Chapter 29. Magnetic Fields

Transcription

1 Chapter 29 Magnetic Fields CHAPTER OUTLNE 29.1 Magnetic Fields and Forces 29.2 Magnetic Force Acting on a Current-Carrying Conductor 29.3 Torque on a Current Loop in a Uniform Magnetic Field 29.4 Motion of a Charged Particle in a Uniform Magnetic Field 29.5 Applications nvolving Charged Particles Moving in a Magnetic Field 29.6 The Hall Effect Magnetic fingerprinting allows fingerprints to be seen on surfaces that otherwise would not allow prints to be lifted. The powder spread on the surface is coated with an organic material that adheres to the greasy residue in a fingerprint. A magnetic brush removes the excess powder and makes the fingerprint visible. (James King-Holmes/Photo Researchers, nc.) 894

2 Many historians of science believe that the compass, which uses a magnetic needle, was used in China as early as the 13th century.c., its invention being of Arabic or ndian origin. The early Greeks knew about magnetism as early as 800.C. They discovered that the stone magnetite (Fe 3 O 4 ) attracts pieces of iron. Legend ascribes the name magnetite to the shepherd Magnes, the nails of whose shoes and the tip of whose staff stuck fast to chunks of magnetite while he pastured his flocks. n 1269 a Frenchman named Pierre de Maricourt found that the directions of a needle near a spherical natural magnet formed lines that encircled the sphere and passed through two points diametrically opposite each other, which he called the poles of the magnet. Subsequent experiments showed that every magnet, regardless of its shape, has two poles, called north (N) and south (S) poles, that exert forces on other magnetic poles similar to the way that electric charges exert forces on one another. That is, like poles (N N or S S) repel each other, and opposite poles (N S) attract each other. The poles received their names because of the way a magnet, such as that in a compass, behaves in the presence of the Earth s magnetic field. f a bar magnet is suspended from its midpoint and can swing freely in a horizontal plane, it will rotate until its north pole points to the Earth s geographic North Pole and its south pole points to the Earth s geographic South Pole. 1 n 1600 William Gilbert ( ) extended de Maricourt s experiments to a variety of materials. Using the fact that a compass needle orients in preferred directions, he suggested that the Earth itself is a large permanent magnet. n 1750 experimenters used a torsion balance to show that magnetic poles exert attractive or repulsive forces on each other and that these forces vary as the inverse square of the distance between interacting poles. Although the force between two magnetic poles is otherwise similar to the force between two electric charges, electric charges can be isolated (witness the electron and proton) whereas a single magnetic pole has never been isolated. That is, magnetic poles are always found in pairs. All attempts thus far to detect an isolated magnetic pole have been unsuccessful. No matter how many times a permanent magnet is cut in two, each piece always has a north and a south pole. 2 The relationship between magnetism and electricity was discovered in 1819 when, during a lecture demonstration, the Danish scientist Hans Christian Oersted found that an electric current in a wire deflected a nearby compass needle. 3 n the 1820s, 1 Note that the Earth s geographic North Pole is magnetically a south pole, whereas its geographic South Pole is magnetically a north pole. ecause opposite magnetic poles attract each other, the pole on a magnet that is attracted to the Earth s geographic North Pole is the magnet s north pole and the pole attracted to the Earth s geographic South Pole is the magnet s south pole. 2 There is some theoretical basis for speculating that magnetic monopoles isolated north or south poles may exist in nature, and attempts to detect them are an active experimental field of investigation. 3 The same discovery was reported in 1802 by an talian jurist, Gian Dominico Romognosi, but was overlooked, probably because it was published in an obscure journal. Hans Christian Oersted Danish Physicist and Chemist ( ) Oersted is best known for observing that a compass needle deflects when placed near a wire carrying a current. This important discovery was the first evidence of the connection between electric and magnetic phenomena. Oersted was also the first to prepare pure aluminum. (North Wind Picture Archives) 895

3 896 CHAPTER 29 Magnetic Fields further connections between electricity and magnetism were demonstrated independently by Faraday and Joseph Henry ( ). They showed that an electric current can be produced in a circuit either by moving a magnet near the circuit or by changing the current in a nearby circuit. These observations demonstrate that a changing magnetic field creates an electric field. Years later, theoretical work by Maxwell showed that the reverse is also true: a changing electric field creates a magnetic field. This chapter examines the forces that act on moving charges and on currentcarrying wires in the presence of a magnetic field. The source of the magnetic field is described in Chapter Magnetic Fields and Forces n our study of electricity, we described the interactions between charged objects in terms of electric fields. Recall that an electric field surrounds any electric charge. n addition to containing an electric field, the region of space surrounding any moving electric charge also contains a magnetic field. A magnetic field also surrounds a magnetic substance making up a permanent magnet. Historically, the symbol has been used to represent a magnetic field, and this is the notation we use in this text. The direction of the magnetic field at any location is the direction in which a compass needle points at that location. As with the electric field, we can represent the magnetic field by means of drawings with magnetic field lines. Figure 29.1 shows how the magnetic field lines of a bar magnet can be traced with the aid of a compass. Note that the magnetic field lines outside the magnet point away from north poles and toward south poles. One can display magnetic field patterns of a bar magnet using small iron filings, as shown in Figure We can define a magnetic field at some point in space in terms of the magnetic force F that the field exerts on a charged particle moving with a velocity v, which we call the test object. For the time being, let us assume that no electric or gravitational fields are present at the location of the test object. Experiments on various charged particles moving in a magnetic field give the following results: Properties of the magnetic force on a charge moving in a magnetic field The magnitude F of the magnetic force exerted on the particle is proportional to the charge q and to the speed v of the particle. The magnitude and direction of F depend on the velocity of the particle and on the magnitude and direction of the magnetic field. When a charged particle moves parallel to the magnetic field vector, the magnetic force acting on the particle is zero. N S At the Active Figures link at you can move the compass around and trace the magnetic field lines for yourself. Active Figure 29.1 Compass needles can be used to trace the magnetic field lines in the region outside a bar magnet.

