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CHAPTER 21 MAGNETIC FORCES AND MAGNETIC FIELDS CONCEPTUAL QUESTIONS _____________________________________________________________________________________________ 1. REASONING AND SOLUTION Magnetic field lines, like electric field lines, never intersect. When a moving test charge is placed in a magnetic field so that its velocity vector has a component perpendicular to the field, the particle will experience a force. That force is perpendicular to both the direction of the field and the direction of the velocity. If it were possible for magnetic field lines to intersect, then there would be a different force associated with each of the two intersecting field lines; the particle could be pushed in two directions. Since the force on a particle always has a unique direction, we can conclude that magnetic field lines can never cross. _____________________________________________________________________________________________ 2. REASONING AND SOLUTION If you accidentally use your left hand, instead of your right hand, to determine the direction of the magnetic force on a positive charge moving in a magnetic field, the direction that you determine will be exactly opposite to the correct direction. _____________________________________________________________________________________________ 3. REASONING AND SOLUTION A charged particle, passing through a certain region of space, has a velocity whose magnitude and direction remain constant. a. If it is known that the external magnetic field is zero everywhere in the region, we can conclude that the electric field is also zero. Any charged particle placed in an electric field will experience a force given by F = qE, where q is the charge and E is the electric field. If the magnitude and direction of the velocity of the particle are constant, then the particle has zero acceleration. From Newton's second law, we know that the net force on the particle is zero. But there is no magnetic field and, hence, no magnetic force. Therefore, the net force is the electric force. Since the electric force is zero, the electric field must be zero. b. If it is known that the external electric field is zero everywhere, we cannot conclude that the external magnetic field is also zero. In order for a moving charged particle to experience a magnetic force when it is placed in a magnetic field, the velocity of the moving charge must have a component that is perpendicular to the direction of the magnetic field. If the moving charged particle enters the region such that its velocity is parallel or antiparallel to the magnetic field, it will experience no magnetic force, even though a magnetic field is present. In the absence of an external electric field, there is no electric force either. Thus, there is no net force, and the velocity vector will not change in any way. _____________________________________________________________________________________________ 4. REASONING AND SOLUTION Suppose that the positive charge in Figure 21.10b were launched from the south pole toward the north pole, in a direction opposite to the magnetic field. Regardless of the strength of the magnetic field, the particle will always reach the north pole. Since the charge is launched directly opposite to the magnetic field, its velocity will be antiparallel to the field and have 118 MAGNETIC FORCES AND MAGNETIC FIELDS no component perpendicular to the field. Therefore, there will be no magnetic force on the particle. The particle will move at constant velocity from the south pole of the magnet to the north pole. _____________________________________________________________________________________________ 5. REASONING AND SOLUTION Since the paths of the particles are perpendicular to the magnetic field, we know that the velocities of the particles are perpendicular to the field. Since the velocity of particle #2 is perpendicular to the magnetic field and it passes through the field undeflected, we can conclude that particle #2 is neutral. Particles #1 and #3 move in circular paths. The figure at the right shows the direction of the (centripetal) magnetic force that acts on the particles. If the F fingers of the right hand are pointed into the page so that the thumb points in the direction of motion of particle #1, the palm of the hand points toward the center #1 of the circular path traversed by the particle. We can conclude, therefore, from RHR-1 that particle #1 is positively charged. If the fingers of the right hand are pointed into the page so that the F thumb points in the direction of motion of particle #3, the palm of the hand points #3 away from the center of the circular path traversed by the particle. We can conclude, therefore, from RHR-1 that particle #3 is negatively charged. _____________________________________________________________________________________________ Chapter 21 Conceptual Questions 6. 119 REASONING AND SOLUTION Three particles have identical #1 charges and masses. They enter a constant magnetic field and follow #2 the paths as shown. The magnitude of the magnetic force on each particle is given by Equation #3 21.1 as F = q0 (v sin θ )B, where q0 is the magnitude of the charge on the particle, v is the speed of the particle and B is the magnitude of the magnetic field. Since the particle paths are perpendicular to the magnetic field, θ = 90° and sin θ = 1, so that F = q0vB . This magnetic force supplies the centripetal force that is necessary for the particles to move on the observed circular paths, so q0 vB = mv2 / r , where m is the mass of the particle and r is the radius of its circular path. Solving for v, we find v = q0 Br / m .Since all three particles have the same charge and mass, we can conclude that the speed of the particle is directly proportional to the radius of the particle's path. Therefore, particle #1 is traveling the fastest, while particle #2 is traveling the