CHAPTER 29—THE MAGNETIC FIELD

ActivPhysics can help with these problems: Activities 13.4, 13.6, 13.7, 13.8

Section 29-2:—The Magnetic Force and Moving Charge

Problem1. (a) What is the minimum magnetic field needed to exert a 5 10 15.4 -N force on an electron moving at

2 1 107. ? m/s (b) What magnetic field strength would be required if the field were at 45° to the electron’s velocity?

Solution(a) From Equation 29-1b, B F e =vsin , which is a minimum when sin 1 (the magnetic field perpendicular to the

velocity). Thus, Bmin ( .4 ) ( . )( . ) . . 5 10 1 6 10 2 1 10 1 61 10 16 115 19 7 3 N C m/s T G.= (b) For °45 ,

B B B ° min minsin .= 45 2 22 7 G.

Problem2. An electron moving at right angles to a 0.10-T magnetic field experiences an acceleration of 6 0 1015. . m/s2 (a) What

is the electron’s speed? (b) By how much does its speed change in 1 10 9 ns ( s ) ?

Solution(a) If the magnetic force is the only one of significance acting in this problem, then F ma e B v sin . Thus, v = = ° ma eB sin ( . )( ) ( . )( . ) sin .42 911 10 6 10 1 6 10 0 1 90 3 1031 15 19 5 kg m/s C T m/s.2 (b) Since F v B» is perpendicular to v, the magnetic force on a charged particle changes its direction, but not its speed.

Problem3. What is the magnitude of the magnetic force on a proton moving at 2 5 105. m/s (a) at right angles; (b) at 30°;

(c) parallel to a magnetic field of 0 50. T?

SolutionFrom Equation 29-1b, F e B F ° v sin , , ( . )( . )( . ) . , so (a) when C m/s T N90 1 6 10 2 5 10 0 5 2 0 1019 5 14

(b) F F e B ° ° ( . ) sin . , sin .2 0 10 30 1 0 10 0 014 14 N N and (c) v

Problem4. A magnetic field of 0 10. T points in the x direction. A charged particle carrying 1 0. C enters the field region moving at 20 m/s. What are the magnitude and direction of the force on the particle when it first enters the field region if it does so moving (a) along the x axis; (b) along the y axis; (c) along the z axis; (d) at 45° to both x and y axes?

SolutionThe magnetic force is qv B v v v ( )( )( . ) , 1 20 0 1 2 C m/s T N where î î is a unit vector in the direction of the velocity of the particle when it first enters the field region. (a) If , . ( ) ,v v v î î b For 0 j the direction of the force

is j î k (along the negative z axis), while (c) . (cos sin ) ,k î î î î k ° ° j j (d) v 45 45 2= so the force is

( ) . 2 N k

Problem5. A particle carrying a 50- C charge moves with velocity v 5 0 3 2. . î k m/s through a uniform magnetic field

B 9 6 7.4 . î j T. (a) What is the force on the particle? (b) Form the dot products F v F B and to show explicitly that the force is perpendicular to both v and B.

2 CHAPTER 29

Solution(a) From Equation 27-2, F v B k q ( )( . ) ( .4 . ) ( )( . . .450 5 3 2 9 6 7 50 10 5 6 7 3 2 96 C m/s T N î î kj j

3 2 6 7 1 072 1 504 1 675 10 3. . ) ( . . . ) î î kj N. (The magnitude and direction can be found from the components, if desired.) (b) The dot products F v F B and are, respectively, proportional to ( . )( ) ( . )( . ) , 1 072 5 1 675 3 2 0 and ( . )( .4) ( . )( . ) , 1 072 9 1 504 6 7 0 since the cross product of two vectors is perpendicular to each factor. (We did not round off the components of F, so that the vanishing of the dot products could be exactly confirmed.)

Problem6. Moving in the x direction, a particle carrying 1 0. C experiences no force. Moving with speed v at 30° to the x axis,

the particle experiences a magnetic force of 2 0. N. What magnetic force would it experience if it moved along the y axis with speed v?

SolutionWe assume that the forces mentioned are due only to the presence of a magnetic field, which is uniform in the region of space considered. The first condition requires that the field be in the x direction, since F1 0 qvî B , implies that î î k k ( ) , .B B B B B B Bx y z y z y zj j 0 0 or The second condition requires that the magnitude of the field

satisfies the equation 2 30 42 N or N. ° F q B q Bv vsin , Therefore, the force on the particle, when its velocity is

v v v, ( ) j j is NF3 4 q B q Bî k k.

Problem7. A proton moving with velocity v1

43 6 10 . m/sj experiences a magnetic force of 7 10 16.4 N.î A second

proton moving on the x axis experiences a magnetic force of 2 8 10 16. N.j Find the magnitude and direction of the magnetic field, and the velocity of the second proton.

SolutionThe magnetic force on the first proton is ( .4 ) ( ) ( ) ( ),7 10 016

1 1 N so î î k k îe B B B e B B Bx y z x z xv vj j

and N7 10 161.4 . e Bzv The force on the second proton is ( . ) ( ) ( ) ( ),2 8 10 16

2 2 N j j je B B e B By z y zv vî k k

so B e B By z z 0 2 8 10 7 10 1 6 10 3 6 10162

16 19 4 and N Therefore, N C m/s. . ( .4 ) ( . )( . )v =B k k

( . ) ,0 128 T k and v2 24 42 8 7 3 6 10 136 10 v =î î î( . .4)( . ) ( . ) . m/s m/s

Problem8. The magnitude of Earth’s magnetic field is a little less than 1 G near Earth’s surface. What is the maximum possible

magnetic force on an electron with kinetic energy of 1 keV? Compare with the gravitational force on the same electron.

SolutionAn electron, moving perpendicularly to the Earth’s magnetic field near the surface, experiences a maximum magnetic

force of e B eB K mv = 2 1 6 10 10 2 10 1 6 10 911 10 3 1019 4 3 19 31 16/ ( . )( ) ( )( . ) ( . )' ' C T eV J/eV kg N. The

weight of an electron near the Earth’s surface is mg ( . )( . )911 10 9 8 9 1031 30 kg m/s N,2 ' a factor of nearly

3 10 14 times smaller.

Problem9. An alpha particle (2 protons, 2 neutrons) is moving with velocity v 150 320 190î k j km /s in a magnetic field

B 0 66 0. î .41 T.j Find the magnitude of the force on the particle.

SolutionThe magnetic force on the alpha particle is (Equation 29-1a): F v BB e 2 2 1 6 10 150 320 19019( . )( C î k)j

( )( . .41) ( . )( .41 . . .41 )10 0 66 0 3 2 10 150 0 320 0 66 190 0 66 190 03 16m/s T N î k k î j j

CHAPTER 1 3

( . . . )24 9 40 1 87 3î kj fN.

