**Unformatted text preview: **6.4 How slow can you go? he Globe of Death, pictured at the start of the
2.2-m-radius vertical loop. To keep control of
wants the normal force on his tires at the top of
r exceed his and the bike’s combined weight.
um speed at which the rider can take the loop? • Today: 181 Apparent Forces in Circular Motion Friday March 1 Solving Strategy 6.1, we’ve chosen the x-axis to point toward the
center of the circle.
SOLVE We will consider the forces at the top point of the loop.
Because the x-axis points downward, Newton’s second law is mv
–Finish chapter 24 ifawe +can.
F =w n=
r
The with learningcatalytics.com: Class
–Second trial minimum acceptable speed occurs when n = w; thus
2w =
session ID is: 9464612mg = mvr
Solving for the speed, we find
–Demos
v = 22gr = 22(9.8 m/s )(2.2 m) = 6.6 m/s
2 ual overview for this problem is shown in
e top of the loop, the normal force of the cage
ownward force. In accordance with Problem- x 2
min g in a vertical loop around the Globe of 2 min • Reminders and new information:
The minimum speed is L 15 mph, which isn’t all that
fast; the bikes can easily reach this speed. But normally several
bikes are in the globe at one time. The big challenge is to keep all
of the riders in the cage moving at this speed in synchrony. The
period for the circular motion at this speed is T = 2pr/v L 2 s,
leaving little room for error!
ASSESS –Chapter 24 homework is available on
MasteringPhysics. Due Monday March 11.
–Create an account at learningcatalytics.com using
access code: ATDFZA7
mportant biological application of circular motion, is used to seps of a liquid with –Text questions to 480-382-3749
different densities. Typically these are different e components of cells, suspended in water. You probably know
uspended in water will eventually settle to the bottom. However,
on due to gravity for extremely small objects such as cells is so
ke days or even months for the cells to settle out. It’s not practical
l samples to separate due to gravity alone.
would go faster if the force of gravity could be increased.
change gravity, we can increase the apparent weight of objects
inning them very fast, and that is what the centrifuge in FIGURE
very high angular velocities, the centrifuge produces centripetal
re thousands of times greater than free-fall acceleration. As the
ly increases gravity to thousands of times its normal value, the
ents settle out and separate by density in a matter of minutes or 1 Circular Motion Revisited A centrifuge.
Recall that whenever the force on an object is perpendicular
The
Analyzing the ultracentrifuge the object will undergo operation of a motion.
to its velocity,
circular
centrifuge.
FIGURE 6.23 r ultracentrifuge produces an extraordinarily large centripetal accelg, where g is the free-fall acceleration due to gravity. What is its frer
hat is the apparent weight of a sample v
with a mass of 0.0030 kg? The high angular velocity requires
a large normal force, which leads
to a large apparent weight. eleration in SI units is 6 4:06 PM Page 807 a = 250,000(9.80 m/s2) = 2.45 *r106 m/s2
fs a
2.45 * 106 m/s2
=
= 5.22 * 103 rad/s
Ar B
0.090 m he diameter, or r = 9.0 cm = 0.090 m. petal acceleration is related to the angular speed by a = v2r. Thus v= r
v Continued 24.5 . Magnetic Fields Exert Forces on Moving Charges r slow down. Suppose a positively charged particle is moving perr
uniform magnetic field B as shown in Figure 24.32. In Chapter 6,
he motion of objects subject to r force that was always perpendicua
v
city. The net result was circular motion at a constant speed. For a
n a circle at the end of a string, the tension force is always perpenr
or a satellite moving in a circular orbit, the Fnet
gravitational force is
r
dicular to v. Now, for a charged particle moving in a magnetic field,
r
tic force that is always perpendicular to v and so causes the particle
r
ircle. Thus a particle moving perpendicular to a uniform magr
dergoes uniform circular motion at constant speed.
