AP Physics C: Electricity & Magnetism Ampere's Law

Hello, everyone, and thanks for joining us here at www.educator.com.0000

I'm Dan Fullerton and in this lesson we are going to talk about in amperes law.0004

Our objectives are going to be to state and apply amperes law.0008

Write symmetry arguments and the right hand rule to relate magnetic field strength to current per plane or cylindrical symmetries.0011

Apply the super position principle to determine the magnetic field produced by combinations of Biot-Savart Law configurations.0017

Let us start by talking about what amperes law is.0028

This provides a much more elegant method of finding the magnetic field due to the current flowing in a wire.0030

The trick is you have to have situations in a plane or in cylindrical symmetry.0036

And here is what the amperes law says.0040

The integral / a closed loop, note now this is not close surface, it is a closed loop.0042

The magnetic field ⋅ted with DL, that path you are taking in that loop = μ₀ × the current that penetrates that loops.0047

A lots of times I will write this as μ₀ I pen, for penetrate.0055

As an example, let us draw a bit of wire coming out toward you.0061

There is our current.0066

We could define some closed path and we will use symmetry.0068

We will make it a circular path in order to make it nice simple and symmetric.0074

Where our magnetic field, we can tell from the right hand rule is going to be in that direction.0078

But we can define DL around that loop, the whole thing is done at some radius R from our amperian loop0083

and then state that the integral / the closed loop.0096

Lots of times I will keep writing down here closed loop of D ⋅ DL = μ₀ I penetrating.0101

D ⋅ DL in this case, that is going to be B and around that loop, that integral of the closed loop is going to be 2 π R = μ₀ I.0112

We can go and say B = μ₀ I / 2 π R, where I is penetrating that is the current that penetrates that loop.0123

Let us put it into practice a little bit more formally with the same basic example problem but state it little differently.0136

Find the magnetic field outside the current carrying wire.0141

We did this with the Biot-Savart Law but my goodness it is much easier with amperes law.0146

We will draw our current, as we look at this end on, there it is I.0151

We have some radius of our wire itself.0158

There is our wire with radius R, and we want to know what the magnetic field is at some point outside that.0164

We will do that at some radius r and I'm going to draw our amperian loop here in red.0170

We will call this r.0180

That 1 r loop we have some small amount of that loop DL and we are going to have the magnetic field B.0184

If we want to know that magnetic field will apply amperes law, integral / the closed loop of B ⋅ DL = μ₀ × whatever current penetrates that loop.0196

The left hand side is going to be B × the length around our entire loop 2 π R = μ₀.0214

In the I that is penetrating our loop, it looks like all of this current I running in the wire0223

is coming through is inside of our amperian loop.0229

That is just going to be our total I like we said before B = μ₀ I/ 2 π R.0234

Nice and slick.0245

What if we want to look inside the wire?0247

We want to know the magnetic field in here.0250

For the wire that is a function of radius, let us draw this again.0253

I'm going to draw my wire first hear in green, there it is with some radius R.0257

We know the current is coming out toward us so I'm going to indicate that in blue.0263

We will draw our amperian loop now inside the wire.0268

There we go, with some radius r.0273

Now when we apply amperes law, the integral over the closed loop of B ⋅ DL = μ₀ I penetrating.0278

The left hand side stays the same, that is going to be B × once around our amperian loop 2 π R, as we integrate that.0294

The right hand side, we still have μ₀ but how much current is actually penetrating?0304

Current is flowing through the entire wire, we need to know the amount that is actually flowing through this section over here.0308

That is going to be a ratio of their areas.0318

That is μ₀ ×, that will be π r², the area of our amperian inside our amperian loop0321

divided by the area of the entire wire, π R² × of course, the current.0330

Then B =, what do we have here.0347

Μ₀ I, we still have a r in the numerator divided by, we have 2 π from the left hand side and we have r² left over from here.0351

Μ₀ I R/ 2 π R².0370

A linear function of radius while you are inside the wire.0375

Let us take a look and see if we can graph that to make more sense of it.0380

As we graph the magnetic field of the current carrying wire as function of the distance from the center of the wire, we will start with our axis.0385

We have our magnetic field strength B.0405

We have our radius r.0409

At some point here, we are going to have our point R right when we go from inside the wire itself to outside the wire.0411

We just found that inside the wire, we had a linear relationship looks kind of like that where0420

we said B was equal to μ₀ I/ 2 π R² × r radial distance.0435

It increases linearly until we get to the edge of the wire.0446

Right at this point, we could plug in R for r to find out that right here our B is equal to μ₀ I/ 2 π R.0451

From there, we have that inverse relationship where B = μ₀ I/ 2 π r.0466

An inverse relationship with the distance from the center of a wire.0477

That is what it would look like as we combine it and put it all together.0481

We can also use amperes law to find the magnetic field in the solenoid, a coil of wire.0486

In this case, we are going to take a look at current flowing this way through the wire with multiple loops.0493

We are going to make the assumption that the magnetic field is 0 outside the solenoid, a reasonable assumption.0500

You look at the cross section, you can see the current coming out up above and in down below.0507

