Lecture 33: Motional EMF, Faraday`s Law

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Electricity and Magnetism
Motional EMF
Faraday’s Law
Lana Sheridan
De Anza College
Nov 16, 2015
Last time
• relativity and electromagnetic fields
Overview
• motional emf
• induction
• Faraday’s law
• Lenz’s law
The strai
form magne
magnetic
forces
on an electron
If a conductor moves
through
a magnetic
field at an angle to the
is moving in
the conductor
balanced.
field, an emf is in
induced
across the are
conductor.
influence of
model, the
S
!
Bin
!
that is direc
S
Fe
influence of
S
accumulate
! " FB
charge sepa
S
E
the electron
S
v
model. The
"
"
on charges r
The electron
There are two ways
Due to
to see
thethis:
magnetic force on
tion for equ
electrons,charges
the ends
1 force on conduction
F =ofq the
v×B
Motional EMFIn steady state, the electric and
2
oppositely
in the rest conductor
frame of thebecome
conductor
there is also an electric field
charged, which establishes an
Motional EMF
In steady state, the electric and
magnetic forces on an electron
in the conductor are balanced.
!
!
S
S
Bin
Fe
!
"
S
FB
S
E
"
"
S
v
Due to the magnetic force on
electrons, the ends of the
The straight conductor o
form magnetic field directed
is moving in a direction per
Once the charge distribution reaches
influence of some external a
equilibrium, the net force on each
charge:model, the electrons in the
that is directed along the
Fnet = q(E + v × B) = 0
influence
of this force, the
accumulate there, leaving a
E = separation,
vB (v ⊥ B) an electr
charge
E
= vB
the electrons
are also desc
`
model.
The charges accumu
E = vB`
on charges remaining in th
The electrons are then desc
tion for equilibrium require
istance. Changing B by a factor of 12, however, causes the
Motional emf in rotating bar (Ex 31.4)
A conducting bar of length `, rotates with a constant angular
speed ω about a pivot at one end. A uniform magnetic field B is
directed perpendicular to the plane of rotation. Find the motional
ing Bar
emf induced between the ends of the bar.
S
Bin
S
v
r
31.10
(Example 31.4)
ucting bar rotating
d a pivot at one end in
rm magnetic field that
endicular to the plane
tion. A motional emf is
d between the ends of
.
dr
!
Pivot
hinking of each small segment as a source of emf, we see
at!
i
Motional emf in rotating bar (Ex 31.4)
Eachin infinitesimal
al emf Induced
a Rotating Bar
slice of the bar, dr, is moving at a different
rotates withspeed
a con- v (r ) = r ω.
pivot at one end. A
directed perpendicas shown in Figure
nduced between the
S
Bin
S
v
r
Figure 31.10 (Example 31.4)
A conducting bar rotating
around a pivot at one end in
a uniform magnetic field that
is perpendicular to the plane
of rotation. A motional emf is
induced between the ends of
the bar.
dr
!
Pivot
bar is different in
n Figure 31.8. Conbar, however. It is a
ving in a magnetic
d in it like the sliding bar. By thinking of each small segment as a source of emf, we see
nd the emfs add.
Using our previous result:
ptualization of the problem, we approach this example as we did in the discussion leadadded feature that the short segments of the bar are traveling in circular paths.
de of the emf induced in a
r having a velocity S
v from
ends of the bar by adding
ments:
ement is related to the
elationship v 5 rv (Eq.
ate:
dE = vB dr
d e 5 Bv dr
e 5 3 Bv dr
e 5 B 3 v dr 5 B v 3
E =
,
0
r dr 5
1
2
2 B v,
=
Z`
ωrB dr
0
1
Bω`2
2 continued
Motional emf and loops
Imagine a loop of wire that moves in a uniform magnetic field,
B, directed into the page.
a
b
S
v
Imagine the loop is composed of a pair of curved rods cut along
the lines shown.
Which way (left or right) the emf directed in the top half? In the
bottom? How do the magnitudes compare?
Motional emf and loops
Imagine a loop of wire that moves in a uniform magnetic field,
B, directed into the page.
a
b
S
v
In this case, part of the loop near a develops a negative charge and
the part near b a positive charge, but overall no steady current
flows around the loop.
Motional emf and loops
Now imagine a loop of wire that moves in a non-uniform
magnetic field falling towards a wire.
a
b
S
v
I
How do the magnitudes of the emfs in the top and bottom
compare?
Motional emf and loops
Now imagine a loop of wire that moves in a non-uniform
magnetic field falling towards a wire.
a
b
S
v
I
How do the magnitudes of the emfs in the top and bottom
compare?
They are not the same! A current can flow.
Motional emf and loops
Now imagine a loop of wire that moves in a non-uniform
magnetic field falling towards a wire. (Quiz 31.3)
a
b
S
v
I
What is the direction of the induced current in the loop of wire?
(A) clockwise
(B) counterclockwise
(C) zero
(D) impossible to determine
Motional emf and loops
Now imagine a loop of wire that moves in a non-uniform
magnetic field falling towards a wire. (Quiz 31.3)
a
b
S
v
I
What is the direction of the induced current in the loop of wire?
(A) clockwise
(B) counterclockwise
←
(C) zero
(D) impossible to determine
Motional emf and loops
What was different in the two cases (uniform vs. non-uniform
field)?
Motional emf and loops
What was different in the two cases (uniform vs. non-uniform
field)?
→ The field at different parts of the loop was different.
→ The magnetic flux through the loop was changing.