4 SECTON 29.1 Magnetic Fields and Forces 897 Henry Leap and Jim Lehman (a) (b) (c) Figure 29.2 (a) Magnetic field pattern surrounding a bar magnet as displayed with iron filings. (b) Magnetic field pattern between opposite poles (N S) of two bar magnets. (c) Magnetic field pattern between like poles (N N) of two bar magnets. When the particle s velocity vector makes any angle 0 with the magnetic field, the magnetic force acts in a direction perpendicular to both v and ; that is, F is perpendicular to the plane formed by v and (Fig. 29.3a). The magnetic force exerted on a positive charge is in the direction opposite the direction of the magnetic force exerted on a negative charge moving in the same direction (Fig. 29.3b). The magnitude of the magnetic force exerted on the moving particle is proportional to sin, where is the angle the particle s velocity vector makes with the direction of. We can summarize these observations by writing the magnetic force in the form F q v (29.1) Vector expression for the magnetic force on a charged particle moving in a magnetic field v F v F q θ F v (a) Figure 29.3 The direction of the magnetic force F acting on a charged particle moving with a velocity v in the presence of a magnetic field. (a) The magnetic force is perpendicular to both v and. (b) Oppositely directed magnetic forces F are exerted on two oppositely charged particles moving at the same velocity in a magnetic field. The dashed lines show the paths of the particles, which we will investigate in Section (b)

5 898 CHAPTER 29 Magnetic Fields F v v F (a) Figure 29.4 Two right-hand rules for determining the direction of the magnetic force F q v acting on a particle with charge q moving with a velocity v in a magnetic field. (a) n this rule, the fingers point in the direction of v, with coming out of your palm, so that you can curl your fingers in the direction of. The direction of v, and the force on a positive charge, is the direction in which the thumb points. (b) n this rule, the vector v is in the direction of your thumb and in the direction of your fingers. The force F on a positive charge is in the direction of your palm, as if you are pushing the particle with your hand. (b) which by definition of the cross product (see Section 11.1) is perpendicular to both v and. We can regard this equation as an operational definition of the magnetic field at some point in space. That is, the magnetic field is defined in terms of the force acting on a moving charged particle. Figure 29.4 reviews two right-hand rules for determining the direction of the cross product v and determining the direction of F. The rule in Figure 29.4a depends on our right-hand rule for the cross product in Figure Point the four fingers of your right hand along the direction of v with the palm facing and curl them toward. The extended thumb, which is at a right angle to the fingers, points in the direction of v. ecause F q v, F is in the direction of your thumb if q is positive and opposite the direction of your thumb if q is negative. (f you need more help understanding the cross product, you should review pages 337 to 339, including Fig ) An alternative rule is shown in Figure 29.4b. Here the thumb points in the direction of v and the extended fingers in the direction of. Now, the force F on a positive charge extends outward from your palm. The advantage of this rule is that the force on the charge is in the direction that you would push on something with your hand outward from your palm. The force on a negative charge is in the opposite direction. Feel free to use either of these two right-hand rules. The magnitude of the magnetic force on a charged particle is Magnitude of the magnetic force on a charged particle moving in a magnetic field F q v sin (29.2) where is the smaller angle between v and. From this expression, we see that F is zero when v is parallel or antiparallel to ( 0 or 180 ) and maximum when v is perpendicular to ( 90 ). There are several important differences between electric and magnetic forces: The electric force acts along the direction of the electric field, whereas the magnetic force acts perpendicular to the magnetic field. The electric force acts on a charged particle regardless of whether the particle is moving, whereas the magnetic force acts on a charged particle only when the particle is in motion.

