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

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## Table of Contents

## Transcription

## Related Books

### The Biot-Savart Law

- A small section of current-carrying wire will create a small magnetic field. If you add up all the little bits of magnetic field created by all the small sections of wire, you will obtain the total magnetic field at a given point. The Biot-Savart Law provides you a method of calculating the magnetic field due to that small section of current-carrying wire.
- Symmetry arguments can be extremely useful in simplifying calculations using the Biot-Savart Law.
- The dl vector is a vector pointing in the direction of positive current flow for a differentially small section of wire.
- The r vector is a vector pointing in the direction from dl to the point of interest.

### The Biot-Savart 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
- Biot-Savart Law 0:24
- Brute Force Method
- Draw It Out
- Diagram
- Example 1 3:43
- Example 2 7:02
- Example 3 14:31

### 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: The Biot-Savart 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 the Biot-Savart Law.*0004

*Our objectives are going to be to deduce the magnitude of direction of the contribution to the magnetic field*0008

*made by a short straight segment of the current carrying wire.*0014

*Apply the expression for the magnitude of the magnetic field on the axis of the circular loop of current.*0018

*Now as we get into the Biot-Savart Law , please understand that this is probably the most difficult topic in the entire E and M course.*0025

*As we go through these derivations, you are probably not going to make sense the first time through.*0034

*Maybe not even the second.*0040

*These take some work some hunkering down, they are not easy concepts.*0041

*Let us start by talking about what the law is.*0046

*It is a brute force method of finding the magnetic field into the length of current carrying wire.*0049

*To draw this out for you, let us assume that we have a current carrying wire here in black.*0055

*What we are going to do is we are going to find the contribution to the magnetic field*0061

*at some point up here due to just a little bit of the wire down here.*0065

*To find the entire magnetic field, we would add up all of these little sections,*0069

*all of the magnetic field contributions to the all of these little sections of wire DL, to get the entire magnetic field.*0074

*It is not a straightforward process.*0083

*It is not an easy formula, it is kind of a brute force method.*0085

*In our following lesson, we will talk about it more streamline method you can use when you have certain symmetry.*0088

*Let us start off with a quick little diagram.*0096

*I'm going to say that we have our little bit of current right there.*0098

*There is current and we want to know the magnetic field strength.*0107

*Let us say somewhere over here, that will be our little bit of magnetic field.*0113

*I’m going to define then a line from my little bit of current to that magnetic field.*0118

*And I could call that entire vector R or I can define the unit factor in that direction R ̂.*0128

*I think that will give us the basis for our analysis.*0136

*If we look at the whole thing, that would be vector R.*0141

*Just that little bit the unit vector in that direction R ̂.*0144

*You will see the Biot-Savart Law written in several different forms.*0147

*One form, the differential of the magnetic field DB is going to be equal to μ₀ I/ 4 π DL × with R ̂ ÷ R².*0150

*There we are using the unit vector R.*0173

*Or same basic diagram, same basic setup, you may also see it written as μ₀ I/ 4 π DL × R vector not R ̂, ÷ R³.*0176

*And those are equivalent because R ̂ is just the R vector ÷ the magnitude of the R vector*0193

*which is going to be R vector / R³ is going to be equivalent to R ̂ / R².*0202

*That is the Biot-Savart Law.*0210

*Now implementing it, using it takes a little bit of finesse, a little bit of practice.*0213

*Let us go through some examples where we actually use it.*0219

*Derive the magnetic field due to a current loop at the center of the loop.*0224

*Let us draw in a loop of current, there we go.*0229

*I have the current moving in that direction, we call that I.*0236

*From the right hand rule, if you look at any pieces here,*0239

*we should be able to see that we are going to have a magnetic field coming out of the center of the loop.*0242

*That is going to be the direction of our magnetic field.*0247

*To do this, let us also define a little bit of our loop of current right there.*0251

*We will call that one DL, going in that direction.*0257

*Our R ̂ vector must be going towards the center, toward where we want to know our magnetic field strength.*0264

*There is R ̂.*0274

*We can write our Biot-Savart Law , DB = μ₀ I/ 4 π DL × R ̂ ÷ R².*0276

*If we want the total magnetic field B, we need the integral of DB, which is the integral of μ₀ I/ 4 π × DL × R ̂ ÷ R².*0293

*We are going to have to integrate that all the way around the loop to add up all those little bits to DL.*0313

*What do we know here? What can we do?*0321

*As I look at DL × R ̂, that is always going to be perpendicular.*0324

*DL × R we know is DL sin θ if we want the magnitude.*0333

*Θ is going to be 90° so sin θ is going to be equal to 1.*0338

*That makes it a little simpler.*0344

*If we pull the constants out here, we can state that B = μ₀.*0346

*That is not a very pretty, let us try that again.*0353

*B = μ₀ I/ 4 π R², those are all constants, we can pull that out.*0356

*We are going to be left with is the integral of DL.*0365

*As we go all the way around a wire.*0374

*As we go all the way around the wire, the integral of DL is just going to be its circumference 2 π R.*0376