Problem10. How much charge would you need to put on a 2.7-g ping-pong ball so the magnetic force when it’s moving at 6 0. m/s

at right angles to Earth’s 44- T magnetic field is equal to its weight? Is this realistic?

SolutionWhen v B , the magnitude of the magnetic force is q Bv . This equals the weight of the ping-pong ball when q mg B =v =( . )( . ) ( )( )2 7 9 8 6 44 100 g m/s m/s T C,2 huge by ordinary standards.

Problem11. A 1.4- C charge moving at 185 m/s experiences a magnetic force FB 2 5 7 0. . î j N in a magnetic field

B 42 15î j mT. What is the angle between the particle’s velocity and the magnetic field?

SolutionEquation 29-1b gives sin ( . ) ( . ) ( .4 )( ) ( ) ( ) . . F q B=v 2 5 7 0 1 185 42 15 0 6442 2 2 2 N C m/s mT Then ° °40 1. or 140 (both are possible since sin sin( )). ° 180

Problem12. A velocity selector uses a 60-mT magnetic field and a 24 kN/C electric field. At what speed will charged particles

pass through the selector undeflected?

SolutionThe condition for zero deflection is v = = E B ( ) ( . )24 0 06 400 kN/C T km/s.

Problem13. A region contains an electric field E î 7 2 8.4 . /j kN C and a magnetic field B 15 36 j k mT. Find the

electromagnetic force on (a) a stationary proton, (b) an electron moving with velocity v 6 1. î Mm/s.

SolutionThe force on a moving charge is given by Equation 29-2 (called the Lorentz force) F E v B q( ). (a) For a stationary proton, q e e and so C kN/C (1.18 ) fN.v F E0 1 6 10 7 2 8 019, ( . )( .4 . ) .448î îj j (b) For the electron, q e and Mm/s,v 6 1. î so the electric force is the negative of the force in part (a) and the magnetic force is ev B ( . )( . ) ( ) . . 1 6 10 6 1 15 36 14 6 35119 C Mm/s mT ( ) fN.î k k j j The total Lorentz force is the sum of

these, or ( . . . ) 118 34 7 14 6î kj fN.

Problem14. A charged particle is moving at right angles to both a 11. kN/C electric field and a 0.75-T magnetic field. If the

magnitude of the electric force on the particle is twice that of the magnetic force, what is the particle’s speed?

SolutionUnder the conditions stated, F qE F q B E BE B 2 2 2 11 2 0 75 733v v = =, ( . ) ( . ) therefore kN/C T m/s.

Section 29-3:–The Motion of Charged Particles in Magnetic Fields

Problem15. What is the radius of the circular path described by a proton moving at 15 km/s in a plane perpendicular to a 400-G

magnetic field?

4 CHAPTER 29

SolutionFrom Equation 29-3, the radius of the orbit is r m eB v= =( . )( ) ( . )( )1 67 10 15 1 6 10 4 1027 19 2 kg km/s C T 3 91. mm. (SI units and data for the proton are summarized in the appendices and inside front cover.)

Problem16. How long does it take an electron to complete a circular orbit at right angles to a 1.0-G magnetic field?

SolutionEquation 29-4 gives a period of T m eB 2 2 911 10 1 6 10 10 35831 19 4 = ( . ) ( . )( ) kg C T ns.

Problem17. Radio astronomers detect electromagnetic radiation at a frequency of 42 MHz from an interstellar gas cloud. If this

radiation is caused by electrons spiraling in a magnetic field, what is the field strength in the gas cloud?

SolutionIf this is electromagnetic radiation at the electron’s cyclotron frequency, Equation 29-5 implies a field strength of B f m e 2 2 42 1 6 10 1 50 10 15 031 19 3 ( ) ( . . .= = MHz)(9.11 10 kg C) T G.

Problem18. A beam of electrons moving in the x direction at 8 7 106. m/s enters a region where a uniform magnetic field of

180 G points in the y direction. How far into the field region does the beam penetrate?

SolutionWhile moving perpendicularly to the magnetic field, the beam is bent into a horizontal circle of radius given by Equation 29-3, r m eB v= =( . )( . ) ( . )( . ) .911 10 8 7 10 1 6 10 0 018 2 7531 6 19 kg m/s C T mm. If the electrons enter normal to the boundary of the field region, r is also the penetration distance, and the beam will travel 180° in a circle before exiting the field region in a direction opposite to which it entered. (However, if the beam enters the field region making an angle with the normal to the boundary, then it penetrates a distance r( sin )1 into the region, travels 180 2° around the circle, and exits at angle on the other side of the normal, as shown, where ° °90 90 .)

Problem 18 Solution.

Problem19. Electrons and protons with the same kinetic energy are moving at right angles to a uniform magnetic field. How do

their orbital radii compare?

SolutionIt is convenient to anticipate the result of Problem 22 for the orbital radius of a non-relativistic charged particle in a plane perpendicular to a uniform magnetic field. From Equation 29-3, r m qB v= . For a non-relativistic particle, K m 1

22v , or

CHAPTER 1 5

v = = 2 2K m r Km qB, therefore . (Note: All quantities can, of course, be expressed in standard SI units, but in many applications, atomic units are more convenient. The conversion factor for electron volts to joules is just the numerical magnitude of the electronic charge, so if K is expressed in MeV, m in MeV =c q2, in multiples of e, and B in teslas, we obtain

rK e m e c

qe B

Km

qB

Km

qB

IKJJ

2 10 10 10 2

3 10

2

300

6 6 2 6

8

( ) ( )

( ) ( ).

=

From this expression, it follows that protons and electrons with the same kinetic energy have radii in the ratio r rp e= m mp e= 1836 43¼ , in the same magnetic field. Heavier particles are more difficult to bend.

Problem20. The Van Allen belts are regions in space where high-energy charged particles are trapped in Earth’s magnetic field. If

the field strength at the Van Allen belts is 0 10. G, what are the period and radius of the spiral path described (a) by a proton with a 1.0-MeV kinetic energy? (b) by a 10-MeV proton?

SolutionThe cyclotron frequency, f qB m =2 , is independent of the particle’s energy, so the orbital period for protons of either

energy is T f 1 2 1 67 10 1 6 10 10 6 5627 19 5= = ( . . ) ( ) . kg C T ms. The radius is given by the result of

Problem 22 (in atomic units, explained in the solution of Problem 19). (a) For a 1 2 300 MeV proton, r Km qB =

2 1 938 300 1 10 145( )( ) ( )( ) .4= km. (b) Since r K» , the orbital radius for a proton with ten times the kinetic energy is

10 14 45 7( .4 ) . km km.