slowest. _____________________________________________________________________________________________ 7. REASONING AND SOLUTION A proton follows the path shown in Figure 21.12. The magnitude of the magnetic force on the proton is given by Equation 21.1 as F = q0 (v sin θ)B, where q0 is the magnitude of the charge on the proton, v is the speed of the proton and B is the magnitude of the magnetic field. Since the proton's path is perpendicular to the magnetic field, θ = 90° and sin θ = 1, so that F = q0vB . This magnetic force supplies the centripetal force that is necessary for the proton 2 to move on the observed circular path, so q0 vB = mprotonv / r , where m is the mass of the particle and r is the radius of its circular path. The magnitude of the magnetic field is, therefore, B = mproton v /(q0r ). If we want an electron to follow exactly the same path, we must adjust the magnetic field. The magnitude of the charge on the electron is the same as that on the proton; however, the electron is negatively charged and the proton is positively charged. As stated in the text, the direction of the force on a negative charge is opposite to that predicted by RHR-1 for a positive charge. Therefore, the direction of the magnetic field must be reversed. In order for the electron to travel with the same speed v in a circular path of the same radius r, the magnitude of the magnetic field must be changed to the value B = melectron v /(q0r) . Thus, the electron will travel in the same path as the proton in Figure 21.12 if the direction of the magnetic field is reversed and the magnitude of the magnetic field is reduced by a factor (melectron / mproton ). _____________________________________________________________________________________________ 120 8. MAGNETIC FORCES AND MAGNETIC FIELDS REASONING AND SOLUTION The drawing shows a top view of four interconnected chambers. A negative charge is fired into chamber 1. By turning on separate magnetic fields in each chamber, the charge is made to exit from chamber 4. 3 4 F F F F a. In each chamber the path of the particle is one1 2 quarter of a circle. The drawing at the right also shows the direction of the centripetal force that v must act on the particle in each chamber in order –q for the particle to traverse the path. The charged particle can be made to move in a circular path by launching it into a region in which there exists a magnetic field that is perpendicular to the velocity of the particle. Using RHR-1, we see that if the palm of the right hand were facing in the direction of F in chamber 1 so that the thumb points along the path of the particle, the fingers of the right hand must point out of the page. This is the direction that the magnetic field must have to make a positive charge move along the path shown in chamber 1. Since the particle is negatively charged, the field must point opposite to that direction or into the page. Similar reasoning using RHR-1, and remembering that the particle is negatively charged, leads to the following conclusions: in region 2 the field must point out of the page, in region 3 the field must point out of the page, and in region 4 the field must point into the page. b. If the speed of the particle is v when it enters chamber 1, it will emerge from chamber 4 with the same speed v. The magnetic force is always perpendicular to the velocity of the particle; therefore, it cannot do work on the particle and cannot change the kinetic energy of the particle, according to the work-energy theorem. Since the kinetic energy is unchanged, the speed remains constant. _____________________________________________________________________________________________ 9. REASONING AND SOLUTION A positive charge moves along a circular path under the influence of a magnetic field. The magnetic field is perpendicular to the plane of the circle, as in Figure 21.12. If the velocity of the particle is reversed at some point along the path, the particle will not retrace its path. If the velocity of the particle is suddenly reversed, then from RHR-1 we see that the force on the particle reverses direction. The particle will travel on a different circle that intersects the point where the direction of the velocity changes. The direction of motion of the particle (clockwise or counterclockwise) will be the same as that in the original circle. This is suggested in the figure below. Chapter 21 Conceptual Questions original path 121 subsequent path F F point of velocity reversal _____________________________________________________________________________________________ 10. REASONING AND SOLUTION A television tube consists of an evacuated tube that contains an electron gun that sends a narrow beam of high-speed electrons toward the screen of the tube (see text Figure 21.40). The inner surface of the screen is covered with a phosphor coating, and when the electrons strike it, they generate a spot of visible light. The electron beam is deflected by the magnetic fields produced by electromagnets placed around the neck of the tube, between the electron gun and the screen. The magnetic fields produced by the electromagnets exert forces on the moving electrons, causing their trajectories to bend and reach different points on the screen, thereby producing the picture. If one end of a bar magnet is placed near a TV screen, the magnetic field of the bar magnet alters the trajectories of the electrons. As a result, the picture becomes distorted. _____________________________________________________________________________________________ 11. REASONING AND SOLUTION When the particle is launched in the x, y plane, its initial velocity will be perpendicular to the magnetic field; therefore, the particle will travel on a circular path in the x, y plane. In order for the charged particle to hit the target, the target must lie on the circular path of the moving particle. This will occur if the particle moves in a counterclockwise circle that passes through the third quadrant of the