AP Physics C: Electricity & Magnetism 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 forces0024

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 velocity0198

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 field0278

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 force0346

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 pointing0464

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 out0663

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 particle0678

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 and0719

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 selector1444

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 of1524

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