chapter 21 magnetic forces and magnetic fields · chapter 21 magnetic forces and magnetic fields...

CHAPTER 21 MAGNETIC FORCES ANDMAGNETIC FIELDS

CONCEPTUAL QUESTIONS_____________________________________________________________________________________________

1. REASONING AND SOLUTION Magnetic field lines, like electric field lines, never intersect.When a moving test charge is placed in a magnetic field so that its velocity vector has a componentperpendicular to the field, the particle will experience a force. That force is perpendicular to both thedirection of the field and the direction of the velocity. If it were possible for magnetic field lines tointersect, then there would be a different force associated with each of the two intersecting field lines;the particle could be pushed in two directions. Since the force on a particle always has a uniquedirection, we can conclude that magnetic field lines can never cross.

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2. REASONING AND SOLUTION If you accidentally use your left hand, instead of your right hand,to determine the direction of the magnetic force on a positive charge moving in a magnetic field, thedirection that you determine will be exactly opposite to the correct direction.

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3. REASONING AND SOLUTION A charged particle, passing through a certain region of space, hasa velocity whose magnitude and direction remain constant.

a. If it is known that the external magnetic field is zero everywhere in the region, we can concludethat the electric field is also zero. Any charged particle placed in an electric field will experience aforce given by F = qE, where q is the charge and E is the electric field. If the magnitude anddirection of the velocity of the particle are constant, then the particle has zero acceleration. FromNewton's second law, we know that the net force on the particle is zero. But there is no magneticfield and, hence, no magnetic force. Therefore, the net force is the electric force. Since the electricforce is zero, the electric field must be zero.

b. If it is known that the external electric field is zero everywhere, we cannot conclude that theexternal magnetic field is also zero. In order for a moving charged particle to experience a magneticforce when it is placed in a magnetic field, the velocity of the moving charge must have a componentthat is perpendicular to the direction of the magnetic field. If the moving charged particle enters theregion such that its velocity is parallel or antiparallel to the magnetic field, it will experience nomagnetic force, even though a magnetic field is present. In the absence of an external electric field,there is no electric force either. Thus, there is no net force, and the velocity vector will not change inany way.

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4. REASONING AND SOLUTION Suppose that the positive charge in Figure 21.10b were launchedfrom the south pole toward the north pole, in a direction opposite to the magnetic field. Regardless ofthe strength of the magnetic field, the particle will always reach the north pole. Since the charge islaunched directly opposite to the magnetic field, its velocity will be antiparallel to the field and have

118 MAGNETIC FORCES AND MAGNETIC FIELDS

no component perpendicular to the field. Therefore, there will be no magnetic force on the particle.The particle will move at constant velocity from the south pole of the magnet to the north pole.

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5. REASONING AND SOLUTION Since the paths of the particles are perpendicular to the magneticfield, we know that the velocities of the particles are perpendicular to the field. Since the velocity ofparticle #2 is perpendicular to the magnetic field and it passes through the field undeflected, we canconclude that particle #2 is neutral.

Particles #1 and #3 move in circularpaths. The figure at the right shows thedirection of the (centripetal) magneticforce that acts on the particles. If thefingers of the right hand are pointed intothe page so that the thumb points in thedirection of motion of particle #1, thepalm of the hand points toward the centerof the circular path traversed by the par-ticle. We can conclude, therefore, fromRHR-1 that particle #1 is positivelycharged. If the fingers of the right handare pointed into the page so that thethumb points in the direction of motion ofparticle #3, the palm of the hand pointsaway from the center of the circular pathtraversed by the particle. We can con-clude, therefore, from RHR-1 that particle#3 is negatively charged.

F

#1

#3

F

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Chapter 21 Conceptual Questions 119

6. REASONING AND SOLUTIONThree particles have identicalcharges and masses. They enter aconstant magnetic field and followthe paths as shown. The mag-nitude of the magnetic force oneach particle is given by Equation21.1 as F = q0(v sin θ)B, whereq0 is the magnitude of the chargeon the particle, v is the speed ofthe particle and B is the magnitudeof the magnetic field. Since theparticle paths are perpendicular tothe magnetic field, θ = 90° and sinθ = 1, so that F = q0vB . Thismagnetic force supplies thecentripetal force that is necessary

#1

#2

#3

for the particles to move on the observed circular paths, so q0vB = mv2 / r , where m is the mass ofthe particle and r is the radius of its circular path. Solving for v, we find v = q0Br / m .Since all threeparticles have the same charge and mass, we can conclude that the speed of the particle is directlyproportional to the radius of the particle's path. Therefore, particle #1 is traveling the fastest, whileparticle #2 is traveling the slowest.

