AP Physics C: Electricity & Magnetism Faraday's Law & Lenz's Law

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

I'm Dan Fullerton and in this lesson we are going to talk about Faraday’s law and Lenz’s law, some of my favorite topics in E and M.0004

Is it really starts to tie the entire course together.0011

Our objectives include recognizing situations in which change in flux through a loop0014

will cause an induced EMF or current in the loop.0018

Calculating the magnitude and direction of the induced EMF in current loop of wire or a conducting bar0021

where the magnitude of a related quantity such as magnetic field or area of the loop0027

as specified is a nonlinear function of time, the change in magnetic fields or areas.0031

Analyzing the forces that act on induced currents to determine the chemical consequences of those forces.0036

Let us talk about what Faraday’s law tells us.0043

Faraday’s law talks about the induced EMF, the induced potential due to a change in magnetic field.0047

It says it is equal in magnitude, the rate of change of the magnetic flux through a surface bounded by a circuit.0054

Moving charges create magnetic fields, change in magnetic fields move charges by creating potential.0060

The direction of the induced current is given by Lenz’s law.0069

You usually use Faraday’s law and Lenz’s law together.0072

Here is Faraday’s law, the induced EMF typically written as script D0075

is the opposite of the time rate of change of the magnetic flux through that surface or the magnetic flux.0080

If you expanded that out, it is the opposite of the derivative of the integral / the open surface of B ⋅ DA.0084

The EMF is the integral/ the closed loop of E ⋅ DL.0094

That is your EMF, that is your potential around a closed loop.0098

If you do not have any change in magnetic flux that actually becomes your Kirchhoff's voltage law.0101

Lenz’s law is this negative sign right there.0108

That is where the Lenz’s law piece comes in.0113

We have our surface, we have magnetic field through it and the area is changing or the magnetic field strength is changing,0116

it leads to an induced potential difference in the circuit.0123

The direction of that, whether it is going clockwise or counterclockwise, that piece comes from Lenz’s law.0127

Let us go to Lenz’s law here.0134

The current induced by a change in magnetic flux creates a magnetic field opposing the change in flux.0136

Or the way I like to think about it is, if something is changing Lenz’s law, the induced current0142

or the induce potential difference wants to counteract that.0147

As an example, if we had this magnet being moved down into this loop of wire so the magnetic flux into this loop of wire is increasing.0151

The induced potential, the induced current wants to oppose that.0162

If it is increasing down into that loop, it wants to create a magnetic field that opposes that up out of the loop.0165

And as I do that with my right hand rule, if the magnetic field strength is coming out of the loop, is what it is trying to oppose,0173

wrap the fingers of your right hand around that and you will see that you will start to get the current in this direction,0181

that counterclockwise direction.0188

You can also look at it from the perspective, let us draw another one over here.0191

We have got a north, I will put a south right beside it.0196

I will draw our loop of wire.0199

If this time, we have our magnetic flux coming through there but now we are moving our magnet away0207

from that loop of wire, Lenz’s law says that it wants it to stay the same.0216

It does not want to reduce.0221

It is going to create a current in the wire that is going to want to maintain that.0223

It is going to increase the magnetic field strength down through the wire.0228

By the right hand rule, that means our current must be flowing that way if we are moving our magnet up out of that loop.0231

That is how Lenz’s law work.0242

It is just a matter of practice.0244

Here we go, if we have a sample like this.0250

If the magnetic field is increasing, we are going to induce a clockwise current.0253

Why? As we take our thumb and point it out of the page, if it is increasing we want to oppose that.0259

To oppose that, we go the opposite direction, wrap the fingers of your right hand and you get clockwise.0265

If that magnetic field strength there is decreasing, Lenz’s law you want to work to maintain where it does not want to decrease.0271

That means you need to add magnetic field coming out of that loop so point your finger coming out of that loop.0278

Wrap the fingers of your right hand around that and you are going to get the counterclockwise current.0283

More examples for Lenz’s law and how you apply it.0291

What we have really done now is, we have put together Maxwell’s equations.0295

We started with Gauss’s law, the integral / the close surface of E ⋅ DA is the enclosed charge ÷ ε₀.0299

This is always true, it is a law.0307

It is useful however in situations of planar, spherical, or cylindrical symmetry.0309

Gauss’s law for magnetism, the integral / close surface of B ⋅ DA is 0, which you can also read as there are no magnetic monopoles.0316

We went to ampere’s law next, the integral / the closed loop of B ⋅ DL = μ₀ I.0326