Problem21. Microwaves in a microwave oven are produced by electrons circling in a magnetic field at a frequency of 2.4 GHz.

(a) What is the magnetic field strength? (b) The electron’s motion takes place inside a special tube called a magnetron. If the magnetron can accommodate electron orbits with a maximum diameter of 2 5. mm, what is the maximum electron energy?

Solution(a) A cyclotron frequency of 2.4 GHz for electrons implies a magnetic field strength of B f m e 2 2 2 ( ) ( .4 )= GHz

( . . ) .911 10 1 6 10 85 931 19 kg C mT= (see Equation 29-5). (b) The kinetic energy of an electron, with the maximum

orbital radius allowed for this magnetron tube (half the diameter), in the field found in part (a), is K reB m ( )2 2=

( . . . ) ( . ) ( . ) ( . ) .1 25 1 6 10 85 9 2 9 11 10 1 62 10 1 6 10 1 0119 2 31 16 19 mm C mT kg J J eV keV = = = (see Problem 22).

The same calculation in atomic units, explaioed in the solution to Problem 19, is ( . . )1 25 10 300 85 9 103 3 2

( . )2 0 511 1 MeV. The electron’s kinetic energy could also be expressed in terms of the cyclotron frequency directly,

K frm m m rf ( ) ( ) ,2 2 22 2 = with the same result.

Problem22. Show that the orbital radius of a charged particle moving at right angles to a magnetic field B can be written

rKm

qB

2,

where K is the kinetic energy in joules, m the particle mass, and q its charge.

SolutionSee solution to Problem 19.

Problem23. Two protons, moving in a plane perpendicular to a uniform magnetic field of 500 G, undergo an elastic head-on

6 CHAPTER 29

collision. How much time elapses before they collide again? Hint: Draw a picture.

SolutionIn an elastic head on collision between particles of equal mass, the particles exchange velocities ( ,v v v v1 2 2 1f i f i and see Section 11-4). Moving in a plane perpendicular to B, each proton describes a different circle of radius r m eB v= , with

period (which is independent of r and v) of T m eB 2 2 1 67 10 1 6 10 5 10 13127 19 2 = =( . ) ( . ) ( ) . kg C T s. After one period, each proton would be back at the site of the collision, and could collide again.

Problem24. Repeat the preceding problem for the case of a proton and an antiproton colliding head-on (a) if they have the same

speed and (b) if they have different speeds. (An antiproton has the same mass as a proton, but carries the opposite charge.)

SolutionA proton and an antiproton have the same cyclotron frequency and would also return to the same point after one period (i.e., 131. s, see preceding solution). (b) If they have different speeds, and therefore different radii, they would not collide at an intervening time. (a) However, if they had the same speed, they would circulate in opposite directions around the same circular orbit, and hence would collide again after half a period, or 0 656. s.

Problem25. A cyclotron is designed to accelerate deuterium nuclei. (Deuterium has one proton and one neutron in its nucleus.)

(a) If the cyclotron uses a 2.0-T magnetic field, at what frequency should the dee voltage be alternated? (b) If the vacuum chamber has a diameter of 0 90. m, what is the maximum kinetic energy of the deuterons? (c) If the magnitude of the potential difference between the dees is 1500 V, how many orbits do the deuterons complete before achieving the energy of part (b)?

Solution(a) The frequency of the accelerating voltage is the cyclotron frequency for deuterons (Equation 29-5), f eB m =2 '

( . )( ) ( . ) .1 6 10 2 2 2 1 67 10 15 219 27 C T kg MHz. (b) We can use the result of Problem 19 (expressed in atomic

units), with the maximum orbital radius equal to the radius of the dees. Thus, K qBr mmax max( ) 300 22= '

( .45) ( ) .4300 1 2 0 2 2 938 192 = MeV. (c) If the deuterons start with essentially zero kinetic energy, and gain

1500 eV each half-orbit, they will make 19.4 MeV 2(1500 eV) 6.48 103= orbits. (Of course, the same results follow in standard SI units.)

Problem26. Without changing the magnetic field, how could the cyclotron of the preceding problem be modified to accelerate

(a) protons and (b) alpha particles (two protons and two neutrons)? What would be the maximum energy achievable with (c) protons and (d) alpha particles?

SolutionThe cyclotron frequency is proportional to q m= . Compared to deuterons, protons have ( ) ( ) ,q m e m e m q mp p p d= = = = 2 2 2¼

and -particles have ( ) ( ) .q m e m q mp d= = = ¼ ¼2 4 Therefore, the frequency of the accelerating voltage should be

(a) doubled for protons, and (b) left unchanged for ’s. The maximum kinetic energy is proportional to q m2= (other

cyclotron design parameters constant). Now ( ) ( ) ( ) ( ) , maxq m e m e m q m q m Kp p p d2 2 2 2 22 4 2= = = = = ¼ ' so will be

doubled for both (c) protons and (d) ’s compared to deuterons.

Problem27. Figure 29-38 shows a simple mass spectrometer, designed to analyze and separate atomic and molecular ions with

different charge-to-mass ratios. In the design shown, ions are accelerated through a potential difference V, after which they enter a region containing a uniform magnetic field. They describe semicircular paths in the magnetic field, and land on a detector a lateral distance x from where they entered the field region, as shown. Show that x is given by

CHAPTER 1 7

xB

V

q m

2 2

( ),

=

where B is the magnetic field strength, V the accelerating potential, and q m= the charge-to-mass ratio of the ion. By counting the number of ions accumulated at different positions x, one can determine the relative abundances of different atomic or molecular species in a sample.

FIGURE 29-38 Problem 27 Solution.

SolutionThe positive ions enter the field region with speed (determined from the work-energy theorem) of 1

22m qVv , or

v = 2V q m( ). They are bent into a semicircle with diameter x r m qB m qB V q m m q V B 2 2 2 2 2 2v= = = = =( ) ( ) ( ) , as shown in Fig. 29-38 (see Equation 29-3).

Problem28. A mass spectrometer like that of the preceding problem has V B 2000 1000 V and G. It is used to analyze a gas

sample suspected of containing Ne, O CO SO , and NO2 2 2, , . Ions are detected at distances of 58 cm, 68 cm, and 87 cm from the entrance to the field region. Which gases are actually present? Assume that all molecules are singly ionized.

