FARADAY’S LAW OF INDUCTION – (Chapter 23)
Look at a conductor moving ACROSS magnetic field lines


Free electrons move perpendicular to B (into page)

 
o Electrons (neg.) feel force F  ev  B

Force on electrons is down
F
v
Bin
o Result is potential difference induced between ends of bar – bottom
more negative
CONCLUSION: potential difference can be induced across a conductor moving
through a magnetic field
Will come back to this example later when we have more theory
SOME DEMONSTRATIONS:

A magnet moving into/out of a loop of
wire induces a current
o See fig. 23.2 in text
Ammeter
N
switch

Starting/stopping current through loop in
one coil (primary) causes pulse of
current in coil (secondary) linked by iron
coil - this is a transformer
o See fig. 23.3 in text
S
iron
Ammeter
battery
primary
secondary
FARADAY’S CONCLUSION from observations like these:

Time-varying magnetic fields can induce currents
To quantify, need to relate induced potential difference to changing magnetic flux
MAGNETIC FLUX: proportional to number of magnetic field lines through area


area element dA : vector perpendicular to area element


 
contribution to magnetic flux from dA is B  dA

 
Total flux through area is  B   B  dA
dA
B

area

Unit for flux is Tm2  Wb (1 Weber)
emf – symbol is 
 Work to push unit charge through a wire (or space)

Units of emf are volts

originally “electromotive force” BUT not a force: long name no longer used

If resistance in loop is R, then current in loop is I   / R
Battery is source of emf
B
FARADAY’S LAW OF INDUCTION

Look at circuit bounding a surface of area A

If magnetic flux  B through surface changes with time:
o induces emf  around circuit

Faraday’s Law of induction relates emf to rate of change of flux:
 
d B
dt

will think about meaning of minus sign later

if circuit is coil of N turns:
  N
d B
dt
Sources of time-varying magnetic flux  B t 


look at loop of area A in field B with angle 


between B and normal A
 
o magnetic flux is:  B  B  A  B Acos 
SO Faraday’s law is   
d B Acos 
dt
Can induce emf by:

varying magnitude B

varying magnitude A

varying angle  (i.e. rotating coil as in a generator)

combination of above
B

A
EXAMPLE (see Exs. 23.1, 23.2) :
A loop of wire is located in a uniform magnetic field that is changing with time.
What is the induced emf?

 B is uniform over the surface of area A bounded by the loop of wire so:
 
 B  B  A  B Acos 
o A and θ are constant.
o B depends on time
dB
 So:    Acos  dt
B

A
emf INDUCED BY MOTION OF CONDUCTOR THROUGH MAG. FIELD (23.2)
1st: Look at straight conductor:
 
 
 Conductor  to B ; v  to B ; v  to length of conductor
 Magnetic force on free electrons in conductor is:

 
F  ev  B  e v B (down)
o lower end → more negative; upper end → more positive
FB
v
Bin
 electrons move UNTIL resulting electric and magnetic forces balance
o Balances when FE  FB so that e E  ev B
o Magnitude of potential difference between conductor ends:
FE
v
V  E l  Bl v

 As long as v is constant
 Top of conductor is positive (here)
FB
Bin
2nd: Complete circuit: Bar sliding on conducting rails connected by resistance R

 Pull bar to right with force FAPP
 Can get direction of I from picture in
previous section
l
R
FB
Bin
I
 What is magnitude of I ?
Use Faraday’s Law:
x
 Magnetic flux through area bounded by circuit is  B  Bl x
d
dx
B
 Induced emf is    dt   Bl d t   Bl v

Bl v
I


 Magnitude of current is
R
R
v
Sliding Bar: Power dissipated (as heat) in resistance is P  I 2 R
 Where does this energy come from?
Look at force needed to pull bar to right at
constant speed?
Current flows up in bar – causes magnetic


force on bar : FB  I l  B  I l B (left)
For constant speed, net force is zero.

 So applied force is: FAPP  I l B (right)
l
FB
R
FAPP
Bin
I
x
Power delivered by applied force is P  FAPP v  I l B v
IR
Bl v
v


I
 But from
Bl
R can write
IR
2
P

I
l
B

I
R
So Power delivered by applied force is
Bl
 Same as power dissipated in resistor – CONSERVATION OF ENERGY
EXAMPLE: emf Induced in a Rotating Bar
A bar of length l is rotating in a plane perpendicular to

magnetic field B with angular speed  . What is the emf
induced across the length of the bar?
ω
v
r
l

Alternating Current Generator: A loop rotating about an axis  to B

 Flux through loop depends on angle  between B
and loop normal vector
N
S
 
o  B  B  A  B Acos 
 If loop is rotating with angular speed  , then  t    t
o Then  B t   B Acos  t
N
 If there are N turns in the loop then induced emf is
d
B



  NAB
N
o
dt
d cos  t
 NAB sin  t
dt
 Result is AC voltage:   NAB sin  t
o Amplitude is  max  NAB
εmax
S
ε
t
Alternating Current Generator (continued): Induced emf is   NAB sin  t
 Induced emf   0 when  t    t  0

o i.e. when loop normal is parallel to B
A
N
S
B

o at this orientation, edges of loop are not cutting any B field lines
 Induced emf    max  NAB when  t    t   / 2

o i.e. when loop normal is perpendicular to B
A
N

o at this orientation, edges of loop are cutting B field lines
 For real AC generator, need “slip rings” to connect to rotating coil
o AC is OK for heat, lights, etc.
 To get DC voltage, need a “rectifier”
B
S
LENZ’S LAW: gives direction of induced current when flux through loop changes
 Rule for “polarity” of induced emf in loop:
o Resulting current will be in direction that tends to produce a magnetic
field that opposes the change in flux that induced the emf
 Example: Bar moving to right increases
magnetic flux through loop
o Lenz’s law → resulting current should cause
field tending to cancel increased flux in loop
v
R
Bin

o SO: B from induced current should be out.
 Means that induced current should be ccw

o Will generate B out of page/screen INSIDE
the loop (by RH-rule)
I
I
R
I
induced B
v
Bin
EXAMPLE:
 Move bar magnet right toward loop
o Increases flux through loop
v
S
N
 By Lenz’s Law, current induced in loop should
oppose increase in flux through loop
o Means that induced current should
generate field pointing to left
 Result: Current in loop generates magnetic dipole
pointing to left
o Acts like magnet with north pole to left
DEMONSTRATIONS:
 Magnetic levitation (jumping rings)
 Eddy current pendulum
Iinduced
Binduced