WEBVTT physics/ap-physics-c-electricity-magnetism/fullerton
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Hello, everyone, and welcome back to www.educator.com.
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I'm Dan Fullerton, and in this lesson we are going to talk about the Biot-Savart Law.
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Our objectives are going to be to deduce the magnitude of direction of the contribution to the magnetic field
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made by a short straight segment of the current carrying wire.
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Apply the expression for the magnitude of the magnetic field on the axis of the circular loop of current.
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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.
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As we go through these derivations, you are probably not going to make sense the first time through.
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Maybe not even the second.
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These take some work some hunkering down, they are not easy concepts.
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Let us start by talking about what the law is.
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It is a brute force method of finding the magnetic field into the length of current carrying wire.
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To draw this out for you, let us assume that we have a current carrying wire here in black.
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What we are going to do is we are going to find the contribution to the magnetic field
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at some point up here due to just a little bit of the wire down here.
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To find the entire magnetic field, we would add up all of these little sections,
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all of the magnetic field contributions to the all of these little sections of wire DL, to get the entire magnetic field.
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It is not a straightforward process.
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It is not an easy formula, it is kind of a brute force method.
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In our following lesson, we will talk about it more streamline method you can use when you have certain symmetry.
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Let us start off with a quick little diagram.
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I'm going to say that we have our little bit of current right there.
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There is current and we want to know the magnetic field strength.
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Let us say somewhere over here, that will be our little bit of magnetic field.
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I’m going to define then a line from my little bit of current to that magnetic field.
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And I could call that entire vector R or I can define the unit factor in that direction R ̂.
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I think that will give us the basis for our analysis.
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If we look at the whole thing, that would be vector R.
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Just that little bit the unit vector in that direction R ̂.
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You will see the Biot-Savart Law written in several different forms.
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One form, the differential of the magnetic field DB is going to be equal to μ₀ I/ 4 π DL × with R ̂ ÷ R².
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There we are using the unit vector R.
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Or same basic diagram, same basic setup, you may also see it written as μ₀ I/ 4 π DL × R vector not R ̂, ÷ R³.
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And those are equivalent because R ̂ is just the R vector ÷ the magnitude of the R vector
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which is going to be R vector / R³ is going to be equivalent to R ̂ / R².
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That is the Biot-Savart Law.
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Now implementing it, using it takes a little bit of finesse, a little bit of practice.
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Let us go through some examples where we actually use it.
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Derive the magnetic field due to a current loop at the center of the loop.
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Let us draw in a loop of current, there we go.
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I have the current moving in that direction, we call that I.
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From the right hand rule, if you look at any pieces here,
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we should be able to see that we are going to have a magnetic field coming out of the center of the loop.
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That is going to be the direction of our magnetic field.
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To do this, let us also define a little bit of our loop of current right there.
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We will call that one DL, going in that direction.
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Our R ̂ vector must be going towards the center, toward where we want to know our magnetic field strength.
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There is R ̂.
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We can write our Biot-Savart Law , DB = μ₀ I/ 4 π DL × R ̂ ÷ R².
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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.
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We are going to have to integrate that all the way around the loop to add up all those little bits to DL.
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What do we know here? What can we do?
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As I look at DL × R ̂, that is always going to be perpendicular.
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DL × R we know is DL sin θ if we want the magnitude.
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Θ is going to be 90° so sin θ is going to be equal to 1.
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That makes it a little simpler.
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If we pull the constants out here, we can state that B = μ₀.
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That is not a very pretty, let us try that again.
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B = μ₀ I/ 4 π R², those are all constants, we can pull that out.
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We are going to be left with is the integral of DL.
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As we go all the way around a wire.
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As we go all the way around the wire, the integral of DL is just going to be its circumference 2 π R.
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Therefore, B is going to be equal to μ₀ I/ 4 π R² × 2 π R,
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which implies with a little bit of simplification that B is going to be equal to μ₀ I / 2R.
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Using the Biot-Savart Law in order to find the magnetic field due to current loop at the center of the loop.
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Let us do a little bit heavier example here.
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Finding the magnetic field due to a long straight current carrying wire.
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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.
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We already talked in our last lesson about what the answer is going to be.
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We know we are going to get μ₀ I/ 2 π R.
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How did we come up with that?
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We should be able to do that using the Biot-Savart Law.
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What we are going to do is I'm going to start over here by defining our little bit of DL
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which is going to be to the right, in that direction.
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There is DL.
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We also are going to define a radius, a distance to point B.
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Let us draw that in here, something like that.
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There is our R vector, or if we want we could define R ̂ as the unit vector in that direction.
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We will call this angle θ and we have got our distance from the wire R.
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As we look at this by the right hand rule, it should be pretty easy to see that
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the direction of the magnetic field is going to be out of the plane of the screen.
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Let me start off by writing the Biot-Savart Law, DB = μ₀ I/ 4 π R² DL × R ̂.
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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.
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That is going to be DX sin θ.