6 SECTON 29.1 Magnetic Fields and Forces 899 The electric force does work in displacing a charged particle, whereas the magnetic force associated with a steady magnetic field does no work when a particle is displaced because the force is perpendicular to the displacement. From the last statement and on the basis of the work kinetic energy theorem, we conclude that the kinetic energy of a charged particle moving through a magnetic field cannot be altered by the magnetic field alone. n other words, when a charged particle moves with a velocity v through a magnetic field, the field can alter the direction of the velocity vector but cannot change the speed or kinetic energy of the particle. From Equation 29.2, we see that the S unit of magnetic field is the newton per coulomb-meter per second, which is called the tesla (T): 1 T 1 N Cm/s The tesla ecause a coulomb per second is defined to be an ampere, we see that 1 T 1 N Am A non-s magnetic-field unit in common use, called the gauss (G), is related to the tesla through the conversion 1 T 10 4 G. Table 29.1 shows some typical values of magnetic fields. Quick Quiz 29.1 The north-pole end of a bar magnet is held near a positively charged piece of plastic. s the plastic (a) attracted, (b) repelled, or (c) unaffected by the magnet? Quick Quiz 29.2 A charged particle moves with velocity v in a magnetic field. The magnetic force on the particle is a maximum when v is (a) parallel to, (b) perpendicular to, (c) zero. Quick Quiz 29.3 An electron moves in the plane of this paper toward the top of the page. A magnetic field is also in the plane of the page and directed toward the right. The direction of the magnetic force on the electron is (a) toward the top of the page, (b) toward the bottom of the page, (c) toward the left edge of the page, (d) toward the right edge of the page, (e) upward out of the page, (f) downward into the page. Table 29.1 Some Approximate Magnetic Field Magnitudes Source of Field Field Magnitude (T) Strong superconducting laboratory magnet 30 Strong conventional laboratory magnet 2 Medical MR unit 1.5 ar magnet 10 2 Surface of the Sun 10 2 Surface of the Earth nside human brain (due to nerve impulses) 10 13

7 900 CHAPTER 29 Magnetic Fields Example 29.1 An Electron Moving in a Magnetic Field An electron in a television picture tube moves toward the front of the tube with a speed of m/s along the x axis (Fig. 29.5). Surrounding the neck of the tube are coils of wire that create a magnetic field of magnitude T, directed at an angle of 60 to the x axis and lying in the xy plane. (A) Calculate the magnetic force on the electron using Equation Solution Using Equation 29.2, we find the magnitude of the magnetic force: F q v sin ( C)( m/s)(0.025 T)(sin 60) N ecause v is in the positive z direction (from the righthand rule) and the charge is negative, F is in the negative z direction. () Find a vector expression for the magnetic force on the electron using Equation Solution We begin by writing a vector expression for the velocity of the electron: v ( î) m/s and one for the magnetic field: (0.025 cos 60 î sin 60 ĵ)t (0.013 î ĵ)t The force on the electron, using Equation 29.1, is F q v v z e 60 y (e)[( î) m/s] [(0.013 î ĵ )T] (e)[( î) m/s] [(0.013 î)t] (e)[( î) m/s] [(0.022 ĵ )T] (e)( m/s)(0.013 T)(î î ) (e)( m/s)(0.022 T)(î ĵ) ( C)( m/s)(0.022 T) ˆk where we have used Equations 11.7a and 11.7b to evaluate î î and î ĵ. Carrying out the multiplication, we find, x F ( N)ˆk F Figure 29.5 (Example 29.1) The magnetic force F acting on the electron is in the negative z direction when v and lie in the xy plane. This expression agrees with the result in part (A). The magnitude is the same as we found there, and the force vector is in the negative z direction Magnetic Force Acting on a Current-Carrying Conductor f a magnetic force is exerted on a single charged particle when the particle moves through a magnetic field, it should not surprise you that a current-carrying wire also experiences a force when placed in a magnetic field. This follows from the fact that the current is a collection of many charged particles in motion; hence, the resultant force exerted by the field on the wire is the vector sum of the individual forces exerted on all the charged particles making up the current. The force exerted on the particles is transmitted to the wire when the particles collide with the atoms making up the wire. efore we continue our discussion, some explanation of the notation used in this book is in order. To indicate the direction of in illustrations, we sometimes present perspective views, such as those in Figure f lies in the plane of the page or is present in a perspective drawing, we use blue vectors or blue field lines with arrowheads. n non-perspective illustrations, we depict a magnetic field

8 SECTON 29.2 Magnetic Force Acting on a Current-Carrying Conductor 901 perpendicular to and directed out of the page with a series of blue dots, which represent the tips of arrows coming toward you (see Fig. 29.6a). n this case, we label the field out. f is directed perpendicularly into the page, we use blue crosses, which represent the feathered tails of arrows fired away from you, as in Figure 29.6b. n this case, we label the field in, where the subscript in indicates into the page. The same notation with crosses and dots is also used for other quantities that might be perpendicular to the page, such as forces and current directions. One can demonstrate the magnetic force acting on a current-carrying conductor by hanging a wire between the poles of a magnet, as shown in Figure 29.7a. For ease in visualization, part of the horseshoe magnet in part (a) is removed to show the end face of the south pole in parts (b), (c), and (d) of Figure The magnetic field is directed into the page and covers the region within the shaded squares. When the current in the wire is zero, the wire remains vertical, as shown in Figure 29.7b. However, when the wire carries a current directed upward, as shown in Figure 29.7c, the wire deflects to the left. f we reverse the current, as shown in Figure 29.7d, the wire deflects to the right. Let us quantify this discussion by considering a straight segment of wire of length L and cross-sectional area A, carrying a current in a uniform magnetic field, as shown in Figure The magnetic force exerted on a charge q moving with a drift velocity v d is q v d. To find the total force acting on the wire, we multiply the force q v d exerted on one charge by the number of charges in the segment. ecause the volume of the segment is AL, the number of charges in the segment is nal, where n is the number of charges per unit volume. Hence, the total magnetic force on the wire of length L is F (q v d )nal We can write this expression in a more convenient form by noting that, from Equation 27.4, the current in the wire is nqv d A. Therefore, F L (29.3) where L is a vector that points in the direction of the current and has a magnitude equal to the length L of the segment. Note that this expression applies only to a straight segment of wire in a uniform magnetic field. out of page: (a) into page: (b) Figure 29.6 (a) Magnetic field lines coming out of the paper are indicated by dots, representing the tips of arrows coming outward. (b) Magnetic field lines going into the paper are indicated by crosses, representing the feathers of arrows going inward. Force on a segment of current-carrying wire in a uniform magnetic field in = 0 in in F q in v d L A (a) (b) Figure 29.7 (a) A wire suspended vertically between the poles of a magnet. (b) The setup shown in part (a) as seen looking at the south pole of the magnet, so that the magnetic field (blue crosses) is directed into the page. When there is no current in the wire, it remains vertical. (c) When the current is upward, the wire deflects to the left. (d) When the current is downward, the wire deflects to the right. (c) (d) Figure 29.8 A segment of a current-carrying wire in a magnetic field. The magnetic force exerted on each charge making up the current is q v d and the net force on the segment of length L is L.