coordinate system. RHR-1 can be used to determine possible trajectories in the following way. Place the fingers of the right hand into the page (direction of the magnetic field) and orient the thumb along one of the coordinate axes (direction of the particle's initial velocity). The palm of the right hand will face the direction in which the force on the charged particle is directed. Since the particle travels on a circle, the direction of the force will point toward the center of the particle's trajectory. The figures below show the results for all four possible cases. Initial velocity in positive x direction Initial velocity in negative x direction F v v Target F 122 MAGNETIC FORCES AND MAGNETIC FIELDS Initial velocity in positive y direction v Initial velocity in negative y direction F F v Clearly, the charged particle can hit the target only if the initial velocity of the particle points either in the negative x direction or the positive y direction. _____________________________________________________________________________________________ 12. REASONING AND SOLUTION Refer to Figure 21.17. We can use RHR-1 with the modification that the direction of the velocity of a positive charge is the same as the direction of the conventional current I. a. If the direction of the current I in the wire is reversed, the thumb of the right hand (direction of I) will be directed into the page, and the palm of the hand (direction of F) will face the left side of the page. Therefore, the wire will be pushed to the left. b. If both the current and the magnetic poles are reversed, then the fingers of the right hand must point toward the top of the page (direction of B), while the thumb will be directed into the page (direction of I). The palm of the hand will face the right (direction of F); therefore, the wire will be pushed to the right. _____________________________________________________________________________________________ 13. REASONING AND SOLUTION According to RHR-1, with the thumb of the right hand pointing in the direction of the current I, and the palm of the hand facing the top of the page (direction of F), the fingers of the right hand point toward the left pole of the magnet. Therefore, the magnetic field in the vicinity of the current generally points from the right pole to the left pole of the horseshoe magnet shown in the drawing in the text. We can deduce, therefore, that the pole on the right is the north pole and the pole on the left is the south pole. _____________________________________________________________________________________________ 14. REASONING AND SOLUTION In Figure 21.31, assume that current I1 is larger than current I2. Consider three regions: Region I lies to the left of both wires, region II lies between the wires, and region III lies to the right of both wires. Let B1 represent the magnetic field due to current I1, and let B2 represent the magnetic field due to current I2. a. The currents point in opposite directions. From RHR-2, the vectors B1 and B2 point in the same direction in region II. Therefore, regardless of the magnitudes of B1 and B2, the resultant of the fields can never be zero between the wires. The fields B1 and B2 point in opposite directions in regions I and III. Therefore, the resultant of B1 and B2 could be zero only in these two regions. The resultant of B1 and B2 will be zero at the point where their magnitudes are equal. The magnitude of the Chapter 21 Conceptual Questions 123 magnetic field at a distance r from a long straight wire is given by Equation 21.5: B = µ0 I /(2πr ), where I is the magnitude of the current in the wire and µ0 is the permeability of free space. If the current I1 is larger than the current I2, the magnitude of B1 will be equal to the magnitude of B2 somewhere in region III. This is because region III is closer to the smaller current I2, and the smaller value for r in Equation 21.5 allows the effect of I2 to offset the effect of the greater and more distant current I1. Therefore, there is a point to the right of both wires where the total magnetic field is zero. b. The currents point in the same direction. From RHR-2, the vectors B1 and B2 point in the same direction in both regions I and III; therefore, the resultant of B1 and B2 cannot be zero in these regions. In region II, the vectors B1 and B2 point in opposite directions; therefore, there is a place where the total magnetic field is zero between the wires. It is located closer to the wire carrying the smaller current I2. _____________________________________________________________________________________________ 15. REASONING AND SOLUTION The drawing shows an endon view of three parallel wires that are perpendicular to the plane of the paper. In two of the wires, the current is directed Current into the paper, while in the remaining wire the current is directed out of the paper. The two outermost wires are held rigidly in place. From Example 9, we know that two parallel currents that point in the same direction attract each other while two parallel currents that point in opposite directions repel each other. Therefore, the middle wire will be attracted to the wire on the left and repelled from the wire on the right. Since each of the fixed wires exerts a force to the left on the middle wire, the net force on the middle wire will be to the left. Thus, the middle wire will move to the left. _____________________________________________________________________________________________ 16. REASONING AND SOLUTION The figure below shows the arrangements of electromagnets and magnets. (a) S N N S S N N S (b) 124 MAGNETIC FORCES AND MAGNETIC FIELDS We can determine the polarity of the electromagnets by using RHR-2. Imagine holding the currentcarrying wire of the electromagnet in the right hand as the wire begins to coil around the iron core. The thumb points in the direction of the current. For the electromagnet in figure (a), the fingers of the right hand wrap around