*Therefore, B is going to be equal to μ₀ I/ 4 π R² × 2 π R,*0384

*which implies with a little bit of simplification that B is going to be equal to μ₀ I / 2R.*0400

*Using the Biot-Savart Law in order to find the magnetic field due to current loop at the center of the loop.*0411

*Let us do a little bit heavier example here.*0420

*Finding the magnetic field due to a long straight current carrying wire.*0424

*Derive the magnetic field strength due to point P located at distance R from an infinitely long current carrying wire using the Biot-Savart Law.*0429

*We already talked in our last lesson about what the answer is going to be.*0437

*We know we are going to get μ₀ I/ 2 π R.*0441

*How did we come up with that?*0444

*We should be able to do that using the Biot-Savart Law.*0446

*What we are going to do is I'm going to start over here by defining our little bit of DL*0449

*which is going to be to the right, in that direction.*0456

*There is DL.*0458

*We also are going to define a radius, a distance to point B.*0460

*Let us draw that in here, something like that.*0466

*There is our R vector, or if we want we could define R ̂ as the unit vector in that direction.*0471

*We will call this angle θ and we have got our distance from the wire R.*0481

*As we look at this by the right hand rule, it should be pretty easy to see that*0488

*the direction of the magnetic field is going to be out of the plane of the screen.*0491

*Let me start off by writing the Biot-Savart Law, DB = μ₀ I/ 4 π R² DL × R ̂.*0499

*As we look at DL × R or R ̂, that is just going to be we got DX and we have got the sin of the angle between them.*0521

*That is going to be DX sin θ.*0533

*DB = μ₀ I DX sin θ/ 4 π R².*0541

*A little bit more work to do here.*0557

*We have got a couple different variables and we got R in here, we have got θ.*0559

*As I look at R, R is just going to be our X coordinate² + R².*0563

*We can write R² = X² + R² and our sin θ as we look at that, sin θ is going to be our opposite/hypotenuse.*0570

*That is going to be our opposite is R, our hypotenuse is going to be √ x² + R².*0587

*Then, DB = μ₀ IDX R / 4 π × X² + R²³/ 2.*0598

*If we want the entire magnetic field, we are going to have to integrate this.*0617

*Which implies then that B = the integral of DB, which is going to be the integral and our variable is X.*0621

*We are going to go from X = -infinity to X = infinity, all the way from the left, all the way to the right.*0629

*It is a very long wire of μ₀ I DX R/ 4 π X² + R²³/2.*0638

*Let us pull that constants we can out of this integration.*0656

*B is going to be equal to μ₀ I R are all constants.*0660

*We can pull out our 4 π in the denominator and we are left with the integral from -infinity to infinity of DX/ X² + R²³/ 2.*0668

*Some of you guys with mad calculus skills may be able to integrate that.*0688

*But that one on my own is a bit beyond me so what I'm going to do is I'm going to go to the front and back of my calculus book,*0693

*look inside the front cover and find a table of integrals in order to integrate that form of DX/ X² + some constant²³/ 2.*0699

*As I do that, let us go on our next page to give ourselves some room.*0713

*I can find that formula and I did.*0717

*On the next page, we will keep writing it, that means that B = μ₀ I R/ 4 π.*0719

*When I use my table of integrals I come up with X/ R² × the quantity R² + X² ^½.*0729

*And that is evaluated from -infinity to infinity.*0743

*Substituting in my limits, my infinity is there.*0748

*We get μ₀ I R/ 4 π ×, we have got infinity/ R² × R² + infinity² ^½ - -infinity / R² × the quantity R² ± infinity² ^½.*0750

*This might make some math teachers roll over in their graves a little bit, that is all right.*0784

*As I look at this, infinity/ an infinity², if we were to look in the limit, √ something + infinity² is something is not going to make a difference.*0790

*√ something + infinity² is infinity.*0800

*Infinity/ infinity is going to give us, this left hand side is going to become 1/ R².*0803

*This right hand side is going to become, we will have -1/ R².*0808

*For the same reason that a - -1, we are just going to end up with another + 1/ R².*0814

*This implies then that B = μ₀ I R/ 4 π 1/ R² + 1/ R² which is μ₀ I R/ 4 π × 2 / R², complies then that B = μ₀ I/ 2 π R.*0821

*The answer that we were expecting.*0853

*There are other ways to actually solve this setup.*0855

*You can actually integrate over θ as you do different things.*0857

*There are lots of ways to go about solving it.*0860

*None of them are overly pretty but I thought that was the most straightforward methods to show you.*0862

*Let us give a shot to one more sample problem.*0867

*Derive the magnetic field due to a current loop at a point out of the plane of the loop that is centered on the loops axis, up here at point P.*0872

*The first thing I'm going to notice is by symmetry we really only need to worry about the magnetic field in the Y direction, the J ̂ direction.*0882