That is a way to find the magnetic field strength when you have situations of symmetry that is a lot simpler than the Biot-Savart.0332

And faraday's law now, the integral/ the closed loop of E ⋅ DL, my voltage is a round the close path is - D / DT,0339

integral over the open surface of B ⋅ DA.0349

Or - the time rate of change of the magnetic flux due to surface.0351

By the way, note the * here for ampere’s law.0358

There is a little bit more to ampere’s law than just this that we are going to get to at the very end of the course.0361

We are going to come back to that and add to it just a touch but you have got the basics for ampere’s law for now.0367

Let us go to some examples, a magnetic field of strength B of T given by 3T² -2T + 10373

is directed out of the plane of the circular loop of wire as shown.0383

A lamp somewhere is part of that loop.0387

Find the generated electro motor force as a function of time.0390

We are going to go right to faraday’s law here.0394

The induce EMF is the opposite of the time rate of change of the magnetic flux which is going to be - D / DT integral / the open surface of B ⋅ DA.0397

Which you will notice that as we do our integral, the area is not changing.0417

The area is constant.0424

Integral of B ⋅ DA with their variable of integration DA is just going to be BA.0426

This becomes - D/ DT × BA, which implies then that B is a function of time but the area is not.0432

Our induced EMF, we can pull our A out of the derivative, is - A × the derivative of B with respect to T,0445

which is - A × the derivative of our function B, which is 3T² -2T + 1.0455

Which implies then that our electromotor force is going to be, our area is - π R² and the derivative of this is going to be 6T-2.0468

There is our function for induced EMF.0485

Alright now that we have that, let us see if we can find the current through the 100 ohm lamp as a function of time.0492

We know the EMF, current is potential ÷ resistance.0497

Ohms law which is - π R² × 6T-2, we just found that in the previous page, ÷ 100 ohms, that easy.0503

How about the direction of our current flow at time T = 5s?0521

Let us take a look, the derivative of B with respect to T at time T is 6T -2.0528

At 5s, that is going to be T = 5s DB DT = 6 × 5s -2 is going to be 28.0534

We have got a positive rate of change of magnetic flux.0549

As we do that, the positive we are getting bigger this way so it wants to oppose that0556

down into this plane of the page, it is going to be clockwise to Lenz’s law.0562

Let us do a second example here.0578

We have got a circuit in which a current carrying rod on rails is moved to the left with some constant velocity V.0580

If the circuit is perpendicular to a constant magnetic field shown here in blue, determine the induced EMF in the circuit.0588

We will use Faraday’s law again, E = – D φ BDT which is - D/ DT.0596

The integral / the open surface of B ⋅ DA which is going to be, B can come out and DA, A is going to be changing but only one piece is.0608

That is going to be - DL and that is coming out we have got this changing X, that will be DX DT.0626

DX DT is the definition of V.0639

Since V = DX DT, we can write this then as E = -B LV, which may be0641

a formula you have memorized if you have a previous algebra based physics course.0652

Best way to get better at these I think is to do a lot of practice.0660

What I'm going to do is a bunch of old AP problems from old AP exams, the free response problems and we will see how they go.0663

Let us start with a 2012 E and M exam free response question number 3.0672

You can download that right up here, take a look at it.0678

Print it out if you want, give it a shot, and then we will come back and do it together.0681

Just like in our example 2, we have one of these situations0686

of a cross bar that can be moving as you have different current running through the system.0689

Taking a look at part A, it asks us to determine the magnitude of the magnetic flux through the loop when the crossbar is at the position shown.0695

That is pretty straightforward, it is not moving.0706

Our total magnetic flux is just the integral / the open surface of B ⋅ DA, which is going to be, we have got B0.0708

The integral / the open surface of DA, if everything is being nice and constant is just going to be our length L × our width or height A to 0.0724

Let us take a look at part B, now the ×bar is released from rest and it slides with negligible friction down without losing electrical contact.0739

Indicate the direction of the current and the ×bar as it falls.0751

We have got a ×bar and as it falls, the flux through the loop is decreasing.0755

Therefore, the induced current wants an inward magnetic flux to oppose that change.0760

Because of that, that requires a clockwise current due to Lenz’s Law and the right hand rule.0765

If this is the top of that loop at clockwise current is going to have current going to the right.0770

To the right would be the direction of current due to Lenz’s law and the right hand rules.0777

And I would explain that in words, it will probably be a great idea do not just leave the drawing there when it asks you to justify your answer.0785

C, find the magnitude of the current and the crossbar as it falls to the function of speed.0793

