Transcript Motional EMF

Motional EMF
AP Physics C
Montwood High School
R. Casao
• Motion EMF is the EMF produced in a conductor
moving through a magnetic field.
• Consider a straight conductor of length l moving
with constant velocity through a uniform magnetic
field directed into the paper.
• The conductor is moving
perpendicular to the field.
• Electrons in the conductor
experience a force along
the conductor given by
F = q·(v x B).
• Under the influence of this
force, the electrons move
to the lower end.
• The lower end of the rod becomes negatively
charged; the upper end of the rod becomes
positively charged.
• An electric field is produced in the conductor as a
result of the charge separation.
• The charge at the ends builds up until the
magnetic force q·v·B is
balanced by the electric
force q·E.
• At this point, the charge
stops flowing and the
condition for equilibrium
requires that: q·E = q·v·B
or E = v·B.
• Since the electric field is constant, the electric field
produced in the conductor is related to the
potential difference across the ends by V = E·l.
• Thus: V = E·l = B·l·v, where the upper end of the
conductor is at a higher potential than the lower
end.
• A potential difference is
maintained as long as
there is motion through
the field.
• If the motion is reversed,
the polarity of the voltage
is also reversed.
• If the moving conductor is part of a closed
conducting path, the changing magnetic flux can
cause an induced current in the closed circuit.
• Consider a circuit consisting of a conducting bar of
length l sliding along two fixed parallel conducting
rails as shown in the figure.
• Assume that the moving
bar has zero resistance
and that the stationary
part of the circuit has a
resistance R.
• A uniform and constant
magnetic field B is applied
perpendicular to the plane
of the circuit.
• As the bar is pulled to the right with a velocity v,
under the influence of an applied force Fapp, free
charges in the bar experience a magnetic force along
the length of the bar.
• This magnetic force sets up an induced current since
the charges are free to move in a closed
conducting path.
• The rate of change of
magnetic flux thru the loop
and the corresponding
induced EMF across the
moving bar are proportional
to the change in the area
of the loop as the bar moves
thru the magnetic field.
• If the bar is pulled to the right with a constant velocity,
the work done by the applied force is dissipated in
the form of joule heating in the circuit’s resistive
element.
• The area of the circuit at any instant is l·x, the
external magnetic flux through
the circuit is Φm = B·l·x,
where x is the width of the
circuit, which changes with
time.
• Using Faraday’s law:
 dΦ m  dB  l  x 
EMF 

dt
dt
dx
EMF  B  l 
 B  l  v
dt
• If the resistance of the circuit is R, the magnitude
of the induced current is:
EMF B  l  v
I

R
R
• The equivalent circuit diagram is shown below.
• Since there is no real
battery in the circuit, the
external force does work
on the conductor, thereby
moving charges thru the
magnetic field.
• This causes the charges to
move along the conductor
with an average drift velocity, and a current is set
up.
• The total work done by the applied force during
some time interval should equal the electrical
energy that the induced EMF supplied in the same
period.
• If the bar moves with constant speed, the work
done must equal the energy dissipated as heat in
the resistor in this time interval.
• As the conductor of length l moves through the
uniform magnetic field B, it experiences a
magnetic force Fm of magnitude I·l·B.
• The direction of the force Fm is opposite the
motion of the bar, or to the left in the figure.
• If the bar is to move with a constant velocity, the
applied force must be equal to and opposite the
magnetic force, or to the right in the figure.
• If the magnetic force acted in the direction of
motion, it would cause the
bar to accelerate once it
was in motion, thereby
increasing its velocity.
• An acceleration would
violate the law of
conservation of energy.
• The power delivered by the applied force is:
P  Fapp  v   I  l  B   v  I  l  B  v
B l  v
I 
R
P=I  V
EMF  B  l  v
B  l  v

B l  v
B l v
P
 l B v 

R
R
R
EMF  B  l  v
2
EMF
P
R
2
2
2
2
• This power is equal to the rate at which energy is
dissipated in the resistor, I2·R.
• This power is also equal to the power I·EMF
supplied by the induced EMF.
• This is a clear demonstration of the conversion of
mechanical energy into electrical energy and
finally into thermal energy (joule heating).
Example: Induced EMF in a Rotating Bar
• A conducting bar of length l rotates with a constant
angular velocity  about a pivot at one end. A
uniform magnetic field B is directed perpendicular
to the plane of rotation. Find the induced EMF
between the ends of the bar.
• Divide the length of the bar into small elements of
length labeled dr, whose tangential velocity is v.
• Tangential velocity v is constant for each element
of length dr; however, the tangential velocity of
each length of dr
increases from the
point O (where it is
0 m/s) to the end
of the bar (where it
is a maximum.
• The EMF induced in each element dr is a
component of the total EMF in the conductor.
• The component of the EMF of each element of
length dr moving perpendicular to a field B is:
dEMF = B·v·dr
• Each element dr of
the bar is moving
perpendicular to B,
so there is an EMF
generated across
each element dr.
• Summing up the EMFs induced across all the
elements, which are in series, gives the total EMF
between the ends of the bar.
• Mathematically: EMF = B·v·dr
• To integrate the expression,
linear speed v is related
to angular speed  by
v = r·.
• Since B and  are
constant:
EMF  B   v  dr
EMF  B   r    dr
l
EMF  B     r  dr
0
• Completing the integration from the pivot point of
the rod at 0 to the end of the rod at l:
l
l
r 
 r 
EMF  B    
  B 

1  1 0
1  1 0
11
2
2
 l 2 02 
l
EMF  B        B   
2
2 2 
2
B l
EMF 
2
Magnetic Force on a Sliding Bar
• A bar of mass m and length l moves on two
frictionless parallel rails in the presence of a
uniform magnetic field directed into the page. The
bar is given an initial velocity vo to the right and
released. Find the velocity of the bar as a function
of time.
– The induced current is
counterclockwise and the
force Fm = -I·l·B, where the
negative sign denotes that
the force is to the left and
retards the motion.
Magnetic Force on a Sliding Bar
– Fm is the only horizontal force acting on the bar and
applying Newton’s second law:
dv
Fx  m  a  m 
 I  l  B
dt
– The induced current is equal to:
Bl v
I
R
• Substituting and rearranging:
dv
m
 I  l  B
dt
2
2
dv
Bl v
- B  l v
m

 l B 
dt
R
R
2
2
dv - B  l  v

dt
Rm
• Write the expression as a differential equation:
dv  B  l

 dt
v
Rm
2
2
• Integrate the equation from v = vo to v = v and
from t = 0 s to t = t.
1
B  l
vo v  dv  0 R  m  dt
2
2
t
B  l
v
  dt
ln vvo 
R m 0
2
2
B  l
t
ln v-ln v o 
  t 0
R m
2
2
v B  l
ln 
  t  0
vo
R m
v
2
t
v B  l
ln 
t
vo
R m
2
2
2
• Exponentiate both sides of the equation:
v
ln
vo
B2 l 2 t
R m
e
e
2 2

B
l t
v
R m
e
vo
vt   v o  e
B2 l 2 t
R m
• The velocity of the bar decreases exponentially
with time under the action of the magnetic
retarding force.