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

### Ampere's Law

- Ampere’s Law provides a method of finding the magnetic field due to current flowing in a wire in situations of planar and cylindrical symmetry.
- The integral of the magnetic field dotted with the differential of wire around a closed loop is proportional to the current penetrating the closed loop.

### Ampere's Law

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 0:00
- Objectives 0:07
- Ampere's Law 0:27
- Finds the Magnetic Field Due to Current Flowing in a Wire in Situations of Planar and Cylindrical Symmetry
- Formula
- Example
- Example 1 2:19
- Example 2 4:08
- Example 3 6:23
- Example 4 8:06
- Example 5 11:43
- Example 6 13:40
- Example 7 17:54

### 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: 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 loop*0083

*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 wire*0223

*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 loop*0321

*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 where*0420

*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 that*0537

*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 penetrating*0662

*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 couple*0693

*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 is*0716

*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 screen*0751

*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 probe*1357

*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 × 10⁻⁶*1484

*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

1 answer

Last reply by: Professor Dan Fullerton

Mon Feb 1, 2016 6:08 AM

Post by Shehryar Khursheed on January 30, 2016

Also, on the questions asking for the slope of the best fit line and using it to find estimations for certain values, do you happen to know the margin of error to still get credit? For example, on the 2011 FRQ, I got .0667 tesla per meter and a permeability of 2*10^-6 for part d. Would I still get credit for these answers which were a little off from college board's official answers?

1 answer

Last reply by: Professor Dan Fullerton

Mon Feb 1, 2016 6:07 AM

Post by Shehryar Khursheed on January 30, 2016

Why isn't there an electric field at point P? I thought that the current traveling in the wire would cause an electric field? This is in the 2011 AP free response, part c.

0 answers

Post by Michael Norton on November 11, 2015

Oh wait I think I get it. Do you put your pointer finger in the direction of the second wire and then your middle finger in the direction from the second wire to the first wire?

1 answer

Last reply by: Professor Dan Fullerton

Mon Nov 23, 2015 7:27 AM

Post by Michael Norton on November 11, 2015

On Example 5, Can you explain how the right hand rule works on this problem and how we know that the magnetic field is going into the screen? Thanks!

0 answers

Post by Professor Dan Fullerton on April 19, 2015

Inside a (with a radius less than a), the penetrating current is zero. If there's no current penetrating, there's no magnetic field.

0 answers

Post by Miras Karazhigitov on April 19, 2015

I didnt get the part in the Example 7, why E at point P is 0? Didn't we justified that the value of the E is (uI)/(4Ð¿b)?