22.2 Motional Emf
THE EMF INDUCED IN A MOVING CONDUCTOR
Conducting
bar not
touching the
plates

F
+

(is equivalent to)
Each charge within the conductor is
moving and experiences a magnetic
force
F  qvB

F
+
V  EL
 vBL
The magnetic force is in fact exactly
equivalent to an electric force exerted by a
parallel plate capacitor moving with the rod.
F  qE, E  vB, V  EL  vBL
The action, on a moving charge, of a magnetic field generated by a magnet at rest
is exactly equivalent to that of an electric field generated in the rest frame (the
frame of reference in which the object in question is at rest) of that charge.
(not in the scope of this class)
1
22.2 Motional Emf
From a different point of view, the bar behaves as if it has a built-in
battery (or DC voltage source). As illustrated on the right side of the
figure, the battery will also cause charges to accumulate at the ends.

F
+

(is equivalent to)
Motional emf when v, B,
and L are mutually
perpendicular
E  vBL
+
-
E

v
The direction of the EMF
(- to the + side of the
battery) is in the same
direction as the magnetic
force given by RHR-1
2
22.2 Motional Emf
The moving bar is in fact exactly equivalent to a battery, that one can
construct a circuit with it to power a light bulb. The diagrams below
show such a circuit made using conducting rails over which the bar
slides while making full electrical contact to form a closed circuit.

I
(is equivalent to)
E  vBL
3
22.2 Motional Emf
Example : Operating a Light Bulb with Motional Emf
In the figure, the conducting rod is
moving with a speed of 5.0m/s
perpendicular to a 0.80T magnetic field.
The rod has a length of 1.6m and a
negligible electrical resistance.
The rails also have a negligible
electrical resistance. The light bulb has
a resistance of 96 ohms. Find
(a) the emf produced by the rod and
(b) the current induced in the circuit.
(a) E  vBL  5.0 m s0.80 T 1.6 m  6.4 V
(b) I 
E 6.4 V

 0.067 A
R 96
4
22.2 Motional Emf
But where does the power and energy
come from to make the light bulb shine?
That hand has to work against the
magnetic force exerted by the field on the
induced current.
RHR-1 gives a magnetic force to the left.


Fhand  - FB ,
In order to keep the rod moving at
constant velocity, the force the hand
exerts on the rod must balance the
magnetic force on the current:


Fhand  FB  ILB
The opposite direction of the force from
the previous page, shown in this figure
below, would violate the principle of
conservation of energy.
A magnetic force produced in the
same direction as the original velocity
of the bar would then accelerate the
bar without any external work: bar
will accelerate and the force would
increase.
THIS DOES NOT HAPPEN
5
22.2 Motional Emf
Example: Conservation of Energy
A conducting rod is free to slide down between two vertical
copper tracks. There is no kinetic friction between the rod and
the tracks. Because the only force on the rod is its weight, it
falls with an acceleration equal to the acceleration of gravity.
Connect a resistor connected between the tops of the tracks.
(a) Does the rod now fall with the acceleration of gravity?
(b) How does the principle of conservation of energy apply?
Answers: With the resistor connected , we now have a closed
circuit. A current will flow: I = E/R = vBL/R.
(a) When released from rest, the rod will initially accelerate
down at g. But it will then encounter a magnetic force
(exerted by the magnetic field on the current) upward with
magnitude FB = ILB=vL2B2/R. So the acceleration will slow
until the rod reaches “terminal” (asymptotic) velocity/speed
when the magnetic force becomes equal to that of gravity:
vL2B2/R=mg  vf = mgR / (L2B2).
(b) Conservation of energy: gravitational potential energy is
being converted to heat dissipated by the resistor:
GPE/t=mg (h/t )=-mg vf =-m2g2R / (L2B2)
Power dissipation in the resistor:
P = If 2R = (vf BL/R)2R = [mg / (LB)]2R = m2g2R / (L2B2).
6
22.3 Magnetic Flux
The induced emf depends on the rate of change of: (a) magnitude of B field,
(b) the direction of B field, (c) size of circuit loop
Generalization: Induced emf is proportional to the rate of change of magnetic flux
 x - xo 
 xL - xo L 
 A - Ao 
BA - BAo
 BL  
 B  
 B 
E  vBL  
t - to
 t - to 
 t - to 
 t - to 
magnetic flux
 B  BA
Induced emf
E 
 -  o 

t - to
t
7
22.3 Magnetic Flux
MORE GENERAL
EXPRESSION FOR
MAGNETIC FLUX
Normal
vector
Normal
vector
The magnetic flux is proportional
to the number of field lines that pass
through a surface.
Directional boundary path. Circulation
sense related to normal vector by RHR-2
 B  BA cos 
8
22.4 Faraday’s Law of Electromagnetic Induction
FARADAY’S LAW OF ELECTROMAGNETIC INDUCTION
The average emf induced in a coil of N loops is
  - o 



E  -N 
 -N

t
 t - to 
Review:
SI Unit of Induced
Emf: volt (V)
Example: The Emf Induced by a Changing Magnetic Field
A coil of wire consists of 20 turns each of which has an area of 0.0015 m2. A
magnetic field is perpendicular to the surface. Initially, the magnitude of the
magnetic field is 0.050T and 0.10s later, it has increased to 0.060 T.
Find the average emf induced in the coil during this time.
BA cos  - Bo A cos 

E  -N
 -N
t
t
0.060 T - 0.050 T
 B - Bo 
2






 - NA cos  

20
0
.
0015
m
cos
0


t
0.10 s


 - 3.0  10 -3 V
9
22.5 Lenz’s Law
LENZ’S LAW
The induced emf resulting from a changing magnetic flux has a
polarity that leads to an induced current whose direction is such that
the induced magnetic field opposes the original flux change.
Change in flux  induced emf  induced current
 induced B field opposes change in flux
Reasoning Strategy
1. Determine whether the magnetic flux that penetrates the coil is
increasing or decreasing (relative to your chosen positive direction
for current).
2. Find what the direction of the induced magnetic field must be so
that it can oppose the change in flux by adding or subtracting from
the original field.
3. Use RHR-2 to determine the direction of the induced current from the
induced field in step 2.
10