For more information, please see full course syllabus of AP Physics C: Electricity & Magnetism

For more information, please see full course syllabus of AP Physics C: Electricity & Magnetism

## Discussion

## Study Guides

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Moving Charges In Magnetic Fields

- Magnetic field strength (B) is a vector quantity. Its units are Tesla (T).
- Magnet fields exert forces on moving charges proportional to the charge, the velocity, and the magnetic field strength. The magnetic force on a moving charge is always perpendicular to both the charge's velocity and the magnetic field.
- F=q(
**v**x**B**)=qvBsinθ - The direction of the magnetic force on a moving charge is given by the right hand rule. Point the fingers of your right hand in the direction of a positive particle’s velocity. Curl your fingers inward in the direction of the magnetic field, and your thumb will point in the direction of the force on the charged particle. For a negatively charged particle, use your left hand.
- A magnetic field can do no work on a moving charged particle, but it can change the particle’s direction.
- A mass spectrometer bends a moving charge using the magnetic force to determine the mass of unknown charged particles.
- A velocity selector utilizes a combination of magnetic and electric fields to allow a charged particle with a specific velocity to pass through in a straight line, while particles of other speeds and charges are deflected.

### Moving Charges In Magnetic Fields

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Objectives
- Magnetic Fields
- Forces on Moving Charges
- Magnetic Force is Always Perpendicular to the Charged Objects Velocity
- Magnetic Force Formula
- Magnitude of That
- Image
- Direction of the Magnetic Force
- Example 1
- Example 2
- Path of Charged Particles in B Fields
- Total Force on a Moving Charged Particle
- Velocity Selector
- Charged Particle in Crosses E and B Fields Can Undergo Constant Velocity Motion
- Particle Can Travel Through the Selector Without Any Deflection
- Mass Spectrometer
- Magnetic Fields Accelerate Moving Charges So That They Travel in a Circle
- Used to Determine the Mass of An Unknown Particle
- Example 3
- Example 4
- Example 5
- Example 6
- Example 7
- Example 8
- Example 9
- Example 10

- Intro 0:00
- Objectives 0:08
- Magnetic Fields 0:57
- Vector Quantity
- Tesla
- Gauss
- Forces on Moving Charges 1:30
- Magnetic Force is Always Perpendicular to the Charged Objects Velocity
- Magnetic Force Formula
- Magnitude of That
- Image
- Direction of the Magnetic Force 3:54
- Right-Hand Rule
- Electron of Negative Charge
- Example 1 4:51
- Example 2 6:58
- Path of Charged Particles in B Fields 8:07
- Magnetic Force Cannot Perform Work on a Moving Charge
- Magnetic Force Can Change Its Direction
- Total Force on a Moving Charged Particle 9:40
- E Field
- B Field
- Lorentz Force
- Velocity Selector 10:33
- Charged Particle in Crosses E and B Fields Can Undergo Constant Velocity Motion
- Particle Can Travel Through the Selector Without Any Deflection
- Mass Spectrometer 12:21
- Magnetic Fields Accelerate Moving Charges So That They Travel in a Circle
- Used to Determine the Mass of An Unknown Particle
- Example 3 13:11
- Example 4 15:01
- Example 5 16:44
- Example 6 17:33
- Example 7 19:12
- Example 8 19:50
- Example 9 24:02
- Example 10 25:21

