Physics For Scientists And Engineers 6E - part 244

SECTION 31.2 • Motional emf

973

31.2 Motional emf

In Examples 31.1 and 31.2, we considered cases in which an emf is induced in a
stationary circuit placed in a magnetic field when the field changes with time. In this
section we describe what is called

motional emf, which is the emf induced in a

conductor moving through a constant magnetic field.

The straight conductor of length ! shown in Figure 31.9 is moving through a

uniform magnetic field directed into the page. For simplicity, we assume that the
conductor is moving in a direction perpendicular to the field with constant velocity
under the influence of some external agent. The electrons in the conductor experi-
ence a force

v " B that is directed along the length !, perpendicular to both

v and B (Eq. 29.1). Under the influence of this force, the electrons move to the
lower end of the conductor and accumulate there, leaving a net positive charge at
the upper end. As a result of this charge separation, an electric field

E is produced

max

Figure 31.7 (Example 31.2) Exponential decrease in the

magnitude of the magnetic field with time. The induced emf

and induced current vary with time in the same way.

Conceptual Example 31.3 Which Bulb Is Shorted Out?

Two bulbs are connected to opposite sides of a circular loop
of wire, as shown in Figure 31.8a. A changing magnetic field
(confined to the smaller circular area shown in the figure)
induces  an  emf  in  the  loop  that  causes  the  two  bulbs  to
light.  When  the  switch  is  closed,  the  resistance-free  wires
connected  to  the  switch  short  out  bulb  2  and  it  goes  out.
What  happens  if  the  wires  containing  the  closed  switch
remain connected at points a and b, but the switch and the
wires are lifted up and moved to the other side of the field,
as in Figure 3.18b? The wire is still connected to bulb 2 as it
was before, so does it continue to stay dark?

Solution When  the  wire  is  moved  to  the  other  side,  even
though the connections have not changed, bulb 1 goes out
and  bulb  2  glows.  The  bulb  that  is  shorted  depends  on
which side of the changing field the switch is positioned! In
Figure  31.8a,  because  the  branch  containing  bulb  2  is  infi-
nitely  more  resistant  than  the  branch  containing  the
resistance-free switch, we can imagine removing the branch
with  the  bulb  without  altering  the  circuit.  Then  we  have  a
simple loop containing only bulb 1, which glows.

When the wire is moved, as in Figure 31.8b, there are

two possible paths for current below points a and b. We can
imagine removing the branch with bulb 1, leaving only a
single loop with bulb 2.

× × × × × × ×

× × × × ×

× × × × × × ×

× × × × ×

Bulb 1

Bulb 2

Switch

(b)

(a)

Figure 31.8 (Conceptual Example 31.3) (a) When the wire

with the switch is located as shown, bulb 2 goes out when the

switch is closed. (b) What happens when the switch and the

wires are moved to the other side of the magnetic field?

BA cos 0 " AB

max

Because AB

max

and a are constants, the induced emf calcu-

lated from Equation 31.1 is

This  expression  indicates  that  the  induced  emf  decays
exponentially  in  time.  Note  that  the  maximum  emf  occurs
at  t " 0,  where

max

aAB

max

. The plot of

versus t is

similar to the B-versus-t curve shown in Figure 31.7.

aAB

max

" %

max

inside  the  conductor.  The  charges  accumulate  at  both  ends  until  the  downward
magnetic  force  qvB on  charges  remaining  in  the  conductor  is  balanced  by  the
upward  electric  force  qE.  At  this  point,  electrons  move  only  with  random  thermal
motion. The condition for equilibrium requires that

qE " qvB

E " vB

The electric field produced in the conductor is related to the potential difference
across the ends of the conductor according to the relationship 'V " E! (Eq. 25.6).
Thus, for the equilibrium condition,

V " E! " B !v

(31.4)

where the upper end of the conductor in Figure 31.9 is at a higher electric potential
than the lower end. Thus,

a potential difference is maintained between the ends

of the conductor as long as the conductor continues to move through the
uniform magnetic field. If the direction of the motion is reversed, the polarity of the
potential difference is also reversed.

A more interesting situation occurs when the moving conductor is part of a closed

conducting path. This situation is particularly useful for illustrating how a changing
magnetic flux causes an induced current in a closed circuit. Consider a circuit consist-
ing of a conducting bar of length ! sliding along two fixed parallel conducting rails, as
shown in Figure 31.10a.