SolutionIf we express the mass of the ions in atomic mass units ( . ),u 1 66 10 27 kg and use the given spectrometer parameters, then the equation for the distances given in Problem 27 becomes

xB

V

q m

V

2 2 2

0 1

2 2000

1 6 10 1 66 1012 919 27( ) .

( )

( . . )( . ) .

= = T

C kg cm

MM

The molecular masses for the ions listed are approximately M 20 2. , 32.0, 28.0, 64.1, and 46.0, respectively. The corresponding distances are x 57 9 72 9. , . , 68.2, 103.2, and 87.4 cm, so the presence of Ne CO and NO+ +

2+, , is

indicated.

Problem29. A mass spectrometer is used to separate the fissionable uranium isotope U-235 from the much more abundant isotope

U-238. To within what percentage must the magnetic field be held constant if there is to be no overlap of these two isotopes? Both isotopes appear as constituents of uranium hexafluoride gas (UF6), and the gas molecules are all singly ionized.

SolutionThe separation of different uranium isotopes in UF6 molecules can be found by differentiation of the result of Problem 27. Keeping the spectrometer parameters fixed, x m dx m dm dx x dm m» »1 2 1

21 2 1

2= = = =, , ( ). and The molecular masses of

the two species are approximately 235 or 238 plus 6 19 which equals 349 or 352, respectively, so 12 3 2 350( )dm m= =¼ 0.43%. For a particular ion, x B dx B dB dx x dB B» » 1 2 and therefore , .= = Thus, variations in B should be less than 0.43% to separate these isotopes.

8 CHAPTER 29

Problem30. An electron is moving in a uniform magnetic field of 0 25. T; its velocity components parallel and perpendicular to the

field are both equal to 31 106. m/s. (a) What is the radius of the electron’s spiral path? (b) How far does it move along the field direction in the time it takes to complete a full orbit about the field direction?

Solution(a) The radius depends only on the perpendicular velocity component, r m eB

v = ( . )( . )911 10 31 1031 6 kg m/s

( . )( . ) .1 6 10 0 25 70 619 C T m. (b) The distance moved parallel to the field is d T Tv|| , where is the cyclotron

period (Equation 29-4). Since v v|| in this case, d m eB v =||( ) ( . )2 2 70 6 444 m m.

Problem31. An electron moving at 3 8 106. m/s enters a region containing a uniform magnetic field B 18 k mT. The electron is

moving at 70° to the field direction, as shown in Fig. 29-39. Find the radius r and pitch p of its spiral path, as indicated in the figure.

FIGURE 29-39 Problem 31.

SolutionThe parallel and perpendicular components of the electron’s velocity (relative to the field direction, or z axis) are v v v v v =|| cos ( . ) cos . sin . . ° ° ° 70 3 8 10 70 130 70 3 57 1136 m/s Mm/s, and Mm/s. Then mmr m eB and

the pitch p T m eB v v =|| ||( ) .2 2 58 mm.

Problem32. A proton in interstellar space describes a spiral path about a 15-mG magnetic field, with velocity component 40 km/s

perpendicular to the field. If the pitch of the helix (see Fig. 29-39) is 8 7. km, what is the proton’s velocity component parallel to the field?

SolutionUsing the expressions for radius and pitch from the previous problem, we find v = =|| . . p T 8 7 43 7 199 km ms km/s.

Section 29-4:—The Magnetic Force on a Current

Problem33. What is the magnitude of the force on a 50-cm-long wire carrying 15 A at right angles to a 500-G magnetic field?

SolutionThe force on a straight current-carrying wire in a uniform magnetic field is (Equation 29-6) F B I . Thus, F I B ° sin ( )( . )( . ) sin . 15 0 5 0 05 90 0 375 A m T N. (The direction is given by the right-hand rule.)

Problem34. A wire coincides with the x axis, carrying 2.4 A in the x direction. The wire passes through a region containing a

CHAPTER 1 9

uniform magnetic field B 0 17 0 32 0 2. . . î kj T. Find a vector expression for the force per unit length on the wire in the magnetic field region.

SolutionA unit length of the current carrying wire is described by I= ( .4 ) ,2 A î on which the magnetic force (Equation 29-6) is

F A ) T N/m.= = ( ) ( .4 ) ( . . . ( . . )I B 2 0 17 0 32 0 21 0 504 0 768î î k kj j

Problem35. A wire carrying 15 A makes a 25° angle with a uniform magnetic field. The magnetic force per unit length of wire is

0 31. N/m. (a) What is the magnetic field strength? (b) What is the maximum force per unit length that could be achieved by reorienting the wire in this field?

SolutionEquation 29-6 gives the magnetic force on a straight current carrying wire in a uniform magnetic field, F B I . (a) From the magnitude of F and the given data, we find B F I ° = =sin ( . ) ( ) sin . 0 31 15 25 48 9 N/m A mT. (b) By placing the wire perpendicular to the field (sin ) 1 a maximum force per unit length of IB ( )( . )15 48 9 A mT 0 734. N/m could be attained/

Problem36. A wire of negligible resistance is bent into a rectangle as shown in Fig. 29-40, and a battery and resistor are connected

as shown. The right-hand side of the circuit extends into a region containing a uniform magnetic field of 38 mT pointing into the page. Find the magnitude and direction of the net force on the circuit.

FIGURE 29-40 Problem 36 Solution.

SolutionThe forces on the upper and lower horizontal parts of the circuit are equal in magnitude, but opposite in direction and thus cancel (see Fig. 29-40), leaving the force on the righthand wire, I B R B ( ) ( )( . )( ) .E= =12 3 0 1 38 15 2 V m mT mN toward the right, as the net force on the circuit.

Problem37. In a high-magnetic-field experiment, a conducting bar carrying 7 5. kA passes through a 30-cm-long region containing

a 22-T magnetic field. If the bar makes a 60° angle with the field direction, what force is necessary to hold it in place?

SolutionThe magnitude of the force necessary to balance the magnetic force on the bar (Equation 29-6) is F I B I B sin ( . )( . )( ) sin . ° 7 5 0 3 22 60 42 9 kA m T kN (nearly 5 tons). The direction of this force is perpendicular to the plane of B in the opposite sense as the magnetic force.

Problem38. A 20-cm-long conducting rod with mass 18 g is suspended by wires of negligible mass, as shown in Fig. 29-41. The

rod is in a region containing a uniform magnetic field of 0 15. T pointing horizontally into the page, as shown. An external circuit supplies current between the support points A and B. (a) What is the minimum current necessary to move the bar to the upper position shown? (b) Which direction should the current flow?

10 CHAPTER 29

FIGURE 29-41 Problem 38 Solution.

