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 4 answersLast reply by: Professor Dan FullertonWed Oct 5, 2016 11:55 AMPost by Jessie Sun on March 17, 2016Hello 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 answerLast reply by: Professor Dan FullertonWed Apr 15, 2015 9:38 AMPost by Luvivia Chang on April 15, 2015Hello Professor Dan FullertonWill 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 answerLast reply by: Professor Dan FullertonTue Mar 24, 2015 6:04 AMPost by Arjun Srivatsa on March 24, 2015How often is the Biot-Savart Law Tested on AP? Is Ampere's Law used much more frequently?

### 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

### 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 field0008

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 field0061

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 vector0193

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 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 DL0449

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 that0488

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

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 field1095

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 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 and1295

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