Fnet r f
4th Proo r v ؉
r ؉ r r F F r
v ؉ r
v The magnetic force is always
r
perpendicular to v, causing the
particle to move in a circle. 3 shows a particle of mass m moving at a speed v in a circle of
und in Chapter 6 that this motion requires a force directed toward
e circle with magnitude
mv 2
r r
v is perpendicular to B.
r B into page F r
ecause the right-hand rule for forces specifies the direction of the
v
positive charge, the force on a negative charge is in the opposite
A negatively charged particle will orbit in the opposite sense from
in Figure 24.32 for a positive charge. ᮤ F5 807 FIGURE 24.32 A charged particle moving
perpendicular to a uniform magnetic
field.
r
v (24.7)
r particle moving in a magnetic field, this force is the force from the
the charged particle with the magnetic field. In Figure 24.32 we
he velocity was perpendicular to the magnetic field, so the angle a
the magnitude of the force on the charged particle due to the magven by Equation 24.6. As this is the force that produces the circular
n equate it to the force in Equation 24.7: F m
Circular Motion of a Charged Particle
r
F must point to the
The radius of the circular orbit is related center ofthe/r towith
to the circleparticle’s
magnitude mv
keep the mass traveling
speed, charge, mass, and the magnetic ﬁeld strength.
around the circle with
r 2 constant speed. mv 2 r F 5 Zq ZvB 5 r FIGURE 24.33 B into page radius of the orbit, we get
r5 mv
Zq ZB (24.8) A particle in circular motion. (a) The velocity can be
broken into components
parallel and perpendicular r
v
to the field. The parallel
component will continue
r
without change.
v' r
vi ends on the ratio of the mass of the particle to its charge, a fact we
The radius also depends on the particle’s speed and the magnetic
r
(b) A top view shows that
Increasing the speed will increase the radius of the circular motion,
the perpendicular
component will change,
ng the field will decrease the radius.
leading to circular
moving perpendicular to a magnetic field moves in a circle. In the
motion.
rt of this section, we saw that a particle moving parallel to a magr
v'
eriences no magnetic force, and so continues in a straight line. A r
r
ituation in which a charged particle’s velocity v is neither parallel
r
(c) The net result is a
icular to the field B is shown in Figure 24.34. The net result is a cirhelical path that spirals
ue to the perpendicular component of the velocity coupled with a
around the field lines.
ity parallel to the field: The charged particle spirals around the
lines in a helical trajectory.
y particles and radiation streaming out from the sun in the solar
ns and electrons as they strike molecules high in the earth’s atmosf these charged particles become trapped in the earth’s magnetic
e 24.35 on the next page shows, the electrons spiral in helical trathe earth’s magnetic field lines. Some of these particles enter the
ar the north and south poles, ionizing gas and creating the ghostly
rora.
FIGURE 24.34 A charged particle in a v F r m r B mv
r=
qB magnetic field follows a helical trajectory. ain mass spectrometer, particles with a charge of 1e
Which particles have a negative charge?
to the spectrometer with a velocity of 2.5 3 105 m/s.
v
v
found to move in a circular path with a radius of
the magnetic field of the spectrometer is 0.050v T,
v
of particles A P T Ethese Magnetic Fields and Forces
810
C H are R 24 . likely to be?
y
ure P24.30, a proton
1.
2.
3.
4.
a 0.40 T magnetic field
These charged particles are traveling in circular orbits with
STOP TO r
THINK 24.4
B
velocities and field directions as noted. Which particles have a negative charge?
cted along the positive
A. 1
B
B
B
B
v
v
e force on the proton is2
B.
r
10215 N, directed out4
F
C.
x
v
v
D.
per. Draw a possible2 and 4
E. 1 and 3
vector for the proton.
he x-component of the
FIGURE P24.30 A.
B.
C.
D.
f the proton?
y
a magnetic moving
0, Theproton isforce on a wire
r
(b)
I
I
ed (a) 5.5 3 105 m/s at
of
B24.6 Magnetic Fields Exert Forces on Currents
2JFK_ch24_v4.qxd 10/3/06 4:06 PM Page 811
of 30° from the x-axis,
r
We have seen that a magnetic field exerts a force on a current. This force is of
v
great practical importance, being responsible for the operation of loudspeakers,
B
in Figure P24.31. A
electric motors, and many other devices. We will find an expression for this force
u = 30°
Fby considering a current as a flow of charged particles, then using our results
field of magnitude
x
F
from the previous section.
pointing in the Lpositive
24.6 . Magnetic Fields Exert Forces on Currents
811
n. What will be the
FIGURE P24.31
The Form of the Magnetic Force on a Current
ate of B If we substitute IL formsB the force equation F 5 qvB, we find that the magnetic
the proton 10 qv in
r r r r r r r r r r r r r r r r r force on a length L of a current-carrying wire is table at the start of Section 24.5, we saw that a magnetic field exerts no force
In the on a charged particle moving parallel to a magnetic field. If a current-carrying
wire is parallel to a magnetic field, we also find that the force on it is zero.