SolutionAn upward magnetic force on the rod equal (in magnitude) to its weight is the minimum force necessary. (a) Since the rod is perpendicular to B I B mg I mg B, ( . . ) ( . . ) . implies N T m A.= =0 018 9 8 0 2 0 15 5 88 (b) The force is upward for current flowing from A to B, consistent with the right hand rule for the cross product.

Problem39. A piece of wire with mass per unit length 75 g/m runs horizontally at right angles to a horizontal magnetic field. A

6.2-A current in the wire results in its being suspended against gravity. What is the magnetic field strength?

SolutionA magnetic force equal in magnitude to the weight of the wire requires that I B mg (since the wire is perpendicular to the field), or B m g I ( )( ) ( )( . ) ( . ) .= = = 75 9 8 6 2 0 119 g/m m/s A T.2

Problem40. A nonuniform magnetic field points out of the page, as shown in Fig. 29-42. The field strength increases at the rate of

2 0. / mT cm as you move to the right. A square wire loop 15 cm on a side lies in a plane perpendicular to the field, and a 2.5-A current circles the loop in the counterclockwise direction. What are the magnitude and direction of the net magnetic force on the loop?

SolutionThe force on the loop is given by Equation 29-8, with the constant current factor outside the integral, and the range of integration split into segments of the square: F B I d( ) .z z z z bot right top left Take the x axis to the right, the y axis up, and the z axis out of the page, as shown in Fig. 29-41. The magnitude of B only depends on x, and its direction is constant, so B B x B x( ) [ ( . ) ] .k k0 0 2 T/m Consideration of symmetrical pairs of path elements, d d and at the same value of x, shows that the net force on the bottom and top segments exactly cancels. The field is constant over the right or left segments, so if we let right left ,j and substitute for B x( ), the force becomes

F B B

I I

I B x B x

I x x

r

r

( ) ( )

[ ( . ) ( . ) ]

( . )( ) ( . )( . ) ( . )

( . ) .

right left

T/m T/m

T/m A m T/m

N

j k

î î

î

0 0

2

2

0 2 0 2

0 2 2 5 0 15 0 2

113 10

CHAPTER 1 11

FIGURE 29-42 Problem 40 Solution.

Problem41. A wire carrying 1 5. A passes through a region containing a 48-mT magnetic field. The wire is perpendicular to the

field and makes a quarter-circle turn of radius 21 cm as it passes through the field region, as shown in Fig. 29-43. Find the magnitude and direction of the magnetic force on this section of wire.

SolutionTake the x-y axes as shown on Fig. 29-43, with the z axis out of the page and the origin at the center of the quarter-circle arc. With measured clockwise from the y axis, d R d ( cos sin )î j as shown, and B B( ).k The magnetic force on the arc of wire is found from Equation 29-8.

F B

I d I R d B

IRB d

IRB IRB

arcz zz

°

°

°

0

90

0

90

0

90

( cos sin ) ( )

( cos sin )

cos sin ( ).

î k

î

î î

j

j

j j

This has magnitude 2 2 1 5 0 21 48 21 A m mT mNIRB ( . )( . )( ) .4 and direction 45° between the positive x and y axes.

FIGURE 29-43 Problem 41 Solution.

Problem42. A wire coincides with the x axis, and carries a current I x 2 0. A in the direction. A nonuniform magnetic field

points in the y direction, given by B B x x0 02( ) ,= j where B x0 00 22 1 0 . . T, m, and x is the x coordinate. Find

the force on the section of wire between x x 1 0 3 5. . m and m.

SolutionEquation 29-8 gives, with d dx î ,

F B FHGIKJ

z z zwire wire m

m

m

m

2

A T/m m m N

Id I dx Bx

xIB

x dx

xIB

x

x ( )

( )( . )

[( . ) ( . ) ] . .

.

.

.

.

î k k

kk

00

2

01 0

3 5 2

02 0

3

02

1 0

3 5

3 3

3

2 0 223

3 5 1 0 6 14

j

Problem43. Apply Equation 29-8 to a closed current loop of arbitrary shape in a uniform magnetic field, and show that the net

force on the loop is zero. Hint: Both I and B are constant as you go around the loop, so you can take them out of the integral. What is the remaining vector integral?

Solution

If both I and B are constants, Equation 29-8, for a closed loop, may be written as F B B Id I d zz ( ) . But the

12 CHAPTER 29

integral is the sum of vectors, d, beginning and ending at the same point, hence d z 0. (This integral relation is a

special case of Equation 8-1 for a constant F.)

Problem44. A rectangular copper strip measures 1 0. mm in the direction of a uniform 2.4-T magnetic field. When the strip carries

a 6.8-A current at right angles to the field, the Hall voltage across the strip is 1 2. V. Find the number density of free electrons in the copper.

SolutionThe geometry in this problem is the same as that in the discussion leading to Equation 29-7, which shows that n IB qV tH = =( . )( .4 ) ( . )( . )( ) .6 8 2 1 6 10 1 2 1 8 50 1019 28 3 A T C V mm m is the number density if conduction electrons are the charge-carriers.

Problem45. The probe in a Hall-effect magnetometer uses a semiconductor doped to a charge-carrier density of 7 5 10 20 3. . m

The probe measures 0 35. mm thick in the direction of the magnetic field being measured, and carries a 2.5-mA current perpendicular to the field. If its Hall potential is 4 5. mV, what is the magnetic field strength?

SolutionIf we assume the charge-carriers are of one type, with charge of magnitude e, then Equation 29-7 and the given data require B nqV t IH = =( . )( . )( . )( . ) ( . ) .7 5 10 1 6 10 4 5 0 35 2 5 75 620 3 19 m C V mm mA mT.

Section 29-5:–A Current Loop in a Magnetic Field

Problem46. Earth has a magnetic dipole moment associated with currents flowing in the planet’s liquid outer core. Suppose that

current flowed in a single loop at the outer edge of the liquid core (radius 3000 km). What current would be needed to give the observed dipole moment of 8 0 1022. ? A m2 (The actual current structure is more complex than a single loop.)

SolutionThe dipole moment of a circular loop is I R2 (Equation 29-10), so I ( . ) ( )8 0 10 3 1022 2 6 2A m m=

2 83 109. A.

Problem47. A single-turn square wire loop 5 0. cm on a side carries a 450-mA current. (a) What is the magnetic moment of the

loop? (b) If the loop is in a uniform 1.4-T magnetic field with its dipole moment vector at 40 ° to the field direction, what is the magnitude of the torque it experiences?