9 902 CHAPTER 29 Magnetic Fields ds Figure 29.9 A wire segment of arbitrary shape carrying a current in a magnetic field experiences a magnetic force. The magnetic force on any segment ds is ds and is directed out of the page. You should use the right-hand rule to confirm this force direction. Now consider an arbitrarily shaped wire segment of uniform cross section in a magnetic field, as shown in Figure t follows from Equation 29.3 that the magnetic force exerted on a small segment of vector length ds in the presence of a field is d F ds (29.4) where d F is directed out of the page for the directions of and ds in Figure We can consider Equation 29.4 as an alternative definition of. That is, we can define the magnetic field in terms of a measurable force exerted on a current element, where the force is a maximum when is perpendicular to the element and zero when is parallel to the element. To calculate the total force F acting on the wire shown in Figure 29.9, we integrate Equation 29.4 over the length of the wire: F b d s a (29.5) where a and b represent the end points of the wire. When this integration is carried out, the magnitude of the magnetic field and the direction the field makes with the vector ds may differ at different points. We now treat two interesting special cases involving Equation n both cases, the magnetic field is assumed to be uniform in magnitude and direction. Case 1. A curved wire carries a current and is located in a uniform magnetic field, as shown in Figure 29.10a. ecause the field is uniform, we can take outside the integral in Equation 29.5, and we obtain b a F b (29.6) ut the quantity ds represents the vector sum of all the length elements from a to b. From the law of vector addition, the sum equals the vector L, directed from a to b. Therefore, Equation 29.6 reduces to F L (29.7) From this we conclude that the magnetic force on a curved current-carrying wire in a uniform magnetic field is equal to that on a straight wire connecting the end points and carrying the same current. a d s ds b L a (a) Figure (a) A curved wire carrying a current in a uniform magnetic field. The total magnetic force acting on the wire is equivalent to the force on a straight wire of length L running between the ends of the curved wire. (b) A current-carrying loop of arbitrary shape in a uniform magnetic field. The net magnetic force on the loop is zero. (b) ds

10 SECTON 29.2 Magnetic Force Acting on a Current-Carrying Conductor 903 Case 2. An arbitrarily shaped closed loop carrying a current is placed in a uniform magnetic field, as shown in Figure 29.10b. We can again express the magnetic force acting on the loop in the form of Equation 29.6, but this time we must take the vector sum of the length elements ds over the entire loop: F d s ecause the set of length elements forms a closed polygon, the vector sum must be zero. This follows from the procedure for adding vectors by the graphical method. ecause ds 0, we conclude that F 0; that is, the net magnetic force acting on any closed current loop in a uniform magnetic field is zero. Quick Quiz 29.4 The four wires shown in Figure all carry the same current from point A to point through the same magnetic field. n all four parts of the figure, the points A and are 10 cm apart. Rank the wires according to the magnitude of the magnetic force exerted on them, from greatest to least. A A 25 (a) (b) A 25 A (c) (d) Figure (Quick Quiz 29.4) Which wire experiences the greatest magnetic force? Quick Quiz 29.5 A wire carries current in the plane of this paper toward the top of the page. The wire experiences a magnetic force toward the right edge of the page. The direction of the magnetic field causing this force is (a) in the plane of the page and toward the left edge, (b) in the plane of the page and toward the bottom edge, (c) upward out of the page, (d) downward into the page. Example 29.2 Force on a Semicircular Conductor A wire bent into a semicircle of radius R forms a closed circuit and carries a current. The wire lies in the xy plane, and a uniform magnetic field is directed along the positive y axis, as shown in Figure Find the magnitude and direction of the magnetic force acting on the straight portion of the wire and on the curved portion. Solution The magnetic force F 1 acting on the straight portion has a magnitude F 1 L 2R because L 2R and

