Two questions:
(1) How to find the force, F on the electric charge, Q excreted by the


field E and/or B?
 
F  qE  qv  B
(2) How fields E and/or B can be created?
Maxwell’s equations
Gauss’s law for electric field
Electric charges create electric field:
E   EA cos  Q /  0
For one not moving
(v<<c) charge:
Ek
Gauss’s law for magnetic field
Magnetic charges do not exist:
B   BA cos  0
Q
r2
Amperes law
Faraday’s law
Electric current creates magnetic field: A changing magnetic flux induces an EMF
 Bl cos  0 I
(As we will see later,
this law should be extended)
 B
 
t
A changing magnetic field
induces an electric field !
1
8. Electromagnetic induction
(Faraday’s law)
2) EMF (review)
1) Flux (review)

 B  BA cos
 B   BA cos
Units: (weber)
 B
 
t
 
 BAcos 
t
A changing magnetic flux
induces an EMF
 B - flux through a closed loop

   El cos
[ΦB] = 1 Wb = 1 T·m2
3) Faraday’s law
W
Fl cos


q
q
 El cos  
 BAcos 
t
A changing magnetic flux induces an
electric field !
This electric field is not a potential field.
The field lines form a closed loops.
- EMF in the closed loop
For N loops:  N  N 1
2
4) Lenz’s law
The direction of any magnetic induction effect is such as to oppose the cause
of the effect
For instance: a current produced by an induced emf moves in a direction so
that its magnetic field opposes the original change in flux

v
S
N
I

B

v
S
N

B

v
I
N
S

B

v
N
S
I

B
I
Example: If a North pole moves toward the loop in the plane of the page,
in what direction is the induced current?
Since the magnet is moving parallel to the loop,
there is no magnetic flux through the loop.
Thus the induced current is zero.
3
Example: In order to change the magnetic flux through the loop,
what would you have to do?
1)
2)
3)
4)
5)
drop the magnet
move the magnet upwards
move the magnet sideways
all of the above
only (1) and (2)
Moving the magnet in any direction would change the
magnetic field through the loop and thus the magnetic flux.
1)
2)
3)
4)
5)
tilt the loop
change the loop area
use thicker wires
all of the above
only (1) and (2)
Since,  B  BAcos changing the area or tilting the loop (which varies
the projected area) would change the magnetic flux through the loop.
4
Example 1: A wire loop is being pulled away from a current-carrying wire.
What is the direction of the induced current in the loop?
1) Clockwise
2) Counterclockwise
3) No induced current
On the right side of the wire the magnetic flux is into the page
and decreasing due to the fact that the loop is being pulled
away. By Lenz’s Law, the induced B field will oppose this
decrease. Thus, the new B field points into the page, which
requires an induced clockwise current to produce such a B field.
I
Example 2: What is the induced current if the wire loop moves down?
1) Clockwise
2) Counterclockwise
3) No induced current
The magnetic flux through the loop is not changing as it moves
parallel to the wire. Therefore, there is no induced current.
I
5
Example1: A wire loop is being pulled through a uniform magnetic field.
What is the direction of the induced current?
1) Clockwise; 2) Counterclockwise; 3) No induced current
x x x x x x x x x
Since the magnetic field is uniform, the magnetic flux through x x x x x x x x x
the loop is not changing. Thus no current is induced.
x x x x x x x x x
x x x x x x x x x
Example2: What is the direction of the induced current if the
x x x x x x x x x
B field suddenly increases while the loop is in the region?
1) Clockwise
2) Counterclockwise
3) No induced current
The increasing B field into the page must be countered by an induced flux out of the
page. This can be accomplished by induced current in the counterclockwise direction
in the wire loop.
Example 3: A wire loop is being pulled through
a uniform magnetic field that suddenly ends.
What is the direction of the induced current?
x x x x x
1) Clockwise 2) Counterclockwise 3) No induced current
x x x x x
The B field into the page is disappearing in the loop, so
it must be compensated by an induced flux also into the
page. This can be accomplished by an induced current
in the clockwise direction in the wire loop.
x x x x x
x x x x x
x x x x x
6
Example: If a coil is shrinking in a magnetic field pointing into the page,
in what direction is the induced current?
1) Clockwise 2) Counterclockwise 3) No induced current
The magnetic flux through the loop is decreasing, so the
induced B field must try to reinforce it and therefore points
in the same direction — into the page. According to the
right-hand rule, an induced clockwise current will generate
a magnetic field into the page.
Example: If a coil is rotated as shown, in a magnetic field
pointing to the left, in what direction is the induced current?
1) Clockwise 2) Counterclockwise 3) No induced current
As the coil is rotated into the B field, the magnetic flux through it increases.
According to Lenz’s Law, the induced B field has to oppose this increase,
thus the new B field points to the right. An induced counterclockwise
current produces just such a B field.
7
Example: Wire #1 (length L) forms a one-turn loop, & a bar magnet is dropped through.
Wire #2 (length 2L) forms a two-turn loop, and the same magnet is dropped through.
Compare the magnitude of the induced currents in these two cases.
1)
2)
3)
4)
I1
I1
I1
I1
>
<
=
=
I2
I2
I2  0
I2 = 0
 B
 B1
 