For simplicity, we assume that the 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

app

, free charges in the bar

experience a magnetic force directed along the length of the bar. This force sets up
an  induced  current  because  the  charges  are  free  to  move  in  the  closed  conducting
path. In this case, the rate of change of magnetic flux through the loop and the corre-
sponding  induced  motional  emf  across  the  moving  bar  are  proportional  to  the
change in area of the loop. If the bar is pulled to the right with a constant velocity, the
work  done  by  the  applied  force  appears  as  internal  energy  in  the  resistor  R.  (See
Section 27.6.)

Because the area enclosed by the circuit at any instant is !x, where x is the position

of the bar, the magnetic flux through that area is

B !x

Using Faraday’s law, and noting that x changes with time at a rate dx/dt " v, we find
that the induced motional emf is

(31.5)

Because the resistance of the circuit is R, the magnitude of the induced current is

(31.6)

The equivalent circuit diagram for this example is shown in Figure 31.10b.

I "

B!v

" %

B!v

" %

(B!x) " %B!

974

CHAPTE R 31 • Faraday’s Law

Figure 31.9 A straight electrical conductor of length ! moving

with a velocity v through a uniform magnetic field B directed

perpendicular to v. Due to the magnetic force on electrons, the

ends of the conductor become oppositely charged. This establishes

an electric field in the conductor. In steady state, the electric and

magnetic forces on an electron in the wire are balanced.

−

–

Active Figure 31.10 (a) A conduct-

ing bar sliding with a velocity v along

two conducting rails under the

action of an applied force F

app

. The

magnetic force F

opposes the

motion, and a counterclockwise

current I is induced in the loop.

(b) The equivalent circuit diagram

for the setup shown in part (a).

At the Active Figures link

at http://www.pse6.com, you

can adjust the applied force,

the magnetic field, and the

resistance to see the effects on

the motion of the bar.

 

(b)

B v

(a)

app

Motional emf

Let us examine the system using energy considerations. Because no battery is in the

circuit, we might wonder about the origin of the induced current and the energy deliv-
ered to the resistor. We can understand the source of this current and energy by noting
that  the  applied  force  does  work  on  the  conducting  bar,  thereby  moving  charges
through  a  magnetic  field.  Their  movement  through  the  field  causes  the  charges  to
move along the bar with some average drift velocity, and hence a current is established.
The change in energy in the system during some time interval must be equal to the
transfer  of  energy  into  the  system  by  work,  consistent  with  the  general  principle  of
conservation of energy described by Equation 7.17.

Let us verify this mathematically. As the bar moves through the uniform magnetic

field

B, it experiences a magnetic force F

of magnitude I!B (see Section 29.2). The

direction of this force is opposite the motion of the bar, to the left in Figure 31.10a.
Because the bar moves with constant velocity, the applied force must be equal in
magnitude and opposite in direction to the magnetic force, or to the right in Figure
31.10a. (If

acted in the direction of motion, it would cause the bar to accelerate,

violating the principle of conservation of energy.) Using Equation 31.6 and the fact
that F

app

I!B, we find that the power delivered by the applied force is

(31.7)

From Equation 27.23, we see that this power input is equal to the rate at which energy
is delivered to the resistor, so that Equation 7.17 is confirmed in this situation.

" "

app

v " (I!B)v "

SECTION 31.2 • Motional emf

975

Example 31.4 Motional emf Induced in a Rotating Bar

A conducting bar of length ! rotates with a constant angular
speed * about a pivot at one end. A uniform magnetic field

B is directed perpendicular to the plane of rotation, as
shown in Figure 31.11. Find the motional emf induced
between the ends of the bar.

Solution Consider a segment of the bar of length dr having
a velocity

v. According to Equation 31.5, the magnitude of

the emf induced in this segment is

Bv dr

Because every segment of the bar is moving perpendicular
to

B, an emf d

of the same form is generated across each

segment. Summing the emfs induced across all segments,
which are in series, gives the total emf between the ends

Interactive

Quick Quiz 31.4

As an airplane flies from Los Angeles to Seattle, it passes

through the Earth’s magnetic field. As a result, a motional emf is developed between
the wingtips. Which wingtip is positively charged? (a) the left wing (b) the right
wing.

Quick Quiz 31.5

In Figure 31.10, a given applied force of magnitude F

app

results in a constant speed v and a power input ". Imagine that the force is increased
so that the constant speed of the bar is doubled to 2v. Under these conditions, the new
force and the new power input are (a) 2F and 2" (b) 4F and 2" (c) 2F and 4" (d) 4F
and 4".