the wire on the left end so that they point, inside the coil, toward the right end. Thus, the right end of the coil must be a north pole. Similar reasoning can be used to identify the north and south poles of the electromagnet in figure (b). The results are shown in the figure above. Since the like poles of two different magnets repel each other and the dissimilar poles of two different magnets attract each other, we can conclude that in both arrangements, the electromagnet is repelled from the permanent magnet at the right. _____________________________________________________________________________________________ 17. REASONING AND SOLUTION The figure below shows the arrangements of electromagnets. (a) S N S N N S S N (b) We can determine the polarity of each electromagnet by using RHR-2. Imagine holding the currentcarrying wire of the electromagnet in the right hand as the wire begins to coil around the iron core. The thumb points in the direction of the current. For the electromagnet on the left in figure (a), the fingers of the right hand wrap around the wire on the left end so that they point, inside the coil, toward the right end. Thus, the right end of the coil must be a north pole. Similar reasoning can be used to identify the north and south poles of the other remaining electromagnets. The results are shown in the figure above. Since the like poles of two different magnets repel each other and the dissimilar poles of two different magnets attract each other, we can conclude that only the arrangement shown in (a) results in attraction. The electromagnets shown in arrangement (b) result in repulsion. _____________________________________________________________________________________________ Chapter 21 Conceptual Questions 18. REASONING AND SOLUTION Refer to Figure North magnetic pole 21.5. If the earth's magnetism is assumed to originate from a large circular loop of current within the earth, the plane of the current loop must be perpendicular to the magnetic axis of the earth, as suggested in the figure at the right. Using RHR-2, the current must flow clockwise when viewed looking down at the loop from the north magnetic pole. 125 Current loop Magnetic axis _____________________________________________________________________________________________ Wire 1 19. REASONING AND SOLUTION The Wire 2 total magnetic field at the point P is the x x B4 B2 resultant of the magnetic field at P due to each individual wire. If the current in all P four wires is directed into the page, then from RHR-2, the magnetic field at P due to the current in wire 1 must point toward wire 3, the magnetic field at P due to the B3 B1 current in wire 2 must point toward wire 1, the magnetic field at P due to the x x current in wire 3 must point toward wire Wire 4 Wire 3 4, and the magnetic field at P due to the current in wire 4 must point toward wire 2, as shown at the right. The magnetic field at a distance r from a long straight wire that carries a current I is = B µ0 I /(2πr ) . Since the current in all four wires has the same magnitude and all four wires are equidistant from the point P, each wire gives rise to a magnetic field at P of the same magnitude. When current is flowing through all four wires, the total magnetic field at P is zero. If the current in any single wire is turned off, the total magnetic field will point toward one of the corners. For example, if the current in wire 1 is turned off, the resultant of B2 and B3 is still zero and the total magnetic field is B4. If the current in wire 2 is turned off, then the total magnetic field is B3, and so on. We could have achieved similar results if the current in all four wires was directed out of the page. If fact, as long as opposite wires along the diagonal have currents in the same direction (for example, wires 1 and 4 with outward currents and wires 2 and 3 with inward currents) the total magnetic field will point toward one of the corners when one of the currents is turned off. _____________________________________________________________________________________________ 126 MAGNETIC FORCES AND MAGNETIC FIELDS 20. REASONING AND SOLUTION You have two bars, one of which is a permanent magnet and the other of which is not a magnet, but is made from a ferromagnetic material like iron. The two bars look exactly alike. a. A third bar (the test bar) that is a permanent magnet can be used to distinguish which of the lookalike bars is the permanent magnet and which is the ferromagnetic bar. Both ends of each of the look-alike bars, in succession, should be brought next to one end of the test bar. The look-alike permanent magnet will be attracted to the test bar when opposite poles are brought together, while it will be repelled from the test bar when like poles are brought together. Both ends of the ferromagnetic bar, however, will be attracted to the test bar. The magnetic field of the test bar will always induce a pole that is opposite in polarity to the pole of the magnet. Ferromagnetism always results in attraction. b. The identities of the look-alike bars can be determined from a third bar that is not a magnet, but is made from a ferromagnetic material. Either end of the look-alike permanent magnet will be attracted to both ends of the ferromagnetic test bar, while the look-alike ferromagnetic bar will have no effect on the ferromagnetic test bar. _____________________________________________________________________________________________ 21. REASONING AND SOLUTION A strong electromagnet picks up a delivery truck carrying cans of soda pop. Inside the truck cans of the soft drink are seen to fly upward and stick to the roof just beneath the electromagnet. We can conclude that the cans must be made from a ferromagnetic material, at least in part. Since aluminum is a nonferromagnetic material, we know that the cans are not entirely made of aluminum. _____________________________________________________________________________________________