11 904 CHAPTER 29 Magnetic Fields R To find the magnetic force F 2 acting on the curved part, we use the results of Case 1. The magnetic force on the curved portion is the same as that on a straight wire of length 2R carrying current to the left. Thus, F 2 L 2R. The direction of F 2 is into the page based on the right-hand rule for the cross product L. ecause the wire lies in the xy plane, the two forces on the loop can be expressed as F 1 2R ˆk Figure (Example 29.2) The net magnetic force acting on a closed current loop in a uniform magnetic field is zero. n the setup shown here, the magnetic force on the straight portion of the loop is 2R and directed out of the page, and the magnetic force on the curved portion is 2R directed into the page. the wire is oriented perpendicular to. The direction of F 1 is out of the page based on the right-hand rule for the cross product L. F 2 2R ˆk The net magnetic force on the loop is F F 1 F 2 2R ˆ k 2R ˆk 0 Note that this is consistent with Case 2, because the wire forms a closed loop in a uniform magnetic field Torque on a Current Loop in a Uniform Magnetic Field (a) F 2 b b 2 a n the preceding section, we showed how a magnetic force is exerted on a currentcarrying conductor placed in a magnetic field. With this as a starting point, we now show that a torque is exerted on a current loop placed in a magnetic field. The results of this analysis will be of great value when we discuss motors in Chapter 31. Consider a rectangular loop carrying a current in the presence of a uniform magnetic field directed parallel to the plane of the loop, as shown in Figure 29.13a. No magnetic forces act on sides and because these wires are parallel to the field; hence, L 0 for these sides. However, magnetic forces do act on sides and because these sides are oriented perpendicular to the field. The magnitude of these forces is, from Equation 29.3, F 2 F 4 a (b) O F 4 Figure (a) Overhead view of a rectangular current loop in a uniform magnetic field. No magnetic forces are acting on sides and because these sides are parallel to. Forces are acting on sides and, however. (b) Edge view of the loop sighting down sides and shows that the magnetic forces F 2 and F 4 exerted on these sides create a torque that tends to twist the loop clockwise. The purple dot in the left circle represents current in wire coming toward you; the purple cross in the right circle represents current in wire moving away from you. The direction of F 2, the magnetic force exerted on wire, is out of the page in the view shown in Figure 29.13a, and that of F 4, the magnetic force exerted on wire, is into the page in the same view. f we view the loop from side and sight along sides and, we see the view shown in Figure 29.13b, and the two magnetic forces F 2 and F 4 are directed as shown. Note that the two forces point in opposite directions but are not directed along the same line of action. f the loop is pivoted so that it can rotate about point O, these two forces produce about O a torque that rotates the loop clockwise. The magnitude of this torque max is max F 2 b 2 F 4 b 2 (a) b 2 (a) b 2 where the moment arm about O is b/2 for each force. ecause the area enclosed by the loop is A ab, we can express the maximum torque as max A ab (29.8) This maximum-torque result is valid only when the magnetic field is parallel to the plane of the loop. The sense of the rotation is clockwise when viewed from side, as indicated in Figure 29.13b. f the current direction were reversed, the force directions would also reverse, and the rotational tendency would be counterclockwise.

12 SECTON 29.3 Torque on a Current Loop in a Uniform Magnetic Field 905 F 2 θ b 2 b sin θ 2 O A θ F 4 Active Figure An end view of the loop in Figure 29.13b rotated through an angle with respect to the magnetic field. f is at an angle with respect to vector A, which is perpendicular to the plane of the loop, the torque is A sin where the magnitude of A is A, the area of the loop. At the Active Figures link at you can choose the current in the loop, the magnetic field, and the initial orientation of the loop and observe the subsequent motion. Now suppose that the uniform magnetic field makes an angle 90 with a line perpendicular to the plane of the loop, as in Figure For convenience, we assume that is perpendicular to sides and. n this case, the magnetic forces F 1 and F 3 exerted on sides and cancel each other and produce no torque because they pass through a common origin. However, the magnetic forces F 2 and F 4 acting on sides and produce a torque about any point. Referring to the end view shown in Figure 29.14, we note that the moment arm of F 2 about the point O is equal to (b/2) sin. Likewise, the moment arm of F 4 about O is also (b/2) sin. ecause F 2 F 4 a, the magnitude of the net torque about O is µ A F 2 b 2 sin b F 4 2 sin a b 2 sin a b 2 sin ab sin A sin where A ab is the area of the loop. This result shows that the torque has its maximum value A when the field is perpendicular to the normal to the plane of the loop ( 90 ), as we saw when discussing Figure 29.13, and is zero when the field is parallel to the normal to the plane of the loop ( 0). A convenient expression for the torque exerted on a loop placed in a uniform magnetic field is A (29.9) where A, the vector shown in Figure 29.14, is perpendicular to the plane of the loop and has a magnitude equal to the area of the loop. We determine the direction of A using the right-hand rule described in Figure When you curl the fingers of your right hand in the direction of the current in the loop, your thumb points in the direction of A. As we see in Figure 29.14, the loop tends to rotate in the direction of decreasing values of (that is, such that the area vector A rotates toward the direction of the magnetic field). The product A is defined to be the magnetic dipole moment (often simply called the magnetic moment ) of the loop: A (29.10) The S unit of magnetic dipole moment is ampere-meter 2 (A m 2 ). Using this definition, we can express the torque exerted on a current-carrying loop in a magnetic field as (29.11) Note that this result is analogous to Equation 26.18, p E, for the torque exerted on an electric dipole in the presence of an electric field E, where p is the electric dipole moment. Figure Right-hand rule for determining the direction of the vector A. The direction of the magnetic moment is the same as the direction of A. Torque on a current loop in a magnetic field Magnetic dipole moment of a current loop Torque on a magnetic moment in a magnetic field