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7. REASONING AND SOLUTION A proton follows the path shown in Figure 21.12. The magnitudeof the magnetic force on the proton is given by Equation 21.1 as F = q0(v sin θ)B, where q0 is themagnitude of the charge on the proton, v is the speed of the proton and B is the magnitude of themagnetic field. Since the proton's path is perpendicular to the magnetic field, θ = 90° and sin θ = 1,so that F = q0vB . This magnetic force supplies the centripetal force that is necessary for the proton

to move on the observed circular path, so q0vB = mprotonv2 / r , where m is the mass of the particle

and r is the radius of its circular path. The magnitude of the magnetic field is, therefore,B = mprotonv /(q0r ) .

If we want an electron to follow exactly the same path, we must adjust the magnetic field. Themagnitude of the charge on the electron is the same as that on the proton; however, the electron isnegatively charged and the proton is positively charged. As stated in the text, the direction of theforce on a negative charge is opposite to that predicted by RHR-1 for a positive charge. Therefore,the direction of the magnetic field must be reversed. In order for the electron to travel with the samespeed v in a circular path of the same radius r, the magnitude of the magnetic field must be changedto the value B = melectronv /(q0r) .

Thus, the electron will travel in the same path as the proton in Figure 21.12 if the direction of themagnetic field is reversed and the magnitude of the magnetic field is reduced by a factor(melectron / mproton ) .

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8. REASONING AND SOLUTION The drawingshows a top view of four interconnectedchambers. A negative charge is fired intochamber 1. By turning on separate magneticfields in each chamber, the charge is made to exitfrom chamber 4.

a. In each chamber the path of the particle is one-quarter of a circle. The drawing at the right alsoshows the direction of the centripetal force thatmust act on the particle in each chamber in orderfor the particle to traverse the path. The chargedparticle can be made to move in a circular path bylaunching it into a region in which there exists amagnetic field that is perpendicular to thevelocity of the particle.

1 2

34

v–q

F F

FF

Using RHR-1, we see that if the palm of the right hand were facing in the direction of F inchamber 1 so that the thumb points along the path of the particle, the fingers of the right hand mustpoint out of the page. This is the direction that the magnetic field must have to make a positivecharge move along the path shown in chamber 1. Since the particle is negatively charged, the fieldmust point opposite to that direction or into the page. Similar reasoning using RHR-1, andremembering that the particle is negatively charged, leads to the following conclusions: in region 2the field must point out of the page, in region 3 the field must point out of the page, and in region 4the field must point into the page.

b. If the speed of the particle is v when it enters chamber 1, it will emerge from chamber 4 with thesame speed v. The magnetic force is always perpendicular to the velocity of the particle; therefore, itcannot do work on the particle and cannot change the kinetic energy of the particle, according to thework-energy theorem. Since the kinetic energy is unchanged, the speed remains constant.

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9. REASONING AND SOLUTION A positive charge moves along a circular path under the influenceof a magnetic field. The magnetic field is perpendicular to the plane of the circle, as in Figure 21.12.If the velocity of the particle is reversed at some point along the path, the particle will not retrace itspath.

If the velocity of the particle is suddenly reversed, then from RHR-1 we see that the force on theparticle reverses direction. The particle will travel on a different circle that intersects the point wherethe direction of the velocity changes. The direction of motion of the particle (clockwise orcounterclockwise) will be the same as that in the original circle. This is suggested in the figurebelow.


F F

original path subsequent path

point ofvelocity reversal

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10. REASONING AND SOLUTION A television tube consists of an evacuated tube that contains anelectron gun that sends a narrow beam of high-speed electrons toward the screen of the tube (see textFigure 21.40). The inner surface of the screen is covered with a phosphor coating, and when theelectrons strike it, they generate a spot of visible light. The electron beam is deflected by themagnetic fields produced by electromagnets placed around the neck of the tube, between the electrongun and the screen. The magnetic fields produced by the electromagnets exert forces on the movingelectrons, causing their trajectories to bend and reach different points on the screen, therebyproducing the picture. If one end of a bar magnet is placed near a TV screen, the magnetic field ofthe bar magnet alters the trajectories of the electrons. As a result, the picture becomes distorted.