Solution(a) Equation 29-10 for the magnetic moment of a loop gives NIA mA cm A m2( )( )( ) . .1 450 5 113 102 3

(b) Equation 29-11 gives the torque on a magnetic dipole moment in a uniform magnetic field, B B sin

( . )( .4 ) sin .113 10 1 40 1 01 103 3 ° A m T N m.2

Problem48. An electric motor contains a 250-turn circular coil 6 2. cm in diameter. If it is to develop a maximum torque of

1 2. N m at a current of 3 3. A, what should be the magnetic field strength?

SolutionThe maximum torque on a plane circular coil follows from Equations 29-10 and 11, max . B NI R B2 In this case,

B ( . ) ( )( . ) ( . )1 2 250 3 3 31 4822 N m A cm mT.=

CHAPTER 1 13

Problem49. A bar magnet experiences a 12-mN m torque when it is oriented at 55° to a 100-mT magnetic field. What is the

magnitude of its magnetic dipole moment?

SolutionEquation 29-11, solved for the magnitude of the dipole moment, gives ° = =B sin ( ) ( . ) sin12 10 0 1 553 N m T

0 146. . A m2

Problem50. A single-turn wire loop 10 cm in diameter carries a 12-A current. It experiences a torque of 0 015. N m when the

normal to the loop plane makes a 25° angle with a uniform magnetic field. What is the magnetic field strength?

SolutionEquations 29-10 and 11 yield a field of magnitude B I R ° = =2 20 015 12 5 5 377sin ( . ) ( ) ( ) sin N m A cm mT.

Problem51. A simple electric motor like that of Fig. 29-36 consists of a 100-turn coil 3 0. cm in diameter, mounted between the

poles of a magnet that produces a 0.12-T field. When a 5.0-A current flows in the coil, what are (a) its magnetic dipole moment and (b) the maximum torque developed by the motor?

Solution(a) From Equation 29-10, the magnetic moment of the coil has magnitude NIA A m100 5 0 031

42( ) ( . )

0 353. . A m2 The direction of , determined from the right-hand rule (see Fig. 29-33), rotates with the coil. (b) The

maximum torque (from Equation 29-11, with sin ) 1 is max ( . )( . ) . B 0 353 0 12 4 24 10 2 A m T N m.2

Problem52. A satellite with rotational inertia 20 kg m2 is in orbit at a height where Earth’s magnetic field strength is 0 18. G. It

has a magnetic torquing system, as described in Example 29-7, that uses a 1000-turn coil 30 cm in diameter. What should be the current in the coil if the magnetic torque is to give the satellite a maximum angular acceleration of 0 0015 2. ? s

SolutionIf we require that the maximum torque from the coil, max B NIAB (from Equations 27-10 and 11), should yield the desired angular acceleration, max ,=Irot then the current necessary is I NAB I NAB max= =rot

( . )( ) ( ) ( . ) ( . ) .1 5 10 20 10 0 3 18 10 23 63 2 3 14

2 5 s kg m m T A.2

Problem53. Nuclear magnetic resonance (NMR) is a technique for analyzing chemical structures and is also the basis of magnetic

resonance imaging used for medical diagnosis. The NMR technique relies on sensitive measurements of the energy needed to flip atomic nuclei upside-down in a given magnetic field. In an NMR apparatus with a 7.0-T magnetic field, how much energy is needed to flip a proton ( .41 ) 1 10 26 A m2 from parallel to antiparallel to the field?

SolutionFrom Equatioo 29-12, the energy required to reverse the orientation of a proton’s magnetic moment from parallel to antiparallel to the applied magnetic field is U B 2 2 1 10 7 0 1 97 1026 25 6 ( .41 )( . ) . A m T J 1.23 10 eV.2 (This amount of energy is characteristic of radio waves of frequency 298 MHz, see Chapter 39.)

Problem54. A wire of length carries a current I. (a) Find an expression for the magnetic dipole moment that results when the

wire is wound into an N-turn circular coil. (b) For what integer value of N is this moment a maximum?

14 CHAPTER 29

Solutioo(a) The number of turns of radius r that can be formed from a wire of length is N r r N = =2 2 , . so The magnitude

of the magnetic dipole moment of such a coil is NI N I N( ) . = =2 42 2 (b) This is clearly a maximum when N is a minimum, and the smallest value of N is, of course, one.

Paired Problems

Problem55. Find the magnetic force on an electron moving with velocity v 8 6 10 4 1 105 5. . î j m/s in a magnetic field

B 0 18 0 64. . j k T.

SolutionFrom Equation 29-1a, F v B e ( . )( . . )( ) ( . . ) ( )( . .1 6 10 8 6 4 1 10 0 18 0 64 16 4 1 0 6419 5 C m/s T f Nî k îj j

8 6 0 64 8 6 0 18 42 0 881 24 8. . . . ) ( . . . ) j jk î k fN. (If necessary, review the cross product of unit vectors; see Fig. 13-9 and use the right hand rule.)

Problem56. A proton moving with velocity v 2 0 10 4 0 105 5. . î j m/s experiences a magnetic force

F 10 5 0 21î k . j fN. What is the z component of the magnetic field?

SolutionEquation 29-1a gives F v B ( ) ( )( ) ( )10 5 21 16 2 4î k î î kj j j f N fC m/se B B Bx y z

( )[ ( ) ].16 4 2 2 4 fA m B B B Bz z y xî k j From either the x or y component, Bz 10 16 4 156 fN fA m mT=( )

( fN) fA m 5 16 2=( ). (The general expression for the cross product in components, with v =vz z x yB F e 0, gives F ey x=v .)

Problem57. Proponents of space-based particle-beam weapons have to confront the effect of Earth’s magnetic field on their beams.

If a beam of protons with kinetic energy 100 MeV is aimed in a straight line perpendicular to Earth’s magnetic field in a region where the field strength is 48 T, what will be the radius of the protons’ circular path?

SolutionIt is simplest to use the result of Problem 22, in atomic units, to find the radius (see solution to Problem 19), r Km qB 2 300 , where r is in meters, K is in MeV, m in MeV/c 2 , q in units of e, and B in teslas. Then

r 2 100 938 300 0 10 30 14( )( ) ( .48 ) .= km. (The mass of a proton in atomic units is approximately 938 MeV/c2.)

Problem58. Electrons are accelerated through a 30-kV potential difference at the rear of a TV tube. The electron beam is initially

headed straight toward the center of the tube. The TV is oriented so the beam is perpendicular to Earth’s magnetic field, in a location where the field strength is 62 T. What will be the radius of the electron beam’s curved path?

SolutionAs in the previous problem, r 2 0 03 0 511 300 62 10 96( . )( . ) ( ) .41= m. (The mass of an electron in atomic units is

approximately 511 2 keV c= .)

