# Mar1ClassNotes - 6.4 How slow can you go he Globe of Death...

• Notes
• 4

This preview shows page 1 out of 4 pages. 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 ...
View Full Document

• • • 