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11. REASONING AND SOLUTION When the particle is launched in the x, y plane, its initial velocitywill be perpendicular to the magnetic field; therefore, the particle will travel on a circular path in thex, y plane. In order for the charged particle to hit the target, the target must lie on the circular path ofthe moving particle. This will occur if the particle moves in a counterclockwise circle that passesthrough the third quadrant of the coordinate system. RHR-1 can be used to determine possibletrajectories in the following way. Place the fingers of the right hand into the page (direction of themagnetic field) and orient the thumb along one of the coordinate axes (direction of the particle'sinitial velocity). The palm of the right hand will face the direction in which the force on the chargedparticle is directed. Since the particle travels on a circle, the direction of the force will point towardthe center of the particle's trajectory. The figures below show the results for all four possible cases.

v

F

v

F

Target

Initial velocity innegative x direction

Initial velocity inpositive x direction


v

F

v

F

Initial velocity innegative y direction

Initial velocity inpositive y direction

Clearly, the charged particle can hit the target only if the initial velocity of the particle points either inthe negative x direction or the positive y direction.

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12. REASONING AND SOLUTION Refer to Figure 21.17. We can use RHR-1 with the modificationthat the direction of the velocity of a positive charge is the same as the direction of the conventionalcurrent I.

a. If the direction of the current I in the wire is reversed, the thumb of the right hand (direction of I)will be directed into the page, and the palm of the hand (direction of F) will face the left side of thepage. Therefore, the wire will be pushed to the left.

b. If both the current and the magnetic poles are reversed, then the fingers of the right hand mustpoint toward the top of the page (direction of B), while the thumb will be directed into the page(direction of I). The palm of the hand will face the right (direction of F); therefore, the wire will bepushed to the right.

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13. REASONING AND SOLUTION According to RHR-1, with the thumb of the right hand pointing inthe direction of the current I, and the palm of the hand facing the top of the page (direction of F), thefingers of the right hand point toward the left pole of the magnet. Therefore, the magnetic field in thevicinity of the current generally points from the right pole to the left pole of the horseshoe magnetshown in the drawing in the text. We can deduce, therefore, that the pole on the right is the northpole and the pole on the left is the south pole.

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14. REASONING AND SOLUTION In Figure 21.31, assume that current I1 is larger than current I2.Consider three regions: Region I lies to the left of both wires, region II lies between the wires, andregion III lies to the right of both wires. Let B1 represent the magnetic field due to current I1, and letB2 represent the magnetic field due to current I2.

a. The currents point in opposite directions. From RHR-2, the vectors B1 and B2 point in the samedirection in region II. Therefore, regardless of the magnitudes of B1 and B2, the resultant of the fieldscan never be zero between the wires. The fields B1 and B2 point in opposite directions in regions Iand III. Therefore, the resultant of B1 and B2 could be zero only in these two regions. The resultantof B1 and B2 will be zero at the point where their magnitudes are equal. The magnitude of the


magnetic field at a distance r from a long straight wire is given by Equation 21.5: B = µ0I /(2πr ) ,

where I is the magnitude of the current in the wire and µ0 is the permeability of free space. If thecurrent I1 is larger than the current I2, the magnitude of B1 will be equal to the magnitude of B2somewhere in region III. This is because region III is closer to the smaller current I2, and the smallervalue for r in Equation 21.5 allows the effect of I2 to offset the effect of the greater and more distantcurrent I1. Therefore, there is a point to the right of both wires where the total magnetic field is zero.

b. The currents point in the same direction. From RHR-2, the vectors B1 and B2 point in the samedirection in both regions I and III; therefore, the resultant of B1 and B2 cannot be zero in theseregions. In region II, the vectors B1 and B2 point in opposite directions; therefore, there is a placewhere the total magnetic field is zero between the wires. It is located closer to the wire carrying thesmaller current I2.

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15. REASONING AND SOLUTION The drawing shows an end-on view of three parallel wires that are perpendicular to theplane of the paper. In two of the wires, the current is directedinto the paper, while in the remaining wire the current is directedout of the paper. The two outermost wires are held rigidly inplace. From Example 9, we know that two parallel

Current

currents that point in the same direction attract each other while two parallel currents that point inopposite directions repel each other. Therefore, the middle wire will be attracted to the wire on theleft and repelled from the wire on the right. Since each of the fixed wires exerts a force to the left onthe middle wire, the net force on the middle wire will be to the left. Thus, the middle wire will moveto the left.

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16. REASONING AND SOLUTION The figure below shows the arrangements of electromagnets andmagnets.

N S

N S

NS

NS

(a)

(b)


We can determine the polarity of the electromagnets by using RHR-2. Imagine holding the current-carrying wire of the electromagnet in the right hand as the wire begins to coil around the iron core.The thumb points in the direction of the current. For the electromagnet in figure (a), the fingers ofthe right hand wrap around the wire on the left end so that they point, inside the coil, toward the rightend. Thus, the right end of the coil must be a north pole. Similar reasoning can be used to identifythe north and south poles of the electromagnet in figure (b). The results are shown in the figureabove. Since the like poles of two different magnets repel each other and the dissimilar poles of twodifferent magnets attract each other, we can conclude that in both arrangements, the electromagnet isrepelled from the permanent magnet at the right.