I think that will set us up but what I'm going to do to help solve this is I'm just going to redraw our wire over here.0513

Their choice of amperian loop is probably not intuitive, unless you have done this a few times.0519

Let us start by drawing our wire.0524

Our solenoid, I should say, loop of wire.0532

We will define the entire length of our wire as L and my amperian loop that0537

I'm going to draw is going to be a rectangle but it is partially outside that solenoid.0545

I'm going to define the width of my amperian loop as l.0555

As I look at this, we have got some nice things going on here.0563

Outside the solenoid, the magnetic field is 0 so we do not have to worry about this part.0567

And let the number the sides 123 and 4.0574

2 does not have any magnetic field that we have to worry about.0580

1 and 3 are parallel to the current so we are not going to have any effect there.0583

We really only have to worry about the section 4, when we do our closed loop.0590

Let us apply amperes law, the integral around a closed loop of B ⋅ DL = μ₀ × I penetrating that loop,0595

which implies then, since I penetrating, how much current is penetrating that loop?0613

We have got the ratio of this length to the total length L / L × the total number of loops N, so L/ L, l/ L × N, tells you the number of loops that are penetrating our amperian loop.0622

And then we have to multiply that by the current that is in each time, so that will be × I.0632

Every time it goes through, we count that as penetrating our loop once.0649

Once we do that, we can take a look and our closed loop since we are only worried about section 4, 1,2, and 3 have no contributions.0654

We have B × l, the length of our section 4 of our wire must equal μ₀ × I penetrating0662

which we said was l/ L, the number of loops M × I or B = M/ L μ₀ I.0673

Nice and slick application of amperes law.0689

Let us take a couple more of these force magnetism type examples before we get into a couple0693

in AP problems to finish out our discussion of amperes law.0700

A rectangular wire loop carrying current I lies a distance R to the right of a long wire carrying current I2.0705

What is the direction of the net force on the loop?0712

The first thing I'm going to do if I want to know the net force on this loop is0716

I'm going to find the magnetic field due to I2 that is going to affect it.0719

If I2 is traveling up the screen, that means using the right hand rule.0723

Our magnetic field over here is going to be in to the plane of the page.0727

Knowing that, I’m then going to look at these different sections and see what we get for force with each section.0738

As I look over here on the right hand section, we have positive current traveling up the screen,0744

bend your fingers of your right hand 90° in the direction of the magnetic field into the plane of the screen0751

and I come up with a force here to the left.0756

At the top is the current as traveling to the left of the screen, I get a force down.0761

To the bottom, as the current is traveling to the right, I get a force up.0767

And these should be balanced because they are symmetric.0773

In this section, as the current travels down I get a force to the right.0776

However, as I look at this we have the strongest force over here because we are closest to the wire, we have got the strongest magnetic field.0785

Therefore, we should draw arrow a little bit bigger than that arrow.0795

I would say that the net force then must be to the right.0799

Let us see, rectangular wire loop current I lies at distance R along wire current,0810

what is the direction of the net force on the loop, very good.0814

Let us take a look at a couple of AP problems.0820

We are going to start with a question from the 2011 E and M exam free response number 3.0823

Take a minute, go find that, google it, go to the address here.0829

Print it out, try it for a couple minutes.0833

Pause it, then come back and we can either check them or if you got stuck, see how we can get to pass that point we got stuck as we go through it.0835

We pull that out and in 2011 E and M question 3, we have a long conducting cylinder with an inner radius A,0842

outer radius B, carrying a current I₀ because it is a uniform current density which they show you.0853

Using amperes law, find the magnitude of the magnetic field in those regions.0859

Let us start with region 1, we will use amperes law, integral/ the closed loop of B ⋅ DL = μ₀ I penetrating.0865

But because we are inside that entire thing, there is no penetrating current.0882

If there is no penetrating current then our magnetic field must be 0 for our inside that, for R less than A.0888

For part 2, this is one we are actually inside that wire or inside that cylinder.0902

Integral/ the closed loop of B ⋅ DL = μ₀ I penetrating.0909

I should mark here closed loop.0918

The left hand side of course is going to be B × 2 π R, our distance as we integrate around our close amperian loop 2 π R = μ₀ I penetrating.0923

But I penetrating is a little tricky here.0936

We have μ₀ and to get the amount that is penetrating, let us take the entire current but how much is actually penetrating our entire loop?0938

That is going to be the ratio of the areas of our cylinder which is going to be π R² - π A².0947

In the numerator we have to divide that by the inner current density of π B² - π A².0956

There is our penetrating current.0972

To solve for B, B = to μ₀ I₀/ 2 π R × the quantity and we can factor out our π here of R² - A²/ B² - A².0977

There is our magnetic field for part 2 inside that cylinder.1006

And for part 3, let us take a look here.1010

When we are at 2B, when the integral/ the closed loop of D ⋅ DL is equal to μ₀ I penetrating.1016

In this case, our R = 2B.1033

The left hand side B × 2 π, our R again is 2 B = μ₀.1037

The current that is penetrating that is that entire current I₀.1044

Therefore, B = μ₀ I₀/ 4 π b.1049

Good start here, let us move on to part B, where it says now we are considering a long solid conducting cylinder.1067