13 906 CHAPTER 29 Magnetic Fields Although we obtained the torque for a particular orientation of with respect to the loop, the equation is valid for any orientation. Furthermore, although we derived the torque expression for a rectangular loop, the result is valid for a loop of any shape. f a coil consists of N turns of wire, each carrying the same current and enclosing the same area, the total magnetic dipole moment of the coil is N times the magnetic dipole moment for one turn. The torque on an N-turn coil is N times that on a oneturn coil. Thus, we write N loop coil. n Section 26.6, we found that the potential energy of a system of an electric dipole in an electric field is given by U p E. This energy depends on the orientation of the dipole in the electric field. Likewise, the potential energy of a system of a magnetic dipole in a magnetic field depends on the orientation of the dipole in the magnetic field and is given by Potential energy of a system of a magnetic moment in a magnetic field U (29.12) From this expression, we see that the system has its lowest energy U min when points in the same direction as. The system has its highest energy U max when points in the direction opposite. Quick Quiz 29.6 Rank the magnitudes of the torques acting on the rectangular loops shown edge-on in Figure 29.16, from highest to lowest. All loops are identical and carry the same current. (a) (b) (c) Figure (Quick Quiz 29.6) Which current loop (seen edge-on) experiences the greatest torque? (Quick Quiz 29.7) Which current loop (seen edge-on) experiences the greatest net force? Quick Quiz 29.7 Rank the magnitudes of the net forces acting on the rectangular loops shown in Figure 29.16, from highest to lowest. All loops are identical and carry the same current. Example 29.3 The Magnetic Dipole Moment of a Coil A rectangular coil of dimensions 5.40 cm 8.50 cm consists of 25 turns of wire and carries a current of 15.0 ma. A T magnetic field is applied parallel to the plane of the loop. (A) Calculate the magnitude of its magnetic dipole moment. Solution ecause the coil has 25 turns, we modify Equation to obtain coil NA (25)( A)( m)( m) () What is the magnitude of the torque acting on the loop? Solution ecause is perpendicular to coil, Equation gives coil ( Am 2 )(0.350 T) Nm Am 2

14 SECTON 29.4 Motion of a Charged Particle in a Uniform Magnetic Field 907 Example 29.4 Satellite Attitude Control Many satellites use coils called torquers to adjust their orientation. These devices interact with the Earth s magnetic field to create a torque on the spacecraft in the x, y, or z direction. The major advantage of this type of attitude-control system is that it uses solar-generated electricity and so does not consume any thruster fuel. f a typical device has a magnetic dipole moment of 250 A m 2, what is the maximum torque applied to a satellite when its torquer is turned on at an altitude where the magnitude of the Earth s magnetic field is T? Solution We once again apply Equation 29.11, recognizing that the maximum torque is obtained when the magnetic dipole moment of the torquer is perpendicular to the Earth s magnetic field: max (250 Am 2 )( T) Nm Example 29.5 The D Arsonval Galvanometer An end view of a D Arsonval galvanometer (see Section 28.5) is shown in Figure When the turns of wire making up the coil carry a current, the magnetic field created by the magnet exerts on the coil a torque that turns it (along with its attached pointer) against the spring. Show that the angle of deflection of the pointer is directly proportional to the current in the coil. Solution We can use Equation to find the torque m that the magnetic field exerts on the coil. f we assume that the magnetic field through the coil is perpendicular to the normal to the plane of the coil, Equation becomes We can substitute this expression for in Equation (1) to obtain (NA) 0 NA Thus, the angle of deflection of the pointer is directly proportional to the current in the loop. The factor NA/ tells us that deflection also depends on the design of the meter. m (This is a reasonable assumption because the circular cross section of the magnet ensures radial magnetic field lines.) This magnetic torque is opposed by the torque due to the spring, which is given by the rotational version of Hooke s law, s, where is the torsional spring constant and is the angle through which the spring turns. ecause the coil does not have an angular acceleration when the pointer is at rest, the sum of these torques must be zero: N S (1) m s Equation allows us to relate the magnetic moment of the N turns of wire to the current through them: NA 0 Coil Spring Figure (Example 29.5) Structure of a moving-coil galvanometer Motion of a Charged Particle in a Uniform Magnetic Field n Section 29.1 we found that the magnetic force acting on a charged particle moving in a magnetic field is perpendicular to the velocity of the particle and that consequently the work done by the magnetic force on the particle is zero. Now consider the special case of a positively charged particle moving in a uniform magnetic field with the initial velocity vector of the particle perpendicular to the field. Let us assume that the direction of the magnetic field is into the page, as in Figure As the particle changes the direction of its velocity in response to the magnetic force, the magnetic force remains perpendicular to the velocity. As we found in Section 6.1, if the force is always perpendicular to the velocity, the path of