Current is potential difference ÷ resistance.0800

It is going to be – D φ B/ DT ÷ our resistance is going to be - D/ DT of, we have B₀ LH all ÷ R.0805

Pulling out our constants, we can pull out - B₀ L/ R.0827

We have got DH DT, but DH DT is just going to be our velocity.0835

Since velocity = DH DT, we can keep going here and write this as current = - B₀ L × our velocity ÷ R.0841

Since it asks for the magnitude, we will state that the magnitude of I = B₀ LV/ R.0859

We will plug in through this one, let us give ourselves some more room on another page.0875

As we go to part D, derive a differential equation to determine the speed of the crossbar as a function of time.0880

The way I would start here is with a free body diagram.0890

We have the magnetic force up and we have gravity down.0894

Let us call down our positive Y direction as we set our axis.0901

As we look at this from the Lenz’s Newton's second law, net force in the Y direction is going to be MG –0906

the magnetic force FB and all of that must be equal to MA.0915

Find the magnetic force though, recall our magnetic force that is going to be the integral of I × DL × B.0923

Or in this case, IL B0.0936

We can write this then as MG – IL B0 = MA.0943

Which implies then, we know that I = B₀ LV/ R, we did that previously.0960

We also know that acceleration is just the time rate of change of the velocity, the derivative of velocity.0970

We can write this as, let us see, we have got MG – I, substituting that in there0978

we are going to get a B₀² L² V/ R and that must be equal to M DV DT.0986

There we have our differential equation.1002

We have velocity and the derivative of velocity in the same equation.1004

If you wanted to, you could go clean this up a bit more, putting it to standard form.1009

It would be fine if you just left it that way, I will do the one last step, in case you want to see it.1012

I might write it as DV DT + B₀² L²/ M RV - G = 0.1017

As you go through some manipulations but you have already done that differential equation once you get to that step.1030

Looking at part E, determine the terminal speed of the crossbar.1037

The trick here is recognizing what it said its terminal speed, its terminal velocity, the magnetic force and the gravitational force must balance.1044

There is no longer any acceleration so the magnetic force must equal the gravitational force MG.1051

Therefore, we said the magnetic force was IL B₀ must equal MG.1059

We figured out that I was B₀ LV/ R.1067

Substituting that in the left hand side, B₀² L² V/ R = MG.1073

Or solving for V which is the terminal velocity, that is just going to be MG R/ B₀² L².1082

Finally part F, if the resistance of the crossbar is increased, does the terminal speed increase? Decrease? Or remain the same?1100

As we look right here, if resistance increases none of these other factors change along with it.1111

It is pretty easy to see that the terminal velocity is going to increase.1117

It asked us to give a physical justification for answer in terms of the forces on the ×bar.1123

To do the physical justification, I would mention something like the terminal velocity increases1128

since currents going to decrease as R increases.1133

Since I = B/ R, current decreases R increases and with decreased current, that magnetic force decreases1137

so the bar has to fall faster to generate enough current to create a magnetic force that is strong enough to balance the force of gravity.1145

Something like that or another physical justification that explains why that is going to increase.1155

And that covers the 2012 question.1162

Let us take a look at 2010 E and M question number 3 in the free response section.1169

Link to download it, print it out, give it a try.1178

Here this looks like some of the sample problems we have been doing.1182

At least parts of it.1185

I got a long straight wire which carries a current I to the right and it varies as a function of time.1188

We also have a rectangular loop that has a light bulb with resistance R in there.1193

Indicate the direction of the current in the loop for part A.1199

As I look at that, the magnetic field through the loop is coming out of the page and it is decreasing.1207

You can see by the function I = I₀ – KT.1217

The induced current must oppose that by creating flux out of the page or out of the screen.1221

By Lenz’s law which by the right hand rule is going to be counterclockwise current.1226

Counterclockwise and then explain your answer in terms of Lenz’s law.1231

Do not just write Lenz’s law, make sure you talk about the flux that are changing.1235

What that is going to do to the current, all of that.1238

Part B, indicate whether the light bulb gets brighter, dimmer, or stays the same brightness over the time period of interest.1243

As we look at that one, I varies with T.1253

The change in flux, the derivative of that is going to be constant.1257

As long as that is constant, the induced current is going to be constant.1261

And brightness depends on power which is I² R.1264

If I is constant and R is constant then the brightness must be constant.1268

That must be constant or I suppose you should write remains the same.1273

And of course, give a detailed explanation to justify your answer.1281

Taking a look at part C, determine the magnetic field at T equal 0 due to current in the long wire, the distance R from the wire.1287

