Transcript Lect16

Faraday's Law
dS
B
B
 
ΦB   B  dS
ε
dΦB
dt
General Review
• Electrostatics
» motion of “q” in external E-field
» E-field generated by Sqi
• Magnetostatics
» motion of “q” and “I” in external B-field
» B-field generated by “I”
• Electrodynamics
» time dependent B-field generates E-field
• AC circuits, inductors, transformers, etc.
» time dependent E-field generates B-field
• electromagnetic radiation - light!
Today...
• Induction Effects
• Faraday’s Law (Lenz’ Law)
• Energy Conservation with induced
currents?
• Faraday’s Law in terms of Electric
Fields
• Cool Applications
1
Lecture 16, Act 1
1A
Inside the shaded region, there is
a magnetic field into the board.
×
B
– If the loop is stationary, the Lorentz
force (on the electrons in the wire)
predicts:
(a) A Clockwise Current
1B
(b) A Counterclockwise Current
(c) No Current
Now the loop is pulled to the right at a velocity v.
– The Lorentz force will now give rise to:
(a) A Clockwise Current
(b) A Counterclockwise Current
(c) No Current, due to Symmetry
×
B
v
Other Examples of Induction
1831: Michael Faraday did the previous experiment, and a few others:
-v
×
B
Move the magnet, not the loop. Here
there is no Lorentz force v  B but
there was still an identical current.
I
This “coincidence” bothered Einstein, and eventually led him to the Special
Theory of Relativity (all that matters is relative motion).
Decrease the strength of B. Now
nothing is moving, but M.F. still saw a
current:
dB
I
dt
×
B
I
Decreasing B↓
Induction Effects
from Currents a
• Switch closed (or opened)
b
 current induced in coil b
• Steady state current in coil a
 no current induced in coil b
• Conclusion:
A current is induced in a loop when:
• there is a change in magnetic field through it
• this can happen many different ways
• How can we quantify this?
Faraday's Law
• Define the flux of the magnetic field through an open
dS
surface as:
 
ΦB   B  dS
B
B
• Faraday's Law:
The emf e induced in a circuit is determined by the time
rate of change of the magnetic flux through that circuit.
dΦB
ε
dt
So what is
this emf??
The minus sign indicates direction of induced current
(given by Lenz's Law).
Electro-Motive Force or emf
time
A magnetic field, increasing in time, passes through the blue loop
An electric field is generated “ringing” the increasing magnetic field
Circulating E-field will drive currents, just like a voltage difference
Loop integral of E-field is the “emf”:
dB
ε  ˜ E  dl  
dt
Note: The loop does not have to be a wire—
the emf exists even in vacuum! When we put a wire there,
the electrons respond to the emf, producing a current.


Disclaimer: In Lect. 5 we claimed  E  dl  0, so we could define a potential
independent of path. This holds only for charges at rest (electrostatics). Forces from
changing magnetic fields are nonconservative, and no potential can be defined!
Escher depiction of nonconservative emf
• Lenz's Law:
Lenz's Law
The induced current will appear in such a direction that it
opposes the change in flux that produced it.
B
B
S
N
v
N
S
v
• Conservation of energy considerations:
Claim: Direction of induced current must be so as to
oppose the change; otherwise conservation of energy
would be violated.
» Why???
• If current reinforced the change, then the change
would get bigger and that would in turn induce a
larger current which would increase the change,
2
etc..
Lecture 16, Acty 2
• A conducting rectangular loop moves with
constant velocity v in the +x direction
through a region of constant magnetic field
B in the -z direction as shown.
• What is the direction of the induced
2A
current in the loop?
(a) ccw
(b) cw
(c) no induced current
• A conducting rectangular loop moves with y
constant velocity v in the -y direction and a
constant current I flows in the +x direction as
shown.
• What is the direction of the induced
2B
current in the loop?
(a) ccw
(b) cw
XXXXXXXXXXXX
XXXXXXXXXXXX
X X X X X X X vX X X X X
XXXXXXXXXXXX
x
I
v
(c) no induced current
x
Lecture 16, Acty 2
• A conducting rectangular loop moves with
constant velocity v in the +x direction
through a region of constant magnetic field
B in the -z direction as shown.
1A
2A – What is the direction of the induced
current in the loop?
(a) ccw
(b) cw
XXXXXXXXXXXX
XXXXXXXXXXXX
X X X X X X X vX X X X X
XXXXXXXXXXXX
x
(c) no induced current
• There is a non-zero flux B passing through the loop since B
is perpendicular to the area of the loop.
• Since the velocity of the loop and the magnetic field are
CONSTANT, however, this flux DOES NOT CHANGE IN
TIME.
• Therefore, there is NO emf induced in the loop; NO current
will flow!!
Induced Current – quantitative
Suppose we pull with velocity v a x x x x x x
coil of resistance R through a
xxxxxx
region of constant magnetic field B.
– Direction of induced current?
xxxxxx
I
w
v
» Lenz’s Law  clockwise!
xxxxxx
x
– What is the magnitude?
» Magnetic Flux: ΦB  B  Area  Bwx
» Faraday’s Law:
ε wBv
dΦB
dx
ε
  Bw
 Bwv  I  
R
R
dt
dt
We must supply a constant force to move the loop
(until it is completely out of the B-field region). The
work we do is exactly equal to the energy dissipated
in the resistor, I2R (see Appendix).
Demo
E-M Cannon
3
v
• Connect solenoid to a source of
alternating voltage.
~
• The flux through the area ^ to axis of
solenoid therefore changes in time.
side view
• A conducting ring placed on top of the
solenoid will have a current induced in it
opposing this change.
F B