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DB = μ₀ I DX sin θ/ 4 π R².
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A little bit more work to do here.
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We have got a couple different variables and we got R in here, we have got θ.
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As I look at R, R is just going to be our X coordinate² + R².
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We can write R² = X² + R² and our sin θ as we look at that, sin θ is going to be our opposite/hypotenuse.
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That is going to be our opposite is R, our hypotenuse is going to be √ x² + R².
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Then, DB = μ₀ IDX R / 4 π × X² + R²³/ 2.
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If we want the entire magnetic field, we are going to have to integrate this.
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Which implies then that B = the integral of DB, which is going to be the integral and our variable is X.
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We are going to go from X = -infinity to X = infinity, all the way from the left, all the way to the right.
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It is a very long wire of μ₀ I DX R/ 4 π X² + R²³/2.
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Let us pull that constants we can out of this integration.
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B is going to be equal to μ₀ I R are all constants.
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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.
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Some of you guys with mad calculus skills may be able to integrate that.
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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,
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look inside the front cover and find a table of integrals in order to integrate that form of DX/ X² + some constant²³/ 2.
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As I do that, let us go on our next page to give ourselves some room.
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I can find that formula and I did.
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On the next page, we will keep writing it, that means that B = μ₀ I R/ 4 π.
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When I use my table of integrals I come up with X/ R² × the quantity R² + X² ^½.
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And that is evaluated from -infinity to infinity.
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Substituting in my limits, my infinity is there.
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We get μ₀ I R/ 4 π ×, we have got infinity/ R² × R² + infinity² ^½ - -infinity / R² × the quantity R² ± infinity² ^½.
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This might make some math teachers roll over in their graves a little bit, that is all right.
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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.
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√ something + infinity² is infinity.
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Infinity/ infinity is going to give us, this left hand side is going to become 1/ R².
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This right hand side is going to become, we will have -1/ R².
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For the same reason that a - -1, we are just going to end up with another + 1/ R².
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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.
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The answer that we were expecting.
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There are other ways to actually solve this setup.
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You can actually integrate over θ as you do different things.
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There are lots of ways to go about solving it.
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None of them are overly pretty but I thought that was the most straightforward methods to show you.
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Let us give a shot to one more sample problem.
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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.
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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.
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That will help and let us see if we can set this up a little bit.
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I’m going to pick some point of our current carrying wire here and we will define our DL.
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There is DL, we have got to draw the line from there to our point P.
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Let us get that all lined up, there we go.
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The magnetic field by the right hand rule is going to be perpendicular to that.
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90° from there, we are going to have our db from that portion of wire.
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We will draw that roughly that way.
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Let us draw a radius over here from the center of the circle to our point DL.
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There is our radius there.
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If the angle between that radius line and our R from the point to P, let us call that φ.
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If that is φ and this must also be φ in geometry.
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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.
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When we start that at 000.
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And I think we are pretty well set up with all the pieces we are going to need.
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If not, we will come back and get them in a few moments.
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We will start by writing Biot-Savart Law, DB = μ₀ I/ 4 π × DL × R ̂ ÷ R², which implies.
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We know that R² = R² + Y² that we found over here.
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DB is going to be equal to μ₀ I/ 4 π.
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Let us see, our DL × R that is just going to be our DL sin θ/ R² + Y² in place of our r².
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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.
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That DB is going to be equal to μ₀ I DL/ 4 π × R² + Y².
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Recognizing again that our magnetic field is going to be in the Y direction, we can state that the total magnetic field
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is going to be the integral of the Y component of DB which is going to be the integral of DB.
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If we want the Y component, if that is our angle θ, the Y component is the adjacent side.
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That is going to be DB cos φ which is cos φ is going to be the adjacent / the hypotenuse.
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Cos is adjacent/hypotenuse.
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Cos φ is going to be the adjacent side which is going to be R ÷ the hypotenuse √ R² + Y².
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That is cos φ and let us go on to our next page to continue that, to give ourselves more room again.
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So then B = μ₀ I/ 4 π.
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We have a RDL in the numerator ÷ R² + Y²³/2.
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We have just got the math piece left, we have really done all the physics.
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Which implies then that B = of course our integral.
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We will pull out our constants, we have got μ₀ IR in the numerator are all constants, μ₀ I R.
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In the denominator, we have 4 π R² + Y²³/ 2.
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None of that is a function of where we are as we integrate around our loop.
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All of that is a constant for this problem.
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R² + Y²³/2 and I'm just left with the integral of DL.
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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.
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Then we can state that B = μ₀ IR/ 4 π R² + Y²³/2 × 2 π R.
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Just finally, one last up to simplify this a little bit.
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D = μ₀ IR, we got our π cancels up there.
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Μ₀ IR, this is 2 π R.
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That R is actually, I wrote that wrong.
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That is 2 π R because we are going around that loop 2 π R.
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Μ₀ μ₀ I R² ÷, 2 ÷ 4 is going to give us ½.
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2 × R