Quick Quiz 31.6

You wish to move a rectangular loop of wire into a region

of uniform magnetic field at a given speed so as to induce an emf in the loop. The
plane of the loop remains perpendicular to the magnetic field lines. In which orienta-
tion should you hold the loop while you move it into the region of magnetic field in
order to generate the largest emf? (a) with the long dimension of the loop parallel to
the velocity vector (b) with the short dimension of the loop parallel to the velocity
vector (c) either way—the emf is the same regardless of orientation.

976

CHAPTE R 31 • Faraday’s Law

Example 31.5 Magnetic Force Acting on a Sliding Bar

The conducting bar illustrated in Figure 31.12 moves on two
frictionless parallel rails in the presence of a uniform mag-
netic field directed into the page. The bar has mass m and
its length is !. The bar is given an initial velocity v

to the

right and is released at t " 0.

(A)

Using Newton’s laws, find the velocity of the bar as a

function of time.

(B)

Show that the same result is reached by using an energy

approach.

Solution (A) Conceptualize this situation as follows. As the
bar  slides  to  the  right  in  Figure  31.12,  a  counterclockwise
current  is  established  in  the  circuit  consisting  of  the  bar,
the  rails,  and  the  resistor.  The  upward  current  in  the  bar
results in a magnetic force to the left on the bar as shown in
the  figure.  As  a  result,  the  bar  will  slow  down,  so  our
mathematical solution should demonstrate this. The text of
part  (A)  already  categorizes  this  as  a  problem  in  using
Newton’s laws. To analyze the problem, we determine from

Equation 29.3 that the magnetic force is F

" %

I!B, where

the  negative  sign  indicates  that  the  retarding  force  is  to
the left.  Because  this  is  the  only horizontal  force  acting
on the  bar,  Newton’s  second  law  applied  to  motion  in  the

Interactive

of the bar:

To integrate this expression, note that the linear speed v of
an element is related to the angular speed * through the
relationship v " r* (Eq. 10.10). Therefore, because B and *
are constants, we find that

What If?

Suppose, after reading through this example, you

come up with a brilliant idea. A Ferris wheel has radial metal-
lic spokes between the hub and the circular rim. These
spokes move in the magnetic field of the Earth, so each

B*!

v dr " B*

r dr "

Bv dr

spoke acts like the bar in Figure 31.11. You plan to use the
emf generated by the rotation of the Ferris wheel to power
the lightbulbs on the wheel! Will this idea work?

Answer The fact that this is not done in practice suggests
that others may have thought of this idea and rejected it. Let
us estimate the emf that is generated in this situation. We
know the magnitude of the magnetic field of the Earth from
Table 29.1, B " 0.5 + 10

T. A typical spoke on a Ferris

wheel might have a length on the order of 10 m. Suppose
the period of rotation is on the order of 10 s. This gives an
angular speed of

Assuming that the magnetic field lines of the Earth are
horizontal at the location of the Ferris wheel and perpendic-
ular to the spokes, the emf generated is

This is a tiny emf, far smaller than that required to operate
lightbulbs.

An additional difficulty is related to energy. Assuming

you could find lightbulbs that operate using a potential dif-
ference on the order of millivolts, a spoke must be part of a
circuit  in  order  to  provide  a  voltage  to  the  bulbs.  Conse-
quently,  the  spoke  must  carry  a  current.  Because  this
current-carrying  spoke  is  in  a  magnetic  field,  a  magnetic
force is exerted on the spoke and the direction of the force
is opposite to its direction of motion. As a result, the motor
of  the  Ferris  wheel  must  supply  more  energy  to  perform
work against this magnetic drag force. The motor must ulti-
mately  provide  the  energy  that  is  operating  the  lightbulbs
and you have not gained anything for free!

2.5 + 10

" 1 mV

B*!

(0.5 + 10

T)(1 s

)(10 m)

* "

10 s

0.63 s

" 1 s

Figure 31.11 (Example 31.4) A conducting bar rotating

around a pivot at one end in a uniform magnetic field that is

perpendicular to the plane of rotation. A motional emf is

induced across the ends of the bar.

At the Interactive Worked Example link at http://www.pse6.com, you can explore the induced emf for different angular
speeds and field magnitudes.

Figure 31.12 (Example 31.5) A conducting bar of length ! on

two fixed conducting rails is given an initial velocity v

to the right.

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