B = μ₀ I/ 2 π R.1299

Since we are at time T equal 0 that means current = I0.1308

Plugging T in for our current equation, therefore B just = I₀/ 2 π R.1314

Let us go on to D and I'm going to go to the next page to make sure we have got plenty of room here.1329

As we look at D, derive an expression for the magnetic flux through the loop as a function of time.1336

The magnetic flux φ B is going to be the integral / the open surface of B ⋅ DA should be the integral from R = D to R = D + A of μ₀ I / 2 π R × B DR.1344

As we cut that up to the little horizontal slices and add them all up to get our total flux.1374

We can pull out our constants from the integral, μ₀ IB/ 2 π.1380

Integral from R = D to D + A of DR/ R, which is going to be the natural log.1390

Μ₀ I B/ 2 π log of D + A / D, which implies then that the total flux φ B is going to be, we got μ₀,1401

we have got B / 2 π log of D + A/ D.1421

We also have our current I, but I is a function of time so let us write that as I₀ – KT.1429

We are off to part E, derive an expression for the power dissipated by the light bulb.1448

Let us see, the resistance is not going to change, that is constant.1458

We are going to need to know either current or voltage, we can use power = V² / R, we will use the induced EMF there.1462

First thing to know that induced EMF that is just going to be the opposite of the time rate of change of the flux with respect to time.1472

Faraday’s law is going to be - the derivative with respect to T of μ₀ B/ 2 π log of D + A / D × I₀ – KT.1480

And that looks like quite a bit but realize the only part of this is a function of T is that - KT over here.1500

What we are going to end up with is our EMF going to be K μ B/ 2 π × the log of D + A/ D.1507

To the find our power, all we have to do is put all that together to be EMF²/ R or we have μ₀ KB / 2 π log of D + A / D,1526

that whole piece squared and then ÷ the resistance R.1545

Let us move on and take a look at another problem.1561

Let us go to the 2009 exam E and M free response number 3.1565

You have not noticed there a lot of these, that means on the E and M test,1571

chances are mighty high you are going to see a Faraday’s law, Lenz’s law, free response question.1575

Taking a look at the 2009 exam free response number 3.1581

We have this square loop of wire with a magnetic field going through it and a couple light bulbs on either side.1587

It looks like the magnetic field is a function of time, derive an expression for the magnitude of the EMF generated in the loop.1593

Right away makes me think Faraday’s law.1602

A, our induced EMF is - time rate of change, our flux derivative of our magnetic flux.1606

Which is - D/ DT of, our flux is just going to be, area that is not changing with respective to time, × B.1616

That will be - A DB DT, which would be- A × the derivative with respect to time of our function1626

for our magnetic field strength AT + B is just going to be - A × a.1638

Our area there is a square, it is just going to be L².1646

Since A = L², we can say that the magnitude of E is just going to be a L².1651

Since we are after magnitude, we do not need worry about our negative sign there.1665

E = a L².1668

Let us take a look here at B.1675

Determine an expression for the current through bulb 2.1678

The current through bulb 2 is just going to be the potential ÷ the resistance, which is going to be,1682

we got a potential drop those are identical.1689

It is going to be half the total potential a L² / 2 × our resistance R0.1692

I think that is it.1700

Indicate on the diagram the direction of the current through bulb 2.1703

As we look at that through bulb 2, it looks like we have an increasing magnetic field in the page.1708

As we have that increasing, we want to oppose that, that means you are going to have by Lenz’s law at counterclockwise current.1715

Through bulb 2, it is going to be heading up, counterclockwise current through our loops.1724

Taking a look at part C then, find an expression for the power dissipated in bulb 1.1731

The power is going to be current × voltage.1743

If we find the total, it is going to be, we got a L² / 2 R0 is the current and our potential is going to be,1749

our total potential is a L² so the potential drop a× our single bulb is going to B ⋅ ÷ 2, since they are identical.1765

I come up with, we have a² down to the 4/ 4 R0.1778

Now we are going to switch things up a bit.1797

It looks like for part D, we have added another identical bulb 3 in parallel with bulb 2,1800

completely outside the magnetic field as shown.1807

How does the brightness of one compared to what was in the previous circuit?1810

Let us see, if we have 2 and 3 in parallel, the equivalent of 2 and 3 over there is going to have less resistance than it had previously.1814

Therefore, the total current in the circuit is going to go up so the current through I1 must go up.1824