 N
t
t
S
N
S
N
Induced emf is twice as large in the wire with 2 loops.
The current is given by Ohm’s law: I = V/R. Since
wire #2 is twice as long as wire #1, it has twice the
resistance, so the current in both wires is the same.
Example: A bar magnet is held above the floor and dropped. In 1, there is nothing
between the magnet and the floor. In 2, the magnet falls through a copper loop.
How will the magnet in case 2 fall in comparison to case 1?
1) it will fall slower; 2) it will fall faster; 3) it will fall the same
When the magnet is falling from above the loop in 2,
the induced current will produce a North pole on top
of the loop, which repels the magnet.
When the magnet is below the loop, the induced
current will produce a North pole on the bottom of the
loop, which attracts the South pole of the magnet.
S
N
S
N
8
Example 1: A coil of 600 turns with area 100 cm2 is placed in a uniform magnetic field.
The angle between the direction of the field and the perpendicular to the loop is 60°.
The field changes at the rate of 0.010 T/s. What is the magnitude of induced emf in
the coil?
N  600
 B   BA cos  NBA cos
  60
B
B
N
A cos
t
t
A  100cm2
B
 0.010T / s
t
 ?
 B
B
 
N
A cos
t
t
  600 0.010T / s   10010 4 m 2 cos60  0.03V
 B   BA cos  NBA cos
Example 2:
B
A
 NB
cos
t
t
N  600
  60
A
 100cm2 / s
t
B  0.010T
 ?
 
 B
A
 NB
cos
t
t
  600 0.010T   10010 4 m 2 / s cos60  0.03V
9
Example 3: A 12.0-cm-diameter wire coil is initially oriented perpendicular to a
1.5 T magnetic field. The loop is rotated so that its plane is parallel to the field
direction in 0.20 s. What is the average induced emf in the loop?
N 1
f   0
 in   90
2 r  12.0cm
A  r 2
B  1.5T
t  0.20s
 ?
 B  BA cos


 B   Bf    Bin   BA cos0  cos90  BA
 
 B BA

t
t
2

1.5T     0.12m / 2

0.20s
 8.5 10 2 V
10
8a. Applications of Faraday’s law
1) Rotating loop:   t
  2f  2 T
I
 B  BAcos  BAcost
B
B
BAcost 
 

t
t
  BA sin t
Example:
N  200
B  0.03T
I
 max
 max  NBA
A  100cm2
  100s 1
 max  ?
 max  200 0.03T   100104 m2  100s 1   6V
11
Example: A generator rotates at 60 Hz in a magnetic field of 0.03 T.
It has 1000 turns and produces voltage that is 120 V at a pick.
What is the area of each turn of the coil?
f  60Hz
B  0.03T
N  1000
 max  NBA
  2f
 max
A
2fNB
 max  120V
A?
120V
A
 102 m 2
2  60Hz  1000 0.03T 
12
8b. EMF induced in a moving conductor
A  lx  lvt

v
l
B
A
 x  v t
 B
BA
 

t
t
  Blv
1) What is polarity of EMF?
2) What would be the direction of the induced
current, if rod slides on a conducting track?
The B field points out of the page. The flux is increasing since the area is increasing.
The induced B field opposes this change and therefore points into the page.
Thus, the induced current runs clockwise according to the right-hand rule.
FE
B

FB
FE  qE

v
FB  qvB
Another method:
qE  qvB    El  Blv
13
Example: A uniform magnetic field B is perpendicular to the area bounded by the
U-shaped conductor and a movable metal rod of length l. The rod is moving along
the conductor at a speed v. The total resistance of the loop is R. What is the
induced emf, the current in the loop, the magnetic force on the moving rod, and
power needed to move the rod?
Given :
B , l , v, R
Find :
, I, F, P
  Blv
 Blv
I
R

R
Blv
B 2l 2 v
F  IlB 
lB 
R
R
B 2l 2 v 2
Pext  Fv 
R
2
2 2 2
Blv
B
l v


2
Pdis  I R  
 R
R
 R 
14
8c. Transformers
B1
Vp  N p
t
 B1
Vs  N s
t
Vs
Ns

Vp N p
Pp  I pV p
Pp  Ps
Is N p

I p Ns
Ps  I sVs
15
Example: What is the voltage
across the lightbulb?
1A
120 V
240 V
120 V
The first transformer has a 2:1 ratio of turns, so the voltage doubles.
But the second transformer has a 1:2 ratio, so the voltage is halved again.
Therefore, the end result is the same as the original voltage.
Example: Given that the intermediate current is 1 A, what is the current
through the lightbulb?
Power in = Power out
240 V  1 A = 120 V  ???
The unknown current is 2 A.
Example: A 6 V battery is connected to one side of a transformer.
Compared to the voltage drop across coil 1, the voltage across coil 2 is:
Batteries provide DC current.
Only a changing magnetic flux induces an EMF.
1
2
Therefore, the voltage across coil 2 is zero.
16