Transcript Slide 1

Physics 1161: Lecture 14
Faraday’s Law
Changing Magnetic Fields create Electric Fields
• Sections 23-1 -- 23-4
Faraday’s Law
• Key to EVERYTHING in E+M
– Generating electricity
– Microphones, Speakers and MP3 Players
– Amplifiers
– Computer disks and card readers
– Ground Fault Interrupters
• Changing B creates new E
Magnetic Flux
B  BA cos
units : Tesla  m  Weber Wb
Faraday’s Law for Coil of N Turns
N   B
 
t
Lenz’s Law
2
Induced current is in a direction so that it opposes the change that produces it.
Faraday’s Law (EMF Magnitude)
Emf = Change in magnetic Flux/Time

 f  i
 

t
t f  ti
Since = B A cos(f, 3 things can
change 
Faraday’s Law (EMF Magnitude)
Emf = Change in magnetic Flux/Time

 f  i
 

t
t f  ti
Since = B A cos(f, 3 things can
change 
1. Area of loop
2. Magnetic field B
3. Angle f between A and B
Lenz’s Law (EMF Direction)
Emf opposes change in flux
• If flux increases:
New EMF makes new field opposite to the
original field (to oppose the increase)
• If flux decreases:
New EMF makes new field in the same direction
as the original field (to oppose the decrease)
Motional EMF circuit
Moving bar acts like battery E = vBL
• Magnitude of current
I = E /R = vBL/R
B
+
V
• Direction of Current
Clockwise (+ charges go down thru bar, up thru bulb)
• Direction of force (F=ILB sin()) on bar due to
magnetic field
To left, slows down
What changes if B
points into page?
Motional EMF, Preflight 14.1
B
L
v
F = q v B sin()
Which of the following statements is true?
• positive charge accumulates at the top of the bar, negative
at the bottom
• since there is not a complete circuit, no charge accumulates
at the bar's ends
• negative charge accumulates at the top of the bar, positive
at the bottom
Motional EMF, Preflight 14.1
B
Moving + charge feels force downwards:
+
F = q v B sin()
v
Velocity
B
Moving + charge still feels force downwards:
+
v
L
+
|ΔV|  |ΔU| = Fd
q
q
EMF = q v B sin() L/q
=vBL
Velocity
Preflight 14.2
• Which bar has the larger
motional emf?
a
E = v B L sin()
v
 is angle between v and B
Case a:
Case b:
v perpendicular to B
b
v
Preflight 14.4, 14.5
Suppose the magnetic field is
reversed so that it now points
OUT of the page instead of IN
as shown in the figure.
To keep the bar moving at the same speed, the force supplied by the
hand will have to:
Increase
Stay the Same
Decrease
To keep the bar moving to the right, the hand will have to supply a
force in the opposite direction:
True
False
Preflight 14.4
Suppose the magnetic field is
reversed so that it now points
OUT of the page instead of IN
as shown in the figure.
To keep the bar moving at the same speed, the force supplied
by the hand will have to:
• Increase
• Stay the Same
• Decrease
F = ILB sin()
B and v still perpendicular (=90),
so F=ILB just like before!
Preflight 14.5
Suppose the magnetic field is
reversed so that it now points
OUT of the page instead of IN
as shown in the figure.
To keep the bar moving to the right, the hand will have to
supply a force in the opposite direction.
• True
• False
Current flows in the opposite direction, so force
from the B field remains the same!
Magnetic Flux
• Count number of field lines through loop.
B
Uniform magnetic field, B, passes
through a plane surface of area A.
Magnetic flux  = B A
B
f
Magnetic flux   B A cos(f)
f is angle between normal and B
Note: The flux can be negative
(if field lines go thru loop in opposite direction)
Preflight 14.7
n B
n
a
b
A = B A cos(0) = BA
B = B A cos(90) = 0
Compare the flux through loops a and b.
1) a>b
2) a< b
Lenz’s Law (EMF Direction)
Emf opposes change in flux
• If flux increases:
New EMF makes new field opposite to the original
field (to oppose the increase)
• If flux decreases:
New EMF makes new field in the same direction as
the original field (to oppose the decrease)
Preflight 14.9
B
B
n
n
a
a
The magnetic field strength through the loop is cut in half
(decreasing the flux). If you wanted to create a second magnetic
field to oppose the change in flux, what would be its direction?
Left
Right
The original flux is to the right and decreasing. To
oppose the change and keep the total field strength the
same, add another field in the same direction.
Which loop has the greatest induced
EMF at the instant shown below?
W
L
1
3
v
2
v
v
33%
33%
33%
1. Loop 1
2. Loop 2
3. Loop 3
1
2
3
Which loop has the greatest induced
EMF at the instant shown below?
33%
W
L
1
33%
33%
3
v
2
v
v
1
1. Loop 1 E 1 = vBL
2. Loop 2 E 2 = 0
3. Loop 3 E 3 = vBW
2
1 moves right - gets 4 more field lines.
2 moves down - gets 0 more field lines.
3 moves down - only gets 2 more lines.
1 is gaining flux fastest!
3
Change Area II
W
W
V
vt
V
L
t=0
0=BLW
I
t
t=BL(W+vt)
 = B A cos(f)
EMF Magnitude:
  t   0 BL (W  vt )  BLW