We got the same potential different.1830

It got the current going up, if I1 is going up.1832

Therefore, we should say that that should get brighter.1836

EMF is going to be the same but 2 and 3 are in parallel so as current through 1 goes up, the whole thing gets brighter.1843

Brightness is related to power.1850

We have got a part E, E says now, we are going to remove that bald 3 but add a wire down the middle.1854

How does the brightness of one compared to what was in the first circuit?1864

If you put that wire right in the middle, you have that change in flux,1869

what you are going to do is each loop is going to create a counterclockwise current.1874

In the middle, there going to each to go in opposite directions, completely cancel each other out since it is symmetric circuit and1878

the whole thing it acts as if you do not have that wire there in the middle at all.1884

It is the exact same circuit function as you had initially so that is going to be the same.1888

Again, make sure you justify your answer, explain how you came to that reasoning.1894

There are couple different path ways you can take to get there.1898

There is the 2009 E and M number 3.1902

For the next one, we are going to go to one of the more challenging questions I have seen on an AP exam.1907

This is going to the 2008 exam free response 3 which is really mostly a Biot-Savart question but1913

it does have some Faraday’s law here at the end.1921

This one is a bit involved.1924

We have got a circular loop of wire in figure 1 that has a radius R and carries some current I and point P is R/ 2 above the center.1927

First off, find the direction of the magnetic field at point P due to the current in the loop.1935

By the right hand rule, that should be heading upwards as you go around the circle.1944

A1 should be up.1949

2 however, calculate the magnitude of the magnetic field B1 at point P.1955

This is going to take a little bit of work, I think.1960

As we look at 2, let us make a diagram of what we have over here.1963

We got our loop of wire, we have got a point P up here, and we have done Biot-Savart problems that are similar.1968

We have got our radius R, we got our current flowing through our loop I.1980

This distance here is R/ 2, let us pick some point over on the side.1988

We will call that our DL and we will draw our line for the distance from P to DL.1995

That is our R vector, where it looks like R is going to be √ of R² + R/ 2², which when we simplify that,2006

that is going to come out to be the magnitude of that is R/ 2 √ φ.2023

There is our R, we will define this angle as φ.2030

I think we have got a pretty good setup.2037

As we come here from here to there, we are going to have our DB in that direction.2039

It is pretty easy to see we do are not going to have to worry about the vertical component of this.2048

Let me just continue up this line a little bit because if that is φ, it is usually helpful to know that that angle there is φ as well.2053

Let us see where we can get with what we have so far.2060

Biot-Savart, DB =, we have got μ₀ I/ 4 π × DL × R vector ÷ R³.2065

We can state that DB = μ₀ I/ 4 π R³.2086

DL × R that is going to be the DL R sin θ.2094

Θ is 90° between those so that is going to be 1, which implies that DB is going to be equal to μ₀ I/ 4 π R² DL.2105

We are only worried about the vertical component, everything else is going to cancel out here.2125

Let us see if we can integrate to find our magnetic field strength.2130

The vertical component here is going to be the integral of DB vertical which will be2137

the integral of DB cos φ which is equal to, as we look at that cos φ is going to be SOHCAHTOA.2143

Adjacent / hypotenuse that is going to be R / r.2160

This is going to be the integral of R / r DB, which implies then that vertical component is going to be the integral of R / r.2168

We already found out the DB is μ₀ I/ 4 π R² DL.2188

It is just going to be, if we pull out all our constants, you have μ₀ IR/ 4 π r².2201

I realize that we could take r³ since we know what r is in terms of R and simplify it.2212

We just leave it like this for now.2217

× the integral around our entire wire of DL is going to be μ₀ IR/ 4 π r³ × 2 π R,2219

which implies the vertical component of our magnetic field is going to be μ₀ I R² / our π2239

are going to cancel out, we are going to get 2 R³.2253

We will substitute in, it is going to be since we know that r is R / 2 × √ φ.2257

I’m going to have for our vertical component, it is going to be equal to μ₀ IR² / 2 ×, let us see.2268

We are going to end up with R/ 2 3 is going to be R³/ 8.2282

The √ of φ³ is going to be φ √ φ, so that is going to be μ₀ I.2290

Our R² can cancel R² to that, over 1/4 × φ, φ/ 4 of √ φ × R.2300

If I bring that 4 to the top, we are going to have a vertical component to our magnetic field of 4 μ₀ I/ φ √ φ R.2316