• There will then be a force on the ring
since it contains a current which is
circulating in the presence of a magnetic
field.
B
F
B
top view
Lecture 16, Act 3
•
3A
For this act, we will predict the results of variants of
the electromagnetic cannon demo which you just
observed.
– Suppose two aluminum rings are used in the demo;
Ring 2 is identical to Ring 1 except that it has a
small slit as shown. Let F1 be the force on Ring 1;
F2 be the force on Ring 2.
Ring 1
Ring 2
(a) F2 < F1
3B
(b) F2 = F1
(c) F2 > F1
– Suppose two identically shaped rings are used in the demo.
Ring 1 is made of copper (resistivity = 1.7X10-8 W-m); Ring 2 is
made of aluminum (resistivity = 2.8X10-8 W-m). Let F1 be the force
on Ring 1; F2 be the force on Ring 2.
(a) F2 < F1
(b) F2 = F1
(c) F2 > F1
Applications of Magnetic Induction
• AC Generator
– Water turns wheel
 rotates magnet
 changes flux
 induces emf
 drives current
• “Dynamic” Microphones
(E.g., some telephones)
– Sound
 oscillating pressure waves
 oscillating [diaphragm + coil]
 oscillating magnetic flux
 oscillating induced emf
 oscillating current in wire
Question: Do dynamic microphones need a battery?
More Applications of Magnetic Induction
• Tape / Hard Drive / ZIP Readout
– Tiny coil responds to change in flux as the magnetic domains
(encoding 0’s or 1’s) go by.
Question: How can your VCR display an image while paused?
• Credit Card Reader
– Must swipe card
 generates changing flux
– Faster swipe  bigger signal
More Applications of Magnetic Induction
• Magnetic Levitation (Maglev) Trains
– Induced surface (“eddy”) currents produce field in opposite
direction
 Repels magnet
 Levitates train
4
S
N
“eddy” current
rails
– Maglev trains today can travel up to 310 mph
 Twice the speed of Amtrak’s fastest conventional train!
– May eventually use superconducting loops to produce B-field
 No power dissipation in resistance of wires!
Lecture 16, Act 4
•
4A
The magnetic field in a region of space of
radius 2R is aligned with the z-direction and
changes in time as shown in the plot.
– What is the direction of the induced electric
field around a ring of radius R at time t=t1?
(a) E is ccw
4B
(b) E = 0
(c) E is cw
y
xxxx
xxxxxx
xxxxxxxx
R
xxxxxx
xxxx
Bz
– What is the relation between the
magnitudes of the induced electric fields ER
at radius R and E2R at radius 2R?
(a) E2R = ER
(b) E2R = 2ER
x
(c) E2R = 4ER
t1
t
Summary
• Faraday’s Law (Lenz’s Law)
– a changing magnetic flux through a loop induces a
current in that loop
dΦB
ε
dt
 
ΦB   B  dS
negative sign indicates that
the induced EMF opposes
the change in flux
• Faraday’s Law in terms of Electric Fields


d B
E

d
l



dt
Appendix: Energy Conservation?
• The induced current gives rise to
F'
a net magnetic force ( F
) on
xxxxxx
the loop which opposes the
motion.
xxxxxx
 wBv 
w2 B2 v
wB 
F  IwB  
R
 R 
F
I 
wBv
R
I
w
xxxxxx
xxxxxx
F'
x
• Agent must exert equal but
opposite force to move the loop with velocity v; therefore,
agent does work at rate P, where
2
2 2
P  Fv 
w B v
R
• Energy is dissipated in circuit at rate P'
2 2 2


wBv
w
B v
2




 R
P I R 
R
 R 
2

P = P' !
v