A wire is perpendicular If the wire carries5 ILB
(24.13)
Fwire
It’s more interesting to consider the wire in Figure 24.40a that is perpendiculato an externally created a current, the magnetic
LINEAR
magnetic field. Magnitude of willforce on a current segment of length L
field the exert a force
r to the magnetic field. (Note that this field is an external magnetic field, created
p. 38
on perpendicular to a magneticCurrents magnet or by other currents; it is not the field of the current I in
the moving charges, by a permanent
Magnetic Fields Exert aForcesofon field
causing deflection
the wire.) The direction of the force on the current is found by considering the
the wire.
Equation 24.13 is a simple result, but remember the two assumptionsin the current. We model current as the flow of positive
force on levitate behind
magneticThe wire must be perpendiculardirection willeach charge the over
field strength and to the field, and the field must be constant it:
(c)
charge, so the right-hand
I
the P24.32?
to the field, the force will
in Figurelength L of the wire. If the wire is at an angle a moving charges.rule for forces applies to currents in the same way it
does for
With your fingers aligned as usual, point your right
Example Levitating a wire
depend on this angle:
thumb in the direction of the current (the direction of the motion of positive
What magnetic ﬁeld strength and direction will a
r
Fwire 5 ILB sin levitate the
(24.14)
r
charges) and your index finger in the direction of B. Your middle finger is then
2.0 g wire B the ﬁgure?The right-hand rule
in
r
It is sometimes useful forces applies.
for to rewrite Equation 24.13 as
r
pointing in the direction of the force F on the wire, as in Figure 24.40c. ConseF
Your thumb should
Fwire
quently, the entire length of wire within the magnetic field experiences a force to
point in the direction
B 5 2.0 g
(24.15)
IL left, perpendicular to both the current direction and the field direction, as
of the current.
the
wire
B-field region shown in Figure 24.40b.
Given
expression, we can
1.5 Athis terms of other units:see that the unit for magnetic field, the tesla, can be
defined in
To find the magnitude of the force, we must express the current I in terms of
N
the motion of charges in the wire. Consider a section of wire of length L in which
1T51 #
FIGURE 24.40 Magnetic force on a
A m is a total charge q moving at a speed v, as seen in Figure 24.41.
there
current-carrying wire.
The current I, by definition, is the rate of flow of charge. That is, the current is
Direction of magnetic ﬁeld? force charge q divided by the time DtThe field of the earth near the to flow out of this section
the
it takes the charge
EXAMPLE 24.6 Finding the magnetic
equator is parallel to the
on a power charges 10 cm
A:
4.32 We consider only the line up in this segment of the wire:
1.
ground and points to the north.
The they move to the right, that will
of the wire. Ashigh-voltage power lines they carry electricity across great
B: very high currents as well. Suppose we have a
down
q
distances carry
all eventually pass the right end.
I5
(24.10)
C: in equator, where the earth’s magnetic field
DC power line near the
Dt
N
q = total charge
is parallel to the ground. The wire carries a current of 850 A to the
D: out
v east. The length of cable between adjacent towers is time the charges
The 400 m. What
؉is the؉ of the earth’s magnetic؉ on this length of wire?
؉ ؉
force ؉
field
r ؉ r B take to move out of the segment of wire is
W E The magnetic force on a wire at an angle to the ﬁeld r B Fwire α
I I L
L = length of
wire in ﬁeld Fwire = ILBsin α Half of the loop of wire is within the magnetic ﬁeld.
In which direction is the net force on the loop? I
A. To the left
B. To the right
C. Into the page
D. Out of the page
E. Up Demo: Force between two wires 2_ch24_pp776-815.qxd 8/17/09 3:02 PM Page 797 Two parallel wires have currents that have the same direction, but differing
magnitude. The current in wire A is I; and the current in wire B is 2I.
Which one of the following statements concerning this situation is true?
a) Both wires repel each other with the same amount of force. 24.6 797 Magnetic Fields Exert Forces on Currents b) Wire
ForcesA attracts wire B with twice the force that wire B attracts wire A.