### AP Physics C: Electricity and Magnetism Online Course

I. Electricity | ||
---|---|---|

Electric Charge & Coulomb's Law | 30:48 | |

Electric Fields | 1:19:22 | |

Gauss's Law | 52:53 | |

Electric Potential & Electric Potential Energy | 1:14:03 | |

Electric Potential Due to Continuous Charge Distributions | 1:01:28 | |

Conductors | 20:35 | |

Capacitors | 41:23 | |

II. Current Electricity | ||

Current & Resistance | 17:59 | |

Circuits I: Series Circuits | 29:08 | |

Circuits II: Parallel Circuits | 39:09 | |

RC Circuits: Steady State | 34:03 | |

RC Circuits: Transient Analysis | 1:01:07 | |

III. Magnetism | ||

Magnets | 8:38 | |

Moving Charges In Magnetic Fields | 29:07 | |

Forces on Current-Carrying Wires | 17:52 | |

Magnetic Fields Due to Current-Carrying Wires | 24:43 | |

The Biot-Savart Law | 21:50 | |

Ampere's Law | 26:31 | |

Magnetic Flux | 7:24 | |

Faraday's Law & Lenz's Law | 1:04:33 | |

IV. Inductance, RL Circuits, and LC Circuits | ||

Inductance | 6:41 | |

RL Circuits | 42:17 | |

LC Circuits | 9:47 | |

V. Maxwell's Equations | ||

Maxwell's Equations | 3:38 | |

VI. Sample AP Exams | ||

1998 AP Practice Exam: Multiple Choice Questions | 32:33 | |

1998 AP Practice Exam: Free Response Questions | 29:55 |

### Transcription: Moving Charges In Magnetic Fields

*Hello, everyone, and welcome back to www.educator.com.*0000

*In this lesson, we are going to talk about moving charges in magnetic fields.*0004

*Our objectives include calculating the magnitude and direction of the magnetic force on moving charges in terms of their charge,*0008

*their velocity, and the magnetic field strength.*0014

*Explaining why the magnetic force cannot perform any work on these charges.*0019

*We will talk about how we deduce the direction of the magnetic field from information about the forces*0024

*experienced by charged particles moving through the fields, getting into our right hand rules.*0029

*We will derive and apply the formula for the radius of the circular path of the charge moving perpendicular to a uniform magnetic field.*0033

*Finally, describing under what conditions particles would move with constant velocity through across the electric and magnetic fields.*0042

*Talking about there is what we are going to call the velocity selector.*0049

*Let us get in there and first of all get a few more details about magnetic fields.*0053

*Magnetic field strength capital B, is a vector quantity.*0058

*We said the units are tesla or a tesla is N-s/ C-m.*0062

*A tesla is a very strong magnetic field as we mentioned before.*0068

*There is also a common nonstandard unit which is called the Gauss.*0073

*Where 1 Gauss is 10⁻⁴ tesla, that should be an exponent 10⁻⁴ T.*0078

*The earth's magnetic field strength is right about ½ Gauss.*0085

*The magnetic force is always perpendicular to the charged objects velocity.*0092

*Because that force is always perpendicular to the velocity,*0096

*the magnetic force on a moving charge cannot be applied in the direction of the object’s displacement.*0100

*You cannot do any work on a moving charge.*0106

*Remember, work is force × displacement.*0109

*It is force dotted with displacement, because the force and the displacement are always going to be perpendicular,*0111

*you cannot do any work with that magnetic field.*0117

*You can, however, change its direction.*0120

*The magnetic force which is F vector and often time you see the subscript M for magnetic,*0124

*or you may see FB for magnetic.*0130

*It is the charge × its velocity × with the magnetic field vector.*0134

*Or the magnitude of that is going to be QVB × sin θ, where θ is the angle between your velocity and your magnetic field.*0140

*If you look over here on the right, we have a positively charge particle Q, moving with some velocity V as shown here.*0150

*It is doing so in the uniform magnetic field.*0157

*And the magnetic field symbol, we have here the dot with a circle around it and it is coming out of the page,*0159

*out of the screen in front of you.*0165

*If instead we saw something like an X with a circle around it, that would indicate it is going into the page or into the screen.*0167

*You are seeing, if you look over here, you are looking at the point of an arrow coming toward you.*0174

*You see the point first.*0180

*If it is going away from you, you see the X, the fletching on the arrow.*0182

*That is what that symbol means, it is showing you a vector in 3 dimensions into or out of the plane.*0185

*In this case, we have a magnetic field coming out of the plane.*0191

*The way we figure out the direction of the force, QV × B.*0194

*If you remember from our math review with × products, point the fingers of your right hand in the direction of the velocity*0198

*and bend your fingers inward in the direction of the magnetic field,*0205

*your thumb is going to point in the direction of the magnetic force on that charged particle.*0209

*In this case, if I use that right hand rule, point the fingers of my right hand in the direction of the velocity.*0215

*Bend them in the direction of the magnetic field, I end up with a force down and into the right on the page.*0221

*Perpendicular to the end, perpendicular to the magnetic field.*0228

*Just covering that again, as we look at this right hand rule, the direction of the force is given by a right hand rule.*0236

*If I say a positively charged particle, you can use your right hand.*0242

*If you are looking at something like an electron, something with a negative charge,*0245

*you can either use a right hand and go in the opposite direction of what your thumb points, or even easier user your left hand if it is negatively charged particle.*0248