Problem59. A 170-mT magnetic field points into the page, confined to a square region as shown in Fig. 29-44. A square

conducting loop 32 cm on a side carrying a 5.0-A current in the clockwise sense extends partly into the field region, as shown. Find the magnetic force on the loop.

CHAPTER 1 15

FIGURE 29-44 Problem 59 Solution.

SolutionThe forces on the upper and lower horizontal parts of the loop in the field region cancel one-another, so the net force is just due to the magnetic force on the right vertical side, which is directed to the right in Fig. 29-43, with magnitude I B ( )( . )( ) .5 0 32 170 0 272 A m mT N. (See Equation 29-6.)

Problem70. Find the force on the circular current loop shown at the right of Fig. 29-44. The loop carries 5 0. A clockwise, has

radius 16 cm, and extends 10 cm into the field region.

SolutionTo find the force, we integrate Equation 29-8 over the part of the loop extending into the field region, with choice of co-ordinate axes and angle as shown in the sketch. Then B B Id IRd ( cos sin ).k îand j The force is

F Bz z I d IRB d IRB

1

22 1 2 1( cos sin ) [ (sin sin ) (cos cos )].j jî î The angular limits are

11

2 1 1 2 12

26 16 180 6 16 1 6 16 ° sin ( ) , sin sin , cos ( ) cos .= = = and so and Finally,

F IRB 2 N,1 6 16 0 2522( ) ( ) .= î î where we used the given values of I, R, and B.

Problem 60 Solution.

Problem61. An old-fashioned analog meter uses a wire coil in a magnetic field to deflect the meter needle. If the coil is 2 0. cm in

diameter and consists of 500 turns of wire, what should be the magnetic field strength if the maximum torque is to be 1 6. N m when the current in the coil is 1 0. mA?

SolutionThe maximum torque on a flat coil, with magnetic moment N R I B2 , max is (see Equations 29-10 and 11).

Thus, B NI R max ( . ) ( )( ) ( )= =2 21 6 500 1 1 102 N m mA cm G.

16 CHAPTER 29

Problem62. A circular wire coil 15 cm in diameter carries a 460-mA current and experiences a 0.020- N m torque when the

normal to the coil makes a 27° angle with a 42-mT magnetic field. How many turns are in the coil?

SolutionCombining Equations 29-10 and 11, we find the magnitude of the torque on a circular current-carrying coil in a uniform magnetic field to be ° B NI R B Nsin sin . ( . ) ( ) ( . ) ( ) sin2 20 02 460 7 5 42 27 129 Thus, N m mA cm mT= turns. (With the numbers given, N is an integer, to an accuracy of better than 0.03%.)

Supplementary Problems

Problem63. Electrons in a TV picture tube are accelerated through a 30-kV potential difference and head straight for the center of

the tube, 40 cm away. If the electrons are moving at right angles to Earth’s 0.50-G magnetic field, by how much do they miss the screen’s exact center?

SolutionThe electrons suffer the maximum deflection when moving perpendicularly to the magnetic field. They are bent into a circle of radius

R Km eB

22 3 10 1 6 10 911 10

1 6 10 5 1011 7

4 19 31

19 5=( )( . )( . )

( . )( ).

eV J/eV kg

C T m.

Geometry gives the maximum deflection from screen center: y R x R ( cos ), sin .1 where = Numerically,

y ( . ){ cos[sin ( .4 . )]} .11 7 1 0 11 7 6 851 m mm.=

Problem 63 Solution.

Problem64. The coil in the loudspeaker of Fig. 29-31 consists of 100 turns of wire, 3 5. cm in diameter. The magnetic field strength

at the coil is 0 64. T. Find the magnitude of the force on the speaker coil when the current in the coil is 2 1. A.

SolutionIn Fig. 29-31(b), the magnetic force on each element of current in the wire has the same magnitude, Id B , perpendicular to the page, so the net force on N turns of diameter d is F IBd NIB dz coil . (The total length of the wire is assumed to be the number of identical turns times the circumference of one turn.) For the values given, F ( )( . )100 2 1 A ( . ) ( . ) .0 64 3 5 14 8 T cm N.

Problem65. A conducting bar with mass 15 0. g and length 22 0. cm is suspended from a spring in a region where a 0.350-T

magnetic field points into the page, as shown in Fig. 29-45. With no current in the bar, the spring length is 26 0. cm. The bar is supplied with current from outside the field region, using wires of negligible mass. When a 2.00-A current flows from left to right in the bar, it rises 1 2. cm from its equilibrium position. Find (a) the spring constant and (b) the unstretched length of the spring.

CHAPTER 1 17

FIGURE 29-45 Problem 65.

SolutionIn equilibrium, the vertical forces on the bar must sum to zero. These include the weight of the bar, mg (negative downward), the upward spring force k k ( )0 (where is the length and 0 the unstretched length of the spring), and when current flows from left to right in the bar, an upward magnetic force of F ILBB (L is the length of the bar). The conditions stated require that k mg( . ) ,26 0 00 cm and k ILB mg( . )24 8 00 cm or k mg( )26 0 cm and k mg ILB( . ) .26 1 20 cm cm These equations can be solved for k and 0 (subtract to eliminate 0 , and divide to eliminate k) with the result k ILB = =( . ) ( )( )( . ) ( . ) .1 2 2 22 0 35 1 2 12 8 cm A cm T cm N/m, and 0 26 1 2 26 0 015 9 8 12 8 24 9 cm cm cm N N/m cm.( )( . ) ( . . ) ( . ) .mg ILB= =

Problem66. In 2 0. s, an electron moves 15 cm in the direction of a 0.10-T magoetic field. If the electron’s velocity components

perpendicular and parallel to the field are equal, (a) what is the length of its actual spiral trajectory and (b) how many orbits about the field direction does it complete?

Solution(a) The velocity component in the direction of B is v =|| 15 2 75 cm s km/s, which is also the numerical value of v ,

the speed perpendicular to B. Thus, the speed along the spiral trajectory is v v v v || || ,2 2 2 and the actual length

traveled in 2 s is 2 75 2 21 2( )( ) . km/s s cm. (b) The cyclotron period is T m eB2 = , so the number of orbits

completed in 2 1 1 6 10 0 1 911 10 5 59 1019 31 3 s is 2 s T s C T kg = = ( )( . )( . ) ( . ) . .

Problem67. A solid disk of mass M and thickness d sits on an incline, as shown in Fig. 29-46. A loop of wire is wrapped around

the disk, running along a diameter and oriented so the loop is parallel to the incline. A uniform magnetic field B points vertically upward. Find an expression for the current I in the loop that will keep it from rolling down the incline.