Between Currents
Because a current produces a other with the and a magnetic of force. a force on a
c) Both wires attract each magnetic field, same amount field exerts
current, it follows that two current-carrying wires will exert forces on each other, as
Ampère discovered. It will be a half the force that wire B to this point to show that the
d) Wire A repels wire B with good check on our results attracts wire A.
experimental results we saw earlier are consistent with our rules for determining magnetic fields repels wire B and determiningforce that currentsattractsmagnetic fields.
e) Wire A from currents with twice the forces on wire B due to wire A.
Suppose we have two parallel wires of length L a distance d apart, each carrying a
current. FIGURE 24.41a shows the currents I1 and I2 in the same direction and
FIGURE 24.41b in opposite directions. We will assume that the wires are sufficiently
long to allow us to use the earlier result, Equation 24.1, for the magnetic field of a
long, straight wire: B = m0 I/2p r.
Let’s look at the situation of Figure 24.41a. There are three steps in our analysis: 13.5 FIGURE 24.41 Forces between currents. (a) Currents in the same direction attract.
Force of
magnetic field
r
B2 on current I1 r Magnetic field B2
created by current I2
I1 B 1. The current I2 in the lower wire creates a magnetic field B2 at the position of the
B
upper wire. This field B2 points out of the page, perpendicular to the current I1 .
B
The magnetic forcedue to the lower wire, that exerts a magnetic force on the
It is this field B2, between two current-carrying wires
upper wire. At the position of the
The magnetic ﬁeld of one wire causesupper wire, whichother wire. distance
a force on the is a constant
r = d from the lower wire, the field has the same value at all points along the
Force of wire. The field is
r
magnetic field
Magnetic field B2
r
r
r
B
B2 on current I1
created by current I2 m0 I2
r
F2 on 1
F2 on 1 (24.13)
B2
B2 = a
, out of the pageb
2pd
I1
I1
B
2. For the upper wire, the current is to the right and the field B2 from the lower
wire points out of the page. Using the right-hand rule for forces, you can see
r
r
F on 1
F2 1
d
d
thatr2the force on theonupper wire is downward, attracting it toward the lower
r
F1
F1 on 2
wire.on 2
3. The magnitude of the force on the upper wire is given by Equation 24.10.
I2
I2
Using the field from Equation 24.13, we compute
r
r
r
B1
r
F1 on 2
F1 on 2
m 0 I2
Force of
Magnetic field B1
magnetic field created by current IFparallel wires = I1 LB2 = I1 L
r
1
2pd L
B on current I
1 r r F2 on 1 F2 on 1 d r r F1 on 2 F1 on 2
I2 Force of
magnetic field
r
B1 on current I2 r Magnetic field B1
created by current I1 (b) Currents in opposite directions repel.
r F2 on 1 r B2 r F2 on 1
I1 d 2 Fparallel wires = m 0 L I1 I2
2pd (24.14) I2
r B1 r F1 on 2 Magnetic force between two parallel current-carrying wires r F1 on 2 L The current in the upper wire exerts an upward-directed magnetic force on
the lower wire with exactly the same magnitude. (You know that this must be
the case: The two forces form a Newton’s third law pair.) You should convince
yourself, using the right-hand rule, that the forces are repulsive and tend to push
the wires apart if the two currents are in opposite directions, as shown in Figure
24.41b. Our rules predict exactly what the experimental results showed: Parallel wires carrying currents in the same direction attract each other; parallel
wires carrying currents in opposite directions repel each other.
EXAMPLE 24.12 Finding the force between wires in jumper cables You may have used a set of jumper cables connected to a running
vehicle to start a car with a dead battery. Jumper cables are a
matched pair of wires, red and black, joined together along their
length. Suppose we have a set of jumper cables in which the two
wires are separated by 1.2 cm along their 3.7 m (12 ft) length.
While starting a car, the wires each carry a current of 150 A, in
opposite directions. What is the force between the two wires?
PREPARE Our first step is to sketch the situation, noting distances and currents, as shown in FIGURE 24.42. Because the currents
in the two wires are in opposite directions, the force between the
two wires is repulsive. FIGURE 24.42 Jumper cables carrying opposite currents. L = 3.7 m
I1 = 150 A
d = 1.2 cm
I2 = 150 A Continued ...

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