One of the tougher pieces of a problem I have seen on AP exam.2339

Let us go to our next page to take a look at part B.2346

For part B, it says determine the magnitude of the net magnetic field B at that point P.2354

We have now that the second ring here.2361

B at point P is just going to be twice what we had before, because the magnetic field due to both of those is going to end up giving you,2364

you are going to add together because they are in the same direction.2373

That is 2, what we had for B vertical is just going to be 8 μ₀ I/ φ √ φ R.2377

That is what we are going to call B net.2388

We have got a square loop of wire between those of length S on each side.2394

We are asked to determine the magnetic flux due to the square loop in terms of B₀ and S.2399

For part C, magnetic flux is the integral of B ⋅ DA is just going to be, in this case just BA.2406

The area is not changing so that is 8 μ₀ I/ φ √ φ R S².2420

It wants it in terms of B net, that is just B net.2431

Simply, that is just B net S².2434

Part B, a square loop is now rotated about an axis in its plane at angular speed ω.2449

In terms B net S and ω, find the induced EMF as a function of time and assume it is horizontal T = 0.2456

Induced EMF, we will use Faraday’s law – D φ BDT.2464

It will be - the derivative with respect to T of what we had for our flux B net S².2470

But it is rotating now at time T = 0, it is horizontal.2482

We can use a cos factor to model that of cos ω T.2484

As we take the derivative of this, we can pull the B and the S is out, those are constants.2499

That is going to be – B net S² × the derivative with respect to T of cos ω T,2507

which implies then that EMF is going to be, derivative of cos is opposite of sign.2518

Those negatives will cancel out.2524

We will have ω × our net magnetic field S² sin ω T.2527

It is the end of that question.2547

Let us go to the 2007 exam now free response number 3.2553

We have seen something similar to this by now as well.2557

We have got another rod on nails with a chrome wire but they bent into a 3 sided rectangle.2562

We have got the moving rod, that is moving to the right.2570

All in the uniform magnetic field in the plane of the screen.2575

As we do all this, indicate the direction of the current induced in the circuit as it moves to the right.2580

As it move to the right, you are increasing the flux into this page or screen so the induced magnetic field has to oppose this out of the screen.2586

The induced current by the right hand rule and Lenz’s law must be counterclockwise.2592

Counterclockwise and justify your answer.2602

I will leave you guys to explain that one in your own papers.2606

Probably it is good to practice that, do not just leave it as counterclockwise.2608

Do not just say Lenz’s law or right hand rule, walk through the explanation.2612

Part B, derive an expression for the magnitude of the induced current as a function of time T.2618

We have already solved this problem before.2627

You might have seen in previous physics courses.2629

The induced EMF is just BLV and we know that the resistance, we can figure out with a little bit of geometry.2631

We have got the resistance prior to length λ × the total length.2640

We are going to have L + the length of the top and bottom pieces which is going to be 2 VT.2647

If we want the current by Ohm’s law, it is V/ R.2655

That is going to be BLV / λ × L + 2 BT.2660

Λ × O + 2 BT should work.2675

Taking a look at parts C now, derive an expression for the magnitude of the magnetic force on the rod as a function of time.2678

The force on that rod is just going to be ILB and we just found I, that is going to be BLV/ λ × L + 2 VT.2687

And we have still got our LB which is going to be B² L² V all ÷ λ × (L + 2 VT).2704

On to part D, we are going to graph so let us give ourselves some more room for part D.2731

On the axis below ,sketch a graph of the external force as a function of time that2738

must be applied to the rod to keep it moving at constant speed.2741

Graph a force vs. Time, label this is our external force, here is our time.2748

The trick here is, in order to move at constant speed, we know that the external force must match,2763

must absolutely oppose the magnetic force so that they are balanced, equal.2770

When we are at equilibrium, we are moving at a constant speed, no acceleration.2774

We know what time 0, that force was to B² LV/ λ because FB at T equal 0 is B² L² V/ λ L.2779

Our L cancel out, B² V/ λ and over time, that is going to steadily decrease.2798

It is going to look something like that, I would expect.2807

Finally E, what do we have for part E?2815

The force pulling the rod is now removed, my goodness.2821

Indicate whether the speed of the rod increases, decreases, or remains the same, and explain why.2824

If there is no external force to keep it moving at the constant speed, the magnetic force is opposing the motion.2831

If it is opposing motion, by Newton’s second law which says that F = MA, the rod must be accelerating2841

but it is accelerating in the direction opposite its velocity, it is slowing down.2855