*You are going to point the fingers of your right hand in the direction of a positive particle’s velocity.*0258

*Curl your fingers inward in the direction of the magnetic field.*0265

*Once you have done that, your thumb is going to point in the direction of the force on that charged particle, as shown in the diagram here.*0269

*Fingers in the direction of the velocity bend inward in the direction of the magnetic field*0278

*and your thumb points in the direction of that magnetic force on the moving charged particle.*0282

*Let us take an example.*0292

*Let us look at the force on an electron.*0294

*An electron moves at 2,000,000 m/ s perpendicular to a magnetic field with the flux density of 2 tesla.*0296

*What is the magnitude of the magnetic force on the electron?*0305

*Starting with what we know, its velocity is 2,000,000 m/ s.*0308

*Our magnetic field strength is 2 tesla, and our charge if it is an electronic is -1.6 × 10 ⁻19 C.*0316

*And because we are worried about the magnitude of the charge, I'm not going to worry about the negatives.*0330

*Our magnetic force which I like to write as FB but you could use FM, is Q the charge × V × B,*0334

*or if we are looking for the magnitude, which implies then that the magnitude of the magnetic force*0346

*on our charged particle is going to be Q VB sin θ, where θ is the angle between V and B.*0353

*It says in the problem that those are perpendicular.*0361

*If θ is 90°, that means its sin θ is sin 90° which is going to be 1.*0363

*That term goes away.*0372

*The magnitude of our magnetic force is our charge 1.6 × 10 ⁻19 C × our velocity 2,000,000 m/ s × our magnetic field strength 2 tesla.*0376

*And I come up with a magnitude of the magnetic force on our moving charge particle of about 6.4 × 10 ⁻13 N.*0394

*We did the magnitude problem here.*0413

*Let us take a look and see if we can figure out one with the direction using our right hand rules.*0415

*The diagram shows a proton, a positively charged particle moving with velocity V about to enter a uniform magnetic field directed in the page.*0419

*The axis are showing the fletching of the arrow, as if the arrow is going away from you.*0428

*The magnetic field is going into the plane of the page of the screen.*0433

*As the proton moves in the magnetic field, determine the direction of the force on the proton.*0439

*We are going to use our right hand rule.*0443

*Positive charge right hand, take you right hand and point the fingers of your right hand in the direction of the velocity.*0446

*From my perspective on the screen, I would point this way.*0452

*The magnetic field is directed into the page and my thumb is going to point in the direction of the force, which for me is toward the top of the screen.*0455

*If you do the same thing, you should come up also with a force that pointing*0464

*toward the top of your screen or the page, depending on how you are looking at this.*0468

*By the right hand rule, we know that the magnetic force on our charged particle is up.*0472

*Note that it is perpendicular to the velocity and perpendicular to the magnetic field again.*0480

*The magnetic force cannot perform work on a moving charge but it can change the charges direction because it always operate perpendicular to the objects velocity.*0488

*It can move it in a circle of the magnetic force is constant.*0498

*Here we have a positively charged particle moving with some velocity V, in the uniform magnetic field that is coming out of the plane of the screen.*0501

*The force on that by the right hand rule, as I do my V × B is toward the center of the circle.*0510

*That is going to provide the centripetal force to allow this object to move in a circular path.*0516

*We can even look at this in a little bit more detail.*0522

*We know the magnitude of that magnetic force is Q VB sin θ.*0524

*The magnetic force in this case must be providing our centripetal force, since the angle between V and B is 90°, sin θ is going to be 1.*0529

*That implies then that the magnetic force Q VB is equal to our centripetal force, MV²/ radius .*0540

*We can solve for the radius equal to MV/ QB.*0550

*It was kind of cool here, note that MV that is the momentum of our particle.*0558

*You can change the radius by changing strength of the magnetic field.*0573

*We have talked about the magnetic force on a moving charge particle.*0581

*We talked about the electrical force, the Coulombic force on a charge particle.*0584

*You have to deal with both of these at the same time.*0588

*The electric field can do work on a moving charge.*0591

*Remember the magnetic field cannot do work on a moving charge.*0594

*But when you put all this together, you get what we call the Lorentz force, the total electromagnetic force on a charge particle.*0596

*And that is going to be the electric field component QE, there is our electric or Coulombic piece + V × B which is QV × B.*0603

*There is our magnetic contribution to the total force on a moving charged particle.*0619