I will do it in the next page to give us a room here.1074

It tells us the magnitude of the magnetic field is given by that function.1079

An experiment is conducted in which we are measuring the magnetic field as a function of R in the data as tabulated.1083

It asks us to graph that question then.1091

I think I skipped part B.1100

We have to get to part B first, we get a cross sectional view.1105

On that cross sectional view, we need to indicate the direction of the field at point P.1108

That it easy, there we go.1114

At point P, if we have current coming out of here by the right hand rule, that is just going to be perpendicular to that radius so there is B.1116

And at point C, an electron is at rest at point B, describe any electromagnetic forces acting on the electron.1123

Let us see, the electric force is Q × the electric field.1134

There is no electric field so there is no electric force.1137

We have the magnetic force = Q V × B and V equal 0.1143

Therefore, there is no magnetic force.1152

There are no electromagnetic forces on it.1155

Let us go to part D in our graph.1162

Let us give ourselves a little bit more room here again too.1165

Drawing our graph, let us label it something like our magnetic field strength in tesla vs. Our distance in meters.1169

I will let you pick the points but I use something like 2 × 10⁻⁴ all the way up to 6 × 10⁻⁴.1190

Here we went 0.002, 0.004, and so on, all the way until you get to about 0.01.1197

Plot your points and draw your best fit line and get something that hopefully looks fairly linear.1205

Once you have done that, use the slope of your line to estimate the value of the permeability μ₀.1218

Slope is rise /run so pick 2 points on the line, not data points to calculate your slope but be careful with the units.1225

And I ended up with something around 0.0635 tesla/ m from my slope.1236

How do we use that to find permeability?1244

We have our equation B = μ₀ I₀ / 2 π B² × R.1248

Fitting this to the form of a line Y = MX.1257

Our slope should be μ₀ I not/ 2 π B².1262

Therefore, if we wanted to just find μ₀, slope = μ₀ I₀ / 2 π B².1268

Let us change up our color a little bit.1282

Which implies then that μ₀ = 2 π B² / I₀ × our slope.1284

We can plug in our values, μ₀ = 2 π × 0.01 m² divided by our I₀ which is 25 amps × our slope 0.0635 tesla / m.1295

And I get about 1.6 × 10⁻⁶ tesla m/ amp.1314

Which is reasonably close to 4 π × 10⁻⁷ which is about 1.26 × 10⁻⁶.1323

We are at least in the ballpark there.1330

That is the 2011 question, let us take a look at the 2005 exam free response number 3.1333

Take a minute, go download it.1341

Print it out, pause the video for a second, give it a try.1344

Come back here and if you had a chance to look at it.1346

It starts out we have this experiment when I do in my classroom all the time.1351

We have got a slinky, we are putting some current through it.1354

Put the variable voltage supply measuring the current and we use a hole probe1357

which is a tool that helps you measure magnetic field of a point.1362

What it is going to have you do for part A is fill in N, the number of turns / m.1365

For the first one for example, when you are at 40 cm, the turns / m, I will let you have 100 total turns divided by 0.4 m, that is 250 turns / m.1371

You just keep filling that out for the 5 different trials to get your data points.1388

Then for part B, what we are going to do is plot the magnetic field vs. N, the turns / m.1395

And when I do a plot like that, I should get the graph that looks roughly linear.1406

And I ended up with points when I do my best fit line somewhere over there in my graph, if I use it carefully.1419

Plot the points, try your best fit line.1428

And then for C, it is going to say obtain the value of permeability using that graph.1432

As we look at what we have there for B, we are going to have to move on.1440

Let us give ourselves a little bit more room for our calculations here.1445

For part C, B = and μ₀ N × our current.1451

Therefore, μ₀ = B/ NI which is going to be the slope divided by the current.1457

Then we can find our slope using points on the line, not data points.1468

Rise / run, I came up with something around 3.81 × 10⁻⁶ tesla m/ turn.1474

And if μ₀ = slope/ I, let us use that μ₀ = slope/ our current which would be our 3.81 × 10⁻⁶1484

divided by our current 3 amps or about 1.27 × 10⁻⁶ tesla m/ amp.1500

As we just mentioned, it is awfully close to 4 π × 10⁻⁷, the accepted value which is 1.26 × 10⁻⁶.1510

For part B, it says using that theoretical value, determine the percent error in the experimental value compared and computed in part C.1518

Our percent error is the absolute value of our actual result - the accepted value 4 × 10⁻⁷ all divided by the accepted, × 100%.1528

The absolute value and this was bigger, we are going to have our 4 π × 10⁻⁷ - 1.27 × 10⁻⁶.1548

I think I did that in a reverse order but it is absolute value, it will work out just fine.1563

Over our 4 π × 10⁻⁷ × 100%, which gives you 1.06%.1568

Pretty good, enough fun experiment to actually try because it usually works out very well.1581

Thank you for watching www.educator.com.1586

We will see again soon, make it a great day everybody.1589