*That will help and let us see if we can set this up a little bit.*0890

*I’m going to pick some point of our current carrying wire here and we will define our DL.*0897

*There is DL, we have got to draw the line from there to our point P.*0906

*Let us get that all lined up, there we go.*0916

*The magnetic field by the right hand rule is going to be perpendicular to that.*0920

*90° from there, we are going to have our db from that portion of wire.*0924

*We will draw that roughly that way.*0931

*Let us draw a radius over here from the center of the circle to our point DL.*0935

*There is our radius there.*0946

*If the angle between that radius line and our R from the point to P, let us call that φ.*0948

*If that is φ and this must also be φ in geometry.*0957

*Our R is going to be equal to, by the Pythagorean Theorem, that will be √ our radius of our loop² + Y coordinate² where P resides.*0964

*When we start that at 000.*0977

*And I think we are pretty well set up with all the pieces we are going to need.*0981

*If not, we will come back and get them in a few moments.*0986

*We will start by writing Biot-Savart Law, DB = μ₀ I/ 4 π × DL × R ̂ ÷ R², which implies.*0990

*We know that R² = R² + Y² that we found over here.*1013

*DB is going to be equal to μ₀ I/ 4 π.*1027

*Let us see, our DL × R that is just going to be our DL sin θ/ R² + Y² in place of our r².*1035

*This implies then, since our angle is going to be 90° here, between our angle θ is going to be 90° so θ = 90° sin θ sin 90° is going to equal 1.*1055

*That DB is going to be equal to μ₀ I DL/ 4 π × R² + Y².*1076

*Recognizing again that our magnetic field is going to be in the Y direction, we can state that the total magnetic field*1095

*is going to be the integral of the Y component of DB which is going to be the integral of DB.*1103

*If we want the Y component, if that is our angle θ, the Y component is the adjacent side.*1110

*That is going to be DB cos φ which is cos φ is going to be the adjacent / the hypotenuse.*1116

*Cos is adjacent/hypotenuse.*1128

*Cos φ is going to be the adjacent side which is going to be R ÷ the hypotenuse √ R² + Y².*1131

*That is cos φ and let us go on to our next page to continue that, to give ourselves more room again.*1145

*So then B = μ₀ I/ 4 π.*1152

*We have a RDL in the numerator ÷ R² + Y²³/2.*1159

*We have just got the math piece left, we have really done all the physics.*1170

*Which implies then that B = of course our integral.*1175

*We will pull out our constants, we have got μ₀ IR in the numerator are all constants, μ₀ I R.*1182

*In the denominator, we have 4 π R² + Y²³/ 2.*1189

*None of that is a function of where we are as we integrate around our loop.*1194

*All of that is a constant for this problem.*1199

*R² + Y²³/2 and I'm just left with the integral of DL.*1202

*Thankfully, after setting all this up, the actual integration is pretty easy because the integral DL around that loop is just going to be 2 π R.*1209

*Then we can state that B = μ₀ IR/ 4 π R² + Y²³/2 × 2 π R.*1218

*Just finally, one last up to simplify this a little bit.*1235

*D = μ₀ IR, we got our π cancels up there.*1239

*Μ₀ IR, this is 2 π R.*1245

*That R is actually, I wrote that wrong.*1254

*That is 2 π R because we are going around that loop 2 π R.*1257

*Μ₀ μ₀ I R² ÷, 2 ÷ 4 is going to give us ½.*1261

*2 × R² + Y²³/ 2.*1269

*I will put that in a 3D box because we are done.*1281

*Hopefully that gets you a good start on Biot-Savart Law.*1285

*It is a tricky law and very tough to implement.*1288

*The concept is simple, actually using it definitely takes some experience and practice.*1290

*Do that and in the next lesson we are going to come up to Amperes law and*1295

*talk about other ways to find the magnetic field when you have a certain symmetry considerations you can use.*1299

*Thank you so much for watching www.educator.com.*1305

*We will see you soon, make it a great day everybody.*1308

4 answers

Last reply by: Professor Dan Fullerton

Wed Oct 5, 2016 11:55 AM

Post by Jessie Sun on March 17, 2016

Hello Professor Fullerton,

Can you please explain why dl cross r hat is equal to dx?

And why dl cross r hat in the previous example is just dlsin(theta)?

1 answer

Last reply by: Professor Dan Fullerton

Wed Apr 15, 2015 9:38 AM

Post by Luvivia Chang on April 15, 2015

Hello Professor Dan Fullerton

Will I be provided with a Table of Integral during an AP physics C test?

If not, what can I do if I derive the functions but cannot finish the integral part? Such as the difficult integral in example 2.

1 answer

Last reply by: Professor Dan Fullerton

Tue Mar 24, 2015 6:04 AM

Post by Arjun Srivatsa on March 24, 2015

How often is the Biot-Savart Law Tested on AP? Is Ampere's Law used much more frequently?