*The velocity selector is a very interesting tool.*0634

*A charged particle in ×, the electric and magnetic fields can undergo constant velocity motion if the velocity, the magnetic field, and the electric field, are all selected perpendicular to each other.*0637

*If you set the velocity equal to the electric field strength ÷ the magnetic field strength,*0649

*the particle can travel through the selector without any deflection whatsoever.*0655

*Particles with any other velocity are diverted off to the sides.*0659

*You put a bunch of charged particles into this machine, the only ones that are going to make it all the way out*0663

*are those that have the exact velocity that you are after, equal to the electric field strength ÷ the magnetic field strength.*0669

*Let us take a look at this analytically for a second.*0676

*In order for this to happen, we had note that the electric field moving up on our particle*0678

*must absolutely balance the magnetic field, the force pulling it toward the opposite direction down.*0684

*The electric force must equal the magnetic force which implies that QE must be equal to QV × B.*0690

*Or looking at magnitudes, QE must equal Q VB.*0708

*We gave it the sin θ piece because the angle between V and B is going to be 90°,*0714

*which implies then that the velocity that makes it through here, diverted where the electric force and*0719

*the magnetic force are absolutely balanced is just going to be E/ V.*0725

*Anything that is different is going to be off in the direction, it is not going to make it through your machine.*0730

*This is a very popular type question on AP tests.*0735

*Let us take a look at mass spectrometer, another tool that uses the magnetic force.*0742

*Magnetic fields accelerate moving charges so that they travel in a circle, in a uniform magnetic field.*0747

*You can use that to determine the mass of an unknown particle.*0752

*You ever charged unknown particle coming in here, we do not know its mass, we can use the magnetic field to apply a force, a centripetal force to spin it.*0755

*Depending on where we detect it here, we can determine what its mass is.*0764

*The further away it is, the more mass it has.*0769

*The closer it is, the smaller the mass with that constant force.*0772

*You can use this in order to determine what sort of masses you have in an unknown conglomeration of materials.*0775

*As long as you can charge and accelerate them, you can then read where they hit here in order to determine their mass.*0783

*A very popular tool.*0789

*Let us take a look at a couple more examples starting with the velocity of the charged particle.*0792

*Let us say we have a particle with a charge of 6.4 × 10 ⁻19 C experiences a force of 2 × 10 ⁻12 N,*0797

*traveling through a 3 tesla magnetic field and angle of 30° to the field.*0805

*What is the particle’s velocity?*0810

*We will start by writing down our givens.*0814

*Our charge 6.4 × 10 ⁻19 C, our force 2 × 10 ⁻12 N, our magnetic field strength is 3 tesla, and our angle is 30°.*0814

*To find the magnetic force that is going to be Q VB sin θ.*0835

*But we are actually after the particles velocity.*0845

*As we are rearranging to find the velocity that is going to be the magnetic force ÷ QB sin θ.*0848

*Now I can substitute in my values.*0860

*This implies then that our velocity is going to be, we have the force of 2 × 10 ⁻12 N ÷ our charge 6.4 ×*0863

*10 ⁻19 C × our magnetic field strength 3 tesla × the sin of our angle 30°.*0875

*I come up with the velocity of about 2.08 × 10⁶ m/ s, 2,000,000 m/ s.*0884

*That particle is moving pretty quick.*0898

*Force on a moving charge.*0903

*We have a charge of 5 µc moving with the velocity of 5,000,000 m / s in the X direction.*0904

*Find the force on the charge due to a magnetic field of ½ tesla in the positive Y direction.*0912

*Let us start with our givens and finds.*0920

*We know our charge is 5 µc or 5 × 10⁻⁶ C.*0922

*Our velocity is 5,000,00 5 × 10⁶ m / s in the X direction.*0929

*I'm just going to write that in unit vector notation I ̂.*0936

*Our magnetic field strength is 0.5 tesla in the positive Y direction which should be J ̂.*0941

*To find our magnetic force that is going to be Q V × B or looking at magnitudes Q VB sin θ.*0949

*But in this problem it is pretty easy to see that they are perpendicular, θ = 90°.*0964

*Sin θ is going to equal 1.*0971

*Therefore, our magnetic force is going to be Q VB which is our charge 5 × 10⁻⁶ C × our velocity 5 × 10⁶ m / s × ½ tesla, or 12.5 N.*0976