15 908 CHAPTER 29 Magnetic Fields v q F r v in q q F F Active Figure When the velocity of a charged particle is perpendicular to a uniform magnetic field, the particle moves in a circular path in a plane perpendicular to. The magnetic force F acting on the charge is always directed toward the center of the circle. At the Active Figures link at you can adjust the mass, speed, and charge of the particle and the magnitude of the magnetic field to observe the resulting circular motion. y q Helical path z x Active Figure A charged particle having a velocity vector that has a component parallel to a uniform magnetic field moves in a helical path. At the Active Figures link at you can adjust the x component of the velocity of the particle and observe the resulting helical motion. v the particle is a circle! Figure shows the particle moving in a circle in a plane perpendicular to the magnetic field. The particle moves in a circle because the magnetic force F is perpendicular to v and and has a constant magnitude qv. As Figure illustrates, the rotation is counterclockwise for a positive charge. f q were negative, the rotation would be clockwise. We can use Equation 6.1 to equate this magnetic force to the product of the particle mass and the centripetal acceleration: F ma c (29.13) That is, the radius of the path is proportional to the linear momentum mv of the particle and inversely proportional to the magnitude of the charge on the particle and to the magnitude of the magnetic field. The angular speed of the particle (from Eq ) is (29.14) The period of the motion (the time interval the particle requires to complete one revolution) is equal to the circumference of the circle divided by the linear speed of the particle: T 2r v F qv mv 2 r mv q v r q m 2 (29.15) These results show that the angular speed of the particle and the period of the circular motion do not depend on the linear speed of the particle or on the radius of the orbit. The angular speed is often referred to as the cyclotron frequency because charged particles circulate at this angular frequency in the type of accelerator called a cyclotron, which is discussed in Section f a charged particle moves in a uniform magnetic field with its velocity at some arbitrary angle with respect to, its path is a helix. For example, if the field is directed in the x direction, as shown in Figure 29.19, there is no component of force in the x direction. As a result, a x 0, and the x component of velocity remains constant. However, the magnetic force q v causes the components v y and v z to change in time, and the resulting motion is a helix whose axis is parallel to the magnetic field. The projection of the path onto the yz plane (viewed along the x axis) is a circle. (The projections of the path onto the xy and xz planes are sinusoids!) Equations to still apply provided that v is replaced by v 2 v y v z 2. r 2m q Quick Quiz 29.8 A charged particle is moving perpendicular to a magnetic field in a circle with a radius r. An identical particle enters the field, with v perpendicular to, but with a higher speed v than the first particle. Compared to the radius of the circle for the first particle, the radius of the circle for the second particle is (a) smaller (b) larger (c) equal in size. Quick Quiz 29.9 A charged particle is moving perpendicular to a magnetic field in a circle with a radius r. The magnitude of the magnetic field is increased. Compared to the initial radius of the circular path, the radius of the new path is (a) smaller (b) larger (c) equal in size.

16 SECTON 29.4 Motion of a Charged Particle in a Uniform Magnetic Field 909 Example 29.6 A Proton Moving Perpendicular to a Uniform Magnetic Field A proton is moving in a circular orbit of radius 14 cm in a uniform 0.35-T magnetic field perpendicular to the velocity of the proton. Find the linear speed of the proton. Solution From Equation 29.13, we have v qr ( C)(0.35 T)(0.14 m) m p kg m/s What f? What if an electron, rather than a proton, moves in a direction perpendicular to the same magnetic field with this same linear speed? Will the radius of its orbit be different? Answer An electron has a much smaller mass than a proton, so the magnetic force should be able to change its velocity much easier than for the proton. Thus, we should expect the radius to be smaller. Looking at Equation 29.13, we see that r is proportional to m with q,, and v the same for the electron as for the proton. Consequently, the radius will be smaller by the same factor as the ratio of masses m e /m p. Example 29.7 ending an Electron eam nteractive n an experiment designed to measure the magnitude of a uniform magnetic field, electrons are accelerated from rest through a potential difference of 350 V. The electrons travel along a curved path because of the magnetic force exerted on them, and the radius of the path is measured to be 7.5 cm. (Fig shows such a curved beam of electrons.) f the magnetic field is perpendicular to the beam, (A) what is the magnitude of the field? Solution Conceptualize the circular motion of the electrons with the help of Figures and We categorize this problem as one involving both uniform circular motion and a magnetic force. Looking at Equation 29.13, we see that we need the speed v of the electron if we are to find the magnetic field magnitude, and v is not given. Consequently, we must find the speed of the electron based on the potential difference through which it is accelerated. Therefore, we also categorize this as a problem in conservation of mechanical energy for an isolated system. To begin analyzing the problem, we find the electron speed. For the isolated electron electric field system, the loss of potential energy as the electron moves through the 350-V potential difference appears as an increase in the kinetic energy of the electron. ecause K i 0 and K f 1, we have 2 m e v 2 Now, using Equation 29.13, we find () K U 0 9: 1 2 m e v 2 (e) V 0 m e v e r 2e V 2( v m e 19 C)(350 V) kg m/s ( kg)( m/s) ( C)(0.075 m) T What is the angular speed of the electrons? Solution Using Equation 29.14, we find that Henry Leap and Jim Lehman Figure (Example 29.7) The bending of an electron beam in a magnetic field. v r m/s m rad/s To finalize this problem, note that the angular speed can be represented as ( rad/s)(1 rev/2 rad) rev/s. The electrons travel around the circle 24 million times per second! This is consistent with the very high speed that we found in part (A). What f? What if a sudden voltage surge causes the accelerating voltage to increase to 400 V? How does this affect the angular speed of the electrons, assuming that the magnetic field remains constant? Answer The increase in accelerating voltage V will cause the electrons to enter the magnetic field with a higher speed v. This will cause them to travel in a circle with a larger radius r. The angular speed is the ratio of v to r. oth v and r increase by the same factor, so that the effects cancel and the angular speed remains the same. Equation is an expression for the cyclotron frequency, which is the same as the angular speed of the electrons. The cyclotron frequency depends only on the charge q, the magnetic field, and the mass m e, none of which have changed. Thus, the voltage surge has no effect on the angular speed. (However, in reality, the voltage surge may also increase the magnetic field if the magnetic field is powered by the same source as the accelerating voltage. n this case, the angular speed will increase according to Equation ) At the nteractive Worked Example link at you can investigate the relationship between the radius of the circular path of the electrons and the magnetic field.