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17. REASONING AND SOLUTION The figure below shows the arrangements of electromagnets.

S N

N S

N

N

S

S

(a)

(b)

We can determine the polarity of each electromagnet by using RHR-2. Imagine holding the current-carrying wire of the electromagnet in the right hand as the wire begins to coil around the iron core.The thumb points in the direction of the current. For the electromagnet on the left in figure (a), thefingers of the right hand wrap around the wire on the left end so that they point, inside the coil,toward the right end. Thus, the right end of the coil must be a north pole. Similar reasoning can beused to identify the north and south poles of the other remaining electromagnets. The results areshown in the figure above. Since the like poles of two different magnets repel each other and thedissimilar poles of two different magnets attract each other, we can conclude that only thearrangement shown in (a) results in attraction. The electromagnets shown in arrangement (b) resultin repulsion.

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18. REASONING ANDSOLUTION Refer to Figure21.5. If the earth's magnetismis assumed to originate from alarge circular loop of currentwithin the earth, the plane ofthe current loop must beperpendicular to the magneticaxis of the earth, as suggestedin the figure at the right. UsingRHR-2, the current must flowclockwise when viewed lookingdown at the loop from the northmagnetic pole.

Current loop

Magnetic axis

North magnetic pole

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19. REASONING AND SOLUTION Thetotal magnetic field at the point P is theresultant of the magnetic field at P due toeach individual wire. If the current in allfour wires is directed into the page, thenfrom RHR-2, the magnetic field at P dueto the current in wire 1 must point towardwire 3, the magnetic field at P due to thecurrent in wire 2 must point toward wire1, the magnetic field at P due to thecurrent in wire 3 must point toward wire4, and the magnetic field at P due to thecurrent in wire 4 must point toward wire2, as shown at the right.

P

Wire 1 Wire 2

Wire 3 Wire 4

B4

B3

B2

B1

x x

xx

The magnetic field at a distance r from a long straight wire that carries a current I isB = µ0I /(2πr ) . Since the current in all four wires has the same magnitude and all four wires areequidistant from the point P, each wire gives rise to a magnetic field at P of the same magnitude.When current is flowing through all four wires, the total magnetic field at P is zero. If the current inany single wire is turned off, the total magnetic field will point toward one of the corners. Forexample, if the current in wire 1 is turned off, the resultant of B2 and B3 is still zero and the totalmagnetic field is B4. If the current in wire 2 is turned off, then the total magnetic field is B3, and soon.

We could have achieved similar results if the current in all four wires was directed out of the page.If fact, as long as opposite wires along the diagonal have currents in the same direction (for example,wires 1 and 4 with outward currents and wires 2 and 3 with inward currents) the total magnetic fieldwill point toward one of the corners when one of the currents is turned off.

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20. REASONING AND SOLUTION You have two bars, one of which is a permanent magnet and theother of which is not a magnet, but is made from a ferromagnetic material like iron. The two barslook exactly alike.

a. A third bar (the test bar) that is a permanent magnet can be used to distinguish which of the look-alike bars is the permanent magnet and which is the ferromagnetic bar. Both ends of each of thelook-alike bars, in succession, should be brought next to one end of the test bar. The look-alikepermanent magnet will be attracted to the test bar when opposite poles are brought together, while itwill be repelled from the test bar when like poles are brought together. Both ends of theferromagnetic bar, however, will be attracted to the test bar. The magnetic field of the test bar willalways induce a pole that is opposite in polarity to the pole of the magnet. Ferromagnetism alwaysresults in attraction.

b. The identities of the look-alike bars can be determined from a third bar that is not a magnet, but ismade from a ferromagnetic material. Either end of the look-alike permanent magnet will be attractedto both ends of the ferromagnetic test bar, while the look-alike ferromagnetic bar will have no effecton the ferromagnetic test bar.

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21. REASONING AND SOLUTION A strong electromagnet picks up a delivery truck carrying cans ofsoda pop. Inside the truck cans of the soft drink are seen to fly upward and stick to the roof justbeneath the electromagnet. We can conclude that the cans must be made from a ferromagneticmaterial, at least in part. Since aluminum is a nonferromagnetic material, we know that the cans arenot entirely made of aluminum.

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chapter 21 magnetic forces and magnetic fields · chapter 21 magnetic forces and magnetic fields...

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