Therefore, our answer must be decreases.2871

Let us take a look now, go to the 2006 exam free response number 3.2883

Here is a kind of a unique one.2893

We now have our circuit that is hanging from the spring.2894

I wonder how they add up some of these but we will deal with it anyhow.2898

Part A, on a diagram of the loop, indicate the direction of the magnetic forces of any that act on each side of the loop.2905

Let us draw the loop in there, we have got our source of potential difference.2912

We got our switch and looks like it is closed.2918

There is our loop up.2925

As we look at the different portions of the loop, its top part is not in the magnetic field so we do not have to worry about that.2928

Our current is going to be flowing this way.2934

Over here on the left hand side, by the right hand rule current flowing toward the top of the screen,2937

positive charges point the fingers of your right hand for the top of the screen and then into the screen the direction of the magnetic field.2944

We should see a force to the left.2950

Similarly on the right side, we are going to see a force to the right.2955

On the bottom, we are going to see a force down.2960

Left, right, and down, that should cover part A.2966

For part B, the switch is open and the loop eventually comes to rest at new equilibrium position2972

at distance X from its former position.2978

Find an expression for the magnitude of the uniform magnetic field.2980

Let us draw a free body diagram there.2985

We have got the force of the spring KX by Hooke’s law up and we have the magnetic force down.2988

If it is coming to rest at new equilibrium position, these two have to be equal.2996

The spring force must equal the magnetic force in magnitude.3001

Therefore, KX the force of the spring must equal the force on our loop of wire in a magnetic field which by now we know as ILB.3006

Therefore, B is going to be equal to KX / IL.3016

The one trick here, our length is W in this problem.3022

I would write this as KX / IW to be consistent with the values and variables that they have given us.3028

Part B, moving on to part C of the question.3038

The spring in the loop place with a loop and the same dimensions of resistance but no battery and no switch.3047

The new loop is pulled upwards at constant speed V0.3053

On the diagram, indicate the direction of the induced current loop as it moves upward.3058

Lenz’s Law problem.3063

We have got our loop right here, we have magnetic field into the plane of the screen.3067

And we are pulling that thing upward for some unknown reason with velocity V0.3080

As we are pulling that up, the flux into the plane of the screen, the magnetic flux is decreasing.3087

It is going to want to oppose that by Lenz’s law by increasing that flux.3095

Point the thumb of your right hand into the plane of the screen.3099

The direction you want increase that by Lenz’s law and you get a clockwise induced current.3101

There is C1, let us take a look at C2.3112

Derive an expression for the magnitude of this current.3118

I could not just leave it without telling you the direction.3121

EMF is - D φ BDT, we will start there.3125

That is going to be - D / DT and our flux is the integral of the B ⋅ DA thought that open surface.3134

That is going to be -, we can pull B out, that is constant for this problem.3145

The derivative with respect to T of the integral of DA.3150

That is going to be – B, the integral of DA is just going to be A, that is not changing with the variable as you integrate / variable DA.3155

That is going to be - B DA DT, which will be - B × the derivative with respect to time of our area which is W our width.3164

Our Y coordinate, Y and W you can come out of there, that is not a constant that is not changing.3178

- W DY DT, but DY DT is just our V0 so then we can say that E, our induced EMF is – BW V0.3184

If we want the current, ohms law I = E / R.3200

I suppose we are after the magnitude of the current so let us put it that way.3206

It is going to be the magnitude of - BW V₀/ R or just BW V₀ / R.3210

Moving on to D, find the power dissipated in the loop as it pulls its constant speed out of the field.3232

Power is current × voltage, we just found our current is BW V₀ ÷ R.3242

We found our potential magnitude of that as BW V₀.3253

When I put all these together, I come up with B² W² V₀² / R.3261

That one was not so bad.3272

We are back to part E, suppose the magnitude of the magnetic field is increased.3276

Does the external force required to pull the loop at speed V₀ increase, decrease, or remain the same?3285

If you have a stronger magnetic field, you are going to get a stronger magnetic force which is equal to ILB.3292

The larger induced current which was BW V₀ / R, as B goes up I induced goes up.3305

Both of those are going to lead to a stronger force applied.3316

All lead to stronger F external.3323

We can say that that is going to increase.3331

That is the end of that question.3341

Let us do one last practice problem here.3343

We will go to the 2013 exam free response number 3.3349

We have got a loop of wire in a magnetic field.3355

It gives us the area, it gives us a nice little plot of the magnetic field strength as a function of time.3358