17 910 CHAPTER 29 Magnetic Fields S N Figure The Van Allen belts are made up of charged particles trapped by the Earth s nonuniform magnetic field. The magnetic field lines are in blue and the particle paths in red. Path of particle Figure A charged particle moving in a nonuniform magnetic field (a magnetic bottle) spirals about the field and oscillates between the end points. The magnetic force exerted on the particle near either end of the bottle has a component that causes the particle to spiral back toward the center. When charged particles move in a nonuniform magnetic field, the motion is complex. For example, in a magnetic field that is strong at the ends and weak in the middle, such as that shown in Figure 29.21, the particles can oscillate back and forth between two positions. A charged particle starting at one end spirals along the field lines until it reaches the other end, where it reverses its path and spirals back. This configuration is known as a magnetic bottle because charged particles can be trapped within it. The magnetic bottle has been used to confine a plasma, a gas consisting of ions and electrons. Such a plasma-confinement scheme could fulfill a crucial role in the control of nuclear fusion, a process that could supply us with an almost endless source of energy. Unfortunately, the magnetic bottle has its problems. f a large number of particles are trapped, collisions between them cause the particles to eventually leak from the system. The Van Allen radiation belts consist of charged particles (mostly electrons and protons) surrounding the Earth in doughnut-shaped regions (Fig ). The particles, trapped by the Earth s nonuniform magnetic field, spiral around the field lines from pole to pole, covering the distance in just a few seconds. These particles originate mainly from the Sun, but some come from stars and other heavenly objects. For this reason, the particles are called cosmic rays. Most cosmic rays are deflected by the Earth s magnetic field and never reach the atmosphere. However, some of the particles become trapped; it is these particles that make up the Van Allen belts. When the particles are located over the poles, they sometimes collide with atoms in the atmosphere, causing the atoms to emit visible light. Such collisions are the origin of the beautiful Aurora orealis, or Northern Lights, in the northern hemisphere and the Aurora Australis in the southern hemisphere. Auroras are usually confined to the polar regions because the Van Allen belts are nearest the Earth s surface there. Occasionally, though, solar activity causes larger numbers of charged particles to enter the belts and significantly distort the normal magnetic field lines associated with the Earth. n these situations an aurora can sometimes be seen at lower latitudes Applications nvolving Charged Particles Moving in a Magnetic Field A charge moving with a velocity v in the presence of both an electric field E and a magnetic field experiences both an electric force qe and a magnetic force q v. The total force (called the Lorentz force) acting on the charge is Lorentz force F q E q v (29.16)

18 SECTON 29.5 Applications nvolving Charged Particles Moving in a Magnetic Field 911 Source in E v Slit qv q qe (a) Active Figure (a) A velocity selector. When a positively charged particle is moving with velocity v in the presence of a magnetic field directed into the page and an electric field directed downward, it experiences a downward electric force qe and an upward magnetic force q v. (b) When these forces balance, the particle moves in a horizontal line through the fields. (b) At the Active Figures link at you can adjust the electric and magnetic fields to try to achieve straight line motion for the charge. Velocity Selector n many experiments involving moving charged particles, it is important that the particles all move with essentially the same velocity. This can be achieved by applying a combination of an electric field and a magnetic field oriented as shown in Figure A uniform electric field is directed vertically downward (in the plane of the page in Fig a), and a uniform magnetic field is applied in the direction perpendicular to the electric field (into the page in Fig a). f q is positive and the velocity v is to the right, the magnetic force q v is upward and the electric force q E is downward. When the magnitudes of the two fields are chosen so that qe qv, the particle moves in a straight horizontal line through the region of the fields. From the expression qe qv, we find that v E (29.17) Only those particles having speed v pass undeflected through the mutually perpendicular electric and magnetic fields. The magnetic force exerted on particles moving at speeds greater than this is stronger than the electric force, and the particles are deflected upward. Those moving at speeds less than this are deflected downward. The Mass Spectrometer A mass spectrometer separates ions according to their mass-to-charge ratio. n one version of this device, known as the ainbridge mass spectrometer, a beam of ions first passes through a velocity selector and then enters a second uniform magnetic field 0 that has the same direction as the magnetic field in the selector (Fig ). Upon entering the second magnetic field, the ions move in a semicircle of radius r before striking a detector array at P. f the ions are positively charged, the beam deflects upward, as Figure shows. f the ions are negatively charged, the beam deflects in Detector array P r E q v Velocity selector 0, in Active Figure A mass spectrometer. Positively charged particles are sent first through a velocity selector and then into a region where the magnetic field 0 causes the particles to move in a semicircular path and strike a detector array at P. At the Active Figures link at you can predict where particles will strike the detector array.