*Let us do another example.*1003

*An electron is projected with an initial velocity of 50,000 m / s parallel to a uniform magnetic field of 25 tesla.*1006

*What is the resulting force on the electron?*1015

*You could go through a lot of math here but there is a key word that simplifies this problem tremendously.*1018

*The velocity is parallel to that uniform magnetic field.*1024

*If they are parallel, there is not going to be any force.*1029

*0 N is on the electron, they have to be perpendicular.*1034

*If you have our sin factor, sin 0 is going to be 0.*1039

*Your force is going to 0.*1044

*Because it is parallel, no force on the electron.*1046

*Example 6, for each of the diagrams below, indicate the direction of the magnetic force on the charged particle.*1054

*Over here, we have an electron going through a magnetic field between a couple of magnets here.*1061

*The first thing I think I'm going to do is draw in the direction of the magnetic field.*1067

*Magnetic fields run from north to south so let us draw those in first.*1071

*We got a negatively charged particle so I'm going to use a left hand rule, instead of the right hand rule.*1076

*I’m going to point the fingers of my left hand in the direction that particle is moving.*1082

*Curl my fingers in the direction of the magnetic field which we just drew in here.*1087

*And when I do that, I find that I get a magnetic force which is into the plane of the page or the plane in the screen.*1092

*There is my magnetic force by the left hand rule because it is a negative charge.*1101

*Let us come over here now and do it for a positive charge.*1107

*It means we can use our right hand rule.*1110

*You have a positive charge, its velocity is moving up.*1112

*Let us label that V.*1116

*Our magnetic field is into the plane of the screen so right hand rule point the fingers of my right hand in the direction the particle is moving.*1119

*Bend them in the direction of the magnetic field and my thumb is going to point in the direction of the force on those,*1126

*which from my perspective is going to be toward the left of the screen again.*1133

*To the left of the screen for this problem, applications of the right hand rule.*1139

*We call them right hand rule, even though you use the left for the negative charges.*1145

*Another example here.*1152

*An electron is released from rest between the poles of 2 bar magnets in a region where the magnitude of the magnetic field strength is 6 tesla as shown below.*1153

*What is the magnetic force on the electron?*1162

*Keywords here are, it is at rest.*1166

*As long as it is at rest, there is no magnetic force on that charged particle.*1170

*It has to be moving, in order to have that magnetic force.*1176

*The magnetic force is going to be 0 because the velocity is 0.*1180

*It is not moving, the only thing you can have on is an electrical force.*1184

*Let us take the example of a proton in orbit.*1191

*We have a proton moving in a circular orbit of radius ½ m perpendicular to a uniform magnetic field of magnitude 0.3 tesla.*1194

*Find the period and frequency of the proton’s orbit, as well as its speed.*1203

*A little more involved here.*1207

*Let us draw a picture of this to begin with.*1209

*There is a nice little happy proton orbit, our radius is 0.5 m.*1213

*As it is going around here at that point, it has an instantaneous velocity in that direction.*1220

*And we have a magnetic field of magnitude 0.3 tesla.*1226

*Find the period and frequency of the proton’s orbit as well as its speed.*1239

*I think where we are going to start here is, let us take a look at how it is moving in a circle.*1244

*We will use that equation we did before where we said the centripetal force has to be equal to the magnetic force,*1251

*because the magnetic force is what is causing that centripetal force.*1258

*The magnetic force is the centripetal force, causing it to move in a circle.*1261

*Centripetal force which is MV²/ R is equal to the magnitude of the magnetic force Q VB.*1266

*And sin θ we do not need because θ is 90° again, they are perpendicular, sin 90 is 1.*1275

*Rearranging here, we can take V²/ V here is going to be Q RB/ M.*1284

*That will simplify nicely.*1294

*This implies then that the left hand side, our velocity is going to be Q RB/ M.*1296

*Our Q we said, it is a proton, it must be 1.6 × 10 ⁻19 C.*1300

*Our radius is ½ m, our magnetic field strength is 0.3 tesla, and we have to divide it by the mass of our particle.*1308

*The mass of the proton you can look up, it is 1.67 × 10 ⁻27 kg.*1320

*Putting all that in my calculator, I come up with 1.44 × 10⁷ m/ s.*1329

*We have found its speed, we are after period and frequency.*1347

*First thing I'm going to do I think, is find out how long it takes to go once around the circle.*1352

*Once I have that, that is really going to be its period.*1356

*I can use the straightforward V = D/ T, because it is moving at a constant velocity,*1359

