

Icoil = 1/Rcoil (Ncoil dΦcoil/dt)
For the above expression:
COIL

Write an expression for
Φcoil in terms of the
magnetic field and the           Solenoid cross section at t = 0.02s

area. Once again, the
pertinent question is
solenoid or coil:
Magnetic field (B)
Area (A)
And what about cos θ?
Setting up the problem

Area (A) – the current is
induced in the coil, due to                 COIL

the flux through the coil;
therefore it’s the area of
Solenoid cross section at t = 0.02s
the coil.
Magnetic field (B) – It’s the
external magnetic field of
the solenoid that induces
a current in the coil
Φcoil = Acoil Bsolenoid
And what is the magnetic
field of a solenoid?
Setting up the problem

Φcoil = Acoil Bsolenoid

COIL

Write an expression for the
magnetic field of a
solenoid. For any              Solenoid cross section at t = 0.02s

variable that has a value
for both coil and solenoid
(e.g I,R,N,r) specify to
which you are referring.
Setting up the problem

Φcoil = Acoil Bsolenoid

COIL

Bsol = (u0NsolIsol)/ℓ
(we weren’t given a length
Solenoid cross section at t = 0.02s
for the coil).
Now we need an
expression for dΦcoil /dt,
so write one.
Setting up the problem

Φcoil = Acoil Bsolenoid

COIL

Bsol = (u0NsolIsol)/ℓ

Solenoid cross section at t = 0.02s

dΦcoil = (Acoil u0Nsol/ℓ)dI/dt
dt
We may actually be ready
to solve this puppy.
Icoil = 1/Rcoil (Ncoil dΦcoil/dt)

COIL

Icoil = 1/Rcoil (Ncoil dΦcoil/dt)
Solenoid cross section at t = 0.02s

Icoil = 3.7 x 10-4A
or .37 mA

Direction: clockwise when
seen from the left
(induces a field to the
right).

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