t
t 0
t
 vBL
EMF Direction: B is out of page and  is increasing
so EMF creates B field (inside loop) going into page.
As current is increasing in the solenoid, what
direction will current be induced in the ring?
1. Same as solenoid
2. Opposite of solenoid
3. No current
33%
1
33%
2
33%
3
As current is increasing in the solenoid, what
direction will current be induced in the ring?
1. Same as solenoid
2. Opposite of solenoid
3. No current
33%
1
33%
2
33%
3
•
•
•
•
 Solenoid current (counter-clockwise)
 B-field (upwards) =>  Flux thru loop
EMF will create opposite B-field (downwards)
Induced loop current must be clockwise
Which way is the magnet moving if it is
inducing a current in the loop as shown?
1. up
2. down
S
N
N
S
50%
1
50%
2
Which way is the magnet moving if it is
inducing a current in the loop as shown?
1. up
2. down
S
N
N
S
50%
1
50%
2
If the resistance in the wire is decreased, what
will happen to the speed of the magnet’s
descent?
S
1. It will fall faster
2. It will fall slower
3. It will maintain
the same speed
N
N
S
33%
1
33%
2
33%
3
If the resistance in the wire is decreased, what
will happen to the speed of the magnet’s
descent?
S
1. It will fall faster
2. It will fall slower
3. It will maintain
the same speed
N
N
S
33%
33%
33%
larger induced current, stronger magnetic field
1
2
3
Change f
A flat coil of wire has A=0.2 m2 and R=10W. At time t=0, it is
oriented so the normal makes an angle f0=0 w.r.t. a
constant B field of 0.12 T. The loop is rotated to an angle
of f=30o in 0.5 seconds. Calculate the induced EMF.
i = B A cos(0)
f = B A cos(30)

f  i
BA(cos(30)  cos(0))
 


t
tf  ti
0.5
 = 6.43x10-3 Volts
What direction is the current induced?
 upwards and decreasing. New field will be in same
direction (opposes change). Current must be counter
clockwise.
B
f
Magnetic Flux Examples
A conducting loop is inside a solenoid (B=monI).
What happens to the flux through the loop when you…
Increase area of solenoid?
Increase area of loop?
Increase current in solenoid?
Rotate loop slightly?
  B A cos(f)
Magnetic Flux Examples
A conducting loop is inside a solenoid (B=monI).
What happens to the flux through the loop when you…
Increase area of solenoid? No
change
Increase area of loop? Increases
Increase current in solenoid?
Increases
Rotate loop slightly? Decreases
  B A cos(f)
Magnetic Flux II
A solenoid (B=monI) is inside a conducting loop. What
happens to the flux through the loop when you…
Increase area of solenoid Increases
Increase area of loop No change
Increase current in solenoid
Increases
  B A cos(f)