*which means that the time it takes is going to be distance it travels ÷ its velocity.*1368

*The distance it travels is its circumference 2 π R or 2 π × 0.5 m.*1373

*We just found its velocity as 1.44 × 10⁷ m / s.*1380

*The time it takes to go once around is right around 2.18 × 10⁻⁷ s, which is its period because period is the time it takes to go once around the circle.*1386

*How about frequency?*1405

*Frequency is just the reciprocal of period which is going to be 1/ 2.18 × 10⁻⁷ s, which is about 4.58 × 10⁶ Hz or 4.58 MHz.*1407

*Proton in orbit, let us do a velocity selector question.*1437

*Find the speed of a charged particle which passes through a velocity selector*1444

*with magnetic field strength of 1 tesla perpendicular to an electric field of 600,000 N/ C.*1447

*I remember the trick in the velocity selector is the electric force and the magnetic force has to balance.*1456

*The electric force must equal the magnetic force.*1462

*They have to have the same magnitude at opposite directions, of course, which implies that QE,*1467

*the magnitude of the electric force must equal the magnitude of the magnetic force Q VB and sin θ again.*1472

*Θ is going to be 90°, sin 90 is 1.*1479

*Therefore, the velocity is just going to be E/ B which will be 600,000 N/ C ÷ 1 tesla,*1482

*which implies then that our velocity is going to be 600,000 m/ s.*1500

*Particle with exactly that velocity is going to make it through that particle selector unhindered.*1511

*Let us do one last question, maybe with a couple of parts in a mass spectrometer.*1517

*We have a proton that is accelerated through a potential differences V before passing into a region of*1524

*uniform magnetic field into the plane of the screen as shown here.*1529

*Find the voltage necessary to give the proton a speed v as it enters the magnetic field region in terms of its mass, its velocity, and its charge Q.*1534

*Let us see, in order to start this, the energy that it has is the work that we do on it, which is going to be the charge × the potential difference.*1545

*Work, which is QV, is going to be manifested, it is going to be its kinetic energy as it goes into the mass spectrometer.*1555

*That is ½ MV², C v.*1564

*Voltage and speed, you got to be careful of these two.*1569

*Which implies then, solving for the voltage, the potential difference is going to be M × velocity²/ 2Q.*1572

*That part was not too bad.*1586

*Let us take a little bit further and determine the expression for the radius of the proton’s motion in the uniform magnetic field region.*1589

*The magnetic force on it is Q VB and that has to equal MV²/ R because it is moving in a circular path,*1601

*which implies that R is going to be equal to MV²/ Q VB, which is equal to MV/ QB, when we cancel out one of those velocity factors.*1611

*Radius in the uniform magnetic field MV/ QB.*1627

*Taking another step, sketch the path of the proton in the magnetic field.*1635

*Let us see, it is a positively charged particle so this looks like your right hand rule problem.*1641

*Point your fingers in the direction that it is moving, to the right of the screen as it enters the magnetic field.*1647

*Curl your fingers into the screen, in the direction of the magnetic field, and you are going to see that initially you have a force up.*1653

*As that moves, that is going to turn it.*1661

*You are going to get this curving path so it travels something like that.*1663

*You are going to see a circular path in that direction.*1671

*Last one, an electric field is applied in the same region as the uniform magnetic field to over here.*1677

*Determine the magnitude and direction of electric field required so that the proton passes through the region in a straight line.*1684

*Now we are setting this up so that we have a velocity selector, so that the particle goes through unhindered.*1690

*The first thing is, we saw the magnetic field.*1698

*The magnetic force is pointed toward the top of our screen, that means that our electric force is going to have to be down.*1701

*Therefore, our electric field is going to have to point down because it is a positive particle.*1710

*In the velocity selector, we have the electric force equal to the magnetic force, which implies that QE = Q VB,*1716

*which implies that the electric field strength has to be equal to V × B.*1728

*Hopefully, that gets you a good start on moving charges in magnetic fields.*1739

*Thank you for watching www.educator.com.*1744

*Make it a great day everyone.*1747

1 answer

Last reply by: Professor Dan Fullerton

Mon Nov 24, 2014 6:08 AM

Post by QuangNguyen VoHuynh on November 24, 2014

I would like to ask the reason why magnetic force cannot perform work? Is it because the force creates a circular motion, so that the displacement is zero?