18 CHAPTER 29

GIGURE 29-46 Problem 67.

SolutiooThe torque of gravity causing the disk to roll down the incline (about the point of contact) is grav ° mgR sin( )180 mgR sin into the page in Fig. 29-46. The magnetic torque on the magnetic moment of the loop, which would cancel this, is mag B sin out of the page for the current shown, where IA A Rd and 2 is the area of the loop. (The magnetic torque is due to a couple and is the same about any point.) In equilibrium, mg I Rd B I mg Bdsin ( ) sin , . 2 2 or = (We assume that static friction between the disk and the incline is sufficient to satisfy the other conditions for equilibrium.)

Problem68. A disk of radius R carries a uniform surface charge density and is rotating with angular frequency Show that its

magnetic dipole moment is 14

4 R . Hint: Divide the disk into concentric rings. Treat each as a loop carrying an infinitesimal current, and integrate over all the loops.

SolutionA ring, of thickness dr and radius r, has charge dq dA r dr 2 . When rotating with period T 2 = , it represents

a current of dI dq T dq r dr = = 2 , and magnetic moment of d dI r r dr 2 3 (directed perpendicular

to the disk, out of the page for the rotation shown in the sketch). The total dipole moment is z 0R d

z 03 1

44R r dr R .

Problem 68 Solution.

Problem69. A 10-turn wire loop measuring 8 0. cm by 16 cm carrying 2 0. A lies in a horizontal plane but is free to rotate about the

axis shown in Fig. 29-47. A 50-g mass hangs from one side of the loop, and a uniform magnetic field points horizontally, as shown. What magnetic field strength is required to hold the loop in its horizontal position?

FIGURE 29-47 Problem 69 Solutioo.

SolutionFor the direction of current shown in Fig. 29-47, the magnetic moment of the loop is downward, and the magnetic torque B is along the axis, out of the page. The gravitational torque r gm is along the axis, into the page. The two torques cancel when B B mgr mgr NIA, ( . )( . )( . ) ( )( . )( . . ) . or kg m/s m A m mT.2 2= =0 05 9 8 0 04 10 2 0 0 8 0 16 76 6

CHAPTER 1 19

Problem70. A closed current loop is made from two semicircular wire arcs of radius R, joined at right angles as shown in

Fig. 29-48. The loop carries a current I and is oriented with the plane of one semicircle perpendicular to a uniform magnetic field B, as shown. Find (a) the magnetic moment of this nonplanar loop and (b) the torque on the loop. Hint: You can think of the loop as a superposition of two semicircular loops, each closed along the dashed line shown (why?).

FIGURE 29-48 Problem 70 Solution.

Solution(a) The whole current loop is equivalent to two semicircular loops, because currents of I cancel along their common boundary. The magnetic moments of each semicircle have equal magnitudes ( ) I R1

22 and are mutually

perpendicular, so the total dipole momeot is tot 1 222 2I R = . The direction of tot is 45° to the

shaded plane in Fig. 29-48, but depends on the sense of circulation of the current. (b) The torque on the loop has magnitude ° tot tot B B I R B I R Bsin ( ) ( )45 2 1 22 1

22= = [the same for either direction of current, since

sin 45 sin(180 )].° ° °45 The direction of is to align tot parallel to B.

Problem71. A circular wire loop of mass m and radius R carries a current I. The loop is hanging horizontally below a cylindrical

bar magnet, suspended by the magnetic force, as shown in Fig. 29-49. If the field lines crossing the loop make an angle with the vertical, show that the strength of the magnet’s field at the loop’s position is B mg RI =2 sin .

FIGURE 29-49 Problem 71 Solution.

SolutionWe assume that the magnetic field has axial symmetry, with vertical axis through the center of the loop’s horizontal plane area. The magnetic field lines intersect the loop at right angles, and if the current circulates clockwise (as seen from the magnet), the magnetic force on an element I d has an upward component dF dF Id By sin sin . This is the

same at any position on the loop, so the net upward force is F IB d RIBy z sin sin . 2 (The horizontal forces on

20 CHAPTER 29

dia-metrically opposite elements of loop cancel, so Fx 0.) When the loop is suspended, the upward magnetic force balances the loop’s weight, so F mg B mg RIy , sin . or =2

Problem72. A square wire loop of mass m carries a current I. It is initially in equilibrium, with its magnetic moment vector aligned

with a uniform magnetic field B. The loop is rotated slightly out of equilibrium about an axis through the centers of two opposite sides and then released. Show that it executes simple harmonic motion with period given by T m IB2 6 = .

SolutionThe torque on the loop, when its normal is displaced by an angle from B, is a restoring torque (in a direction to align with B, and so opposite to the sense of increasing ), of value B I d dt Isin ,rot rot where 2 2= is the rotational inertia of the loop about the axis of rotation specified. (In particular, two sides of the loop are parallel to the axis and two sides are perpendicularly bisected by the axis, so I m a m a marot 2 21

412

2 14

112

2 16

2( )( ) ( )( ) . ) For small angular

displacements, sin , ( ) , ¼ ¼ so rotd dt B I2 2= = which represents simple harmonic motion with B I=rot

( ) ( ) , .Ia B ma IB m T m IB2 16

2 6 2 2 6= = = = and period

Problem73. Early models pictured the electron in a hydrogen atom as being in a circular orbit of radius 5 29 10 11. m about the

stationary proton, held in orbit by the electric force. Find the magnetic dipole moment of such an atom. This quantity is called the Bohr magneton and is typical of atomic-sized magnetic moments. Hint: The full electron charge passes any given point in the orbit once per orbital period. Use this fact to calculate the average current.

SolutionOne electronic charge passes a given point on the orbit every period of revolution, so the magnitude of the average current corresponding to the electron’s orbital motion is I q t e r = = =v( ).2 (The current circulates opposite to the orbital motion, since the electron is negatively charged.) In the simplest version of the Bohr model for the hydrogen atom, the electron moves in a circular orbit, around a fixed proton, under the influence of the Coulomb force, so that

m r ke rv = =2 2 2 , or v= =r ke mr 2 3 . Thus, I e r e k mr ( )( ) ( ) .= v= = =2 22 3 The magnetic dipole moment associated

with this orbital atomic current (called a Bohr magneton) has magnitude B I r e kr m 2 12

2 12

19 21 6 10= ¼ ( . ) C

( )( . ) ( . ) . ,9 10 5 29 10 911 10 9 25 109 2 11 31 24 N m C m kg A m2 2= = ¼ and is typical of the size of atomic

magnetic dipole moments in general.