There is something funky about that graph too.3366

If you look at the X axis, the time in seconds.3368

We have even intervals of 4, 8, 12, 16 but at the last interval looks like it should be 4s, but it is only 2 to get the 183371

so that is not drawn to scalar, somebody made a mistake when they drew up the question.3380

Do not think that is going to affect us though but we will keep an eye for that.3384

Alright part A1, derive an expression for the magnitude of the induced EMF in the loop as a function of time from 0 to 8s.3389

In that curvy part of the graph where we are given that function for magnetic field strength.3399

We will go to Faraday’s law E = – D φ B DT, which is - the derivative with respect to time of AB which,3405

since A is a constant, will be - A DB DT which is - A × the derivative with respect to time of our function3419

for B given to us on that graph is 1.8 E⁰.05 T.3433

This implies then that E = - A and the derivative of that is going to be, the derivative of E ⁺U is E ⁺UDU.3443

We will have our 1.8 E⁰.05 T × -0.05.3454

Putting all that together that is going to be our negative area 0.25 m².3465

We got our 1.8 E⁰.05 T, we have got our -0.05, which implies then that our total induce EMF is going to be 0.0225 E⁰.05 T.3471

There is our induced EMF as a function of time from 0 to 8s.3496

I guess that is A2, calculate the magnitude of the induced current at 4s.3502

Current is E/ R V/ R by Ohm’s law, is just going to be 0.0225 E⁰.05 T/ 12.3512

Since, we said T is = 4, I plugged into my calculator and I come up with a value of about 0.00154 amps or 1.54 milliamps.3528

Let us take a look now at part B here.3549

Give ourselves some room.3553

It looks like it is a graph in question.3554

We will draw our graph here first.3557

We have current in amps vs. Time in seconds.3574

We got 4, 8, 12, 16, and then they have another evenly spaced.3582

I’m going to draw mine right and say that is 18.3588

And 4, 8, 12, 16, 18s for our graph.3591

Sketch a graph of the induced current of loop as a function of time assuming a counterclockwise current is positive.3597

This is counterclockwise up here, this is clockwise.3605

If I were to draw that graph, I'm going to start with the easy parts first.3610

Recognizing that where we do not have any change in magnetic flux it is going to be 0 so from 8 to 12 s, it is got to be 0.3615

We do not have any DB DT is 0.3623

And beyond 16s, 16 to 18 is 0.3626

Now, let us also look in the region from 0 to 8s.3631

It is going to be a clockwise current Lenz’s law and it is going to go something like that.3638

And from 12 to 16s, we have got a strong negative slope but it is constant so we should have a strong negative or clockwise current right there.3646

That should lineup with 12 and 16s.3659

I would do that for our graph, part 2 B2 says for the time interval from 12 to 16s, justify the direction of the current that you have indicated.3663

By Lenz’s law, the magnetic field is in the page and it is decreasing.3675

The induced current creates a field into the page that is opposing that change which is clockwise by the right hand rule.3678

There is part 2, write that out somehow in your own words, of course.3688

For part C, let us take a look at C here.3692

Calculate the total energy dissipated in the loop during the first 8s.3698

Energy is going to be the integral of power with respect to time so that is going to be the integral of E²/ R DT3704

to be the integral from T = 0 to 8s of 0.0225 E⁰.05 T all of that squared ÷ our 12 ohms DT.3717

Pulling out our constants, that will be 0.0225²/ 12 integral from 0 to 8s of E⁰.1 T DT.3735

This implies then that our energy is going to be, all of that comes out to about 4.22 × 10⁻⁵.3752

We get to have that E ⁺U DU, we need to have a DU there so it is going to be -0.1.3764

We got to pullout a 1/ -0.1 infinity integral sign, integral from 0 to 8 of E⁰.1 T × -0.1 DT.3771

We can fit the form E ⁺U DU integral of E ⁺U DU is just E U.3787

That is going to be negative 4.22 × 10⁻⁴, as we multiply those together,3792

× E⁰.1 T which we are going to evaluate from 0 to 8s, which implies then that E = -4.22 × 10⁻⁴.3802

E⁰.8 - E⁰, E⁰ is 1 so this is just going to be -4.22 × 10⁻⁴.3821

E⁰.8 is 0.449 – E⁰ is 01, put that all into my calculator and I come up with about 2.32 × 10⁻⁴ J.3834

I think that finishes out our look at Faraday’s law and Lenz’s law with a bunch of sample problems.3855

Keep practicing these, you will see a lot of the same patterns and forms come up again and again and again.3864

Thank you so much for watching www.educator.com.3869

Make it a great day everyone.3872