Definition of magnetic flux
Reminder:(30.18)
Magnetic
Flux
S
field B that makes an
s case is
(30.19)
a, then u 5 908 and the
the plane as in Figure
maximum value).
a weber Magnetic
(Wb); 1 Wb
5
flux
The magnetic flux
e plane is a
magnetic
r to the plane.
S
dA
S
B
u
S
dA
Figure 30.19 The magnetic
flux through an area element dA
S
S
is B ? d A 5 B dA cos u, where
Sa magnetic field through
of
d A is a vector perpendicular to
Z
the surface.
a surface A is
ΦB = B · dA
Units: Tm2
If the surface is a flat plane and B is uniform, that just reduces to:
ΦB = B · A
n a magnet is moved
When the magnet is
When the magnet is held
d a loop of wire
moved away from the
Changing
emf there is no
ected
to a sensitive flux andstationary,
loop, the ammeter sh
induced current in the
eter, the ammeter
that the induced curr
theof wire there is no
that a current
is opposite
that shown
When ais magnet is atloop,
rest even
near when
a loop
potential
inside
the loop.
ed in the
loop.
part a .
difference
across the magnet
ends ofisthe
wire.
I
I
N
S
N
b
S
N
c
S
When the magnet is
toward a loop of wire
stationary, there is no
connected
to
a
sensitive
Changing flux and emf
induced current in th
ammeter, the ammeter
loop,
eventhe
when the
shows
a current
is
When the north pole of
the that
magnet
is moved
towards the
loop,
magnet
is
inside
the
induced
in
the
loop.
magnetic flux increases.
I
N
.1 A simple experiment
at a current is induced
hen a magnet is moved
away from the loop.
a
A current flows clockwise in the loop.
S
N
b
S
the magnet is held
moved away from the
ary,
there
is
no
Changing flux andloop,
emfthe ammeter shows
d current in the
that the induced current
ven when
the the north pole of
is opposite
that shown
in away from the loop,
When
the magnet
is moved
a
t is inside
the
loop.
part
.
the magnetic flux decreases.
I
N
S
N
S
c
A current flows counterclockwise in the loop.
Faraday’s Law
Faraday’s Law
If a conducting loop experiences a changing magnetic flux through
the area of the loop, an emf EF is induced in the loop that is
directly proportional to the rate of change of the flux, ΦB with
time.
Faraday’s Law for a conducting loop:
EF = −
dΦB
dt
Faraday’s Law
Faraday’s Law
If a conducting loop experiences a changing magnetic flux through
the area of the loop, an emf EF is induced in the loop that is
directly proportional to the rate of change of the flux, ΦB with
time.
Faraday’s Law for a conducting loop:
EF = −
dΦB
dt
Note: If there is a changing flux, there will be an induced emf,
B
however, if dΦ
dt = 0 there could still be emf in a wire from other
effects. (Enet = EF + Eother )
Faraday’s Law
Faraday’s Law for a coil of N turns:
EF = −N
dΦB
dt
if ΦB is the flux through a single loop.
Changing Magnetic Flux
The magnetic flux might change for any of several reasons:
• the magnitude of B can change with time,
• the area A enclosed by the loop can change with time, or
• the angle θ between the field and the normal to the loop can
change with time.
Additional examples, video, and practice available at
Lenz’s Law
Lenz’s Law
30-4 Lenz’s Law
An induced current has a direction such that the
Faraday
proposed
histhe
law of induc
magnetic fieldSoon
dueafter
to the
current
opposes
devised a rule for determining the direction of an
change in the magnetic flux that induces the
current.
An induced current has a direction such that the ma
opposes the change in the magnetic flux that induces t
S
The magnet's motion
creates a magnetic
dipole that opposes
the motion.
N
N
µ
Furthermore, the direction of an induced emf is th
a feel for Lenz’s law, let us apply it in two different
where the north pole of a magnet is being moved to
1. Opposition
to Pole
Movement.
The approach
Basically, Lenz’s
law let’s us
interpret
the minus
Fig. 30-4 increases the magnetic flux through
sign in the equation we write to represent
current in the loop. From Fig. 29-21, we know t
Faraday’s Law. netic dipole with a south pole and a north po
:
moment !
is directed from south to north.
dΦB
i
increase
being
E: = − caused by the approaching mag
thus ! ) must dt
face toward the approaching no
S
30-4). Then the curled – straight right-hand rul
Fig. 30-4 Lenz’s law at work. As the
the current induced in the loop must be counte
magnet is 1moved toward the loop, a current
If we
Figure from Halliday, Resnick, Walker, 9th
ed. next pull the magnet away from th
result
Faraday’s and Lenz’s Laws
in the
What about this case? We found the current should flow
counterclockwise.
he sysonsisefore,
wire
in
to
S
v
I
Figure 31.12 (Quick Quiz 31.3)
result
Faraday’s and Lenz’s Laws
in the
What about this case? We found the current should flow
counterclockwise.
he sysonsisefore,
wire
in
to
S
v
I
Figure 31.12
The flux from the wire is into the
page and
increasing.
(Quick
Quiz
31.3)
The field from the current is out of the page.
There is an upward resistive force on the ring.
Summary
• motional emf
• Faraday’s law
• Lenz’s law
• applications
Next Test this Friday, Nov 20.
Homework
Serway & Jewett:
• NEW: Ch 31, Obj. Qs: 1, 3, 5, 7; Conc. Qs: 3, 5; Problems:
1, 5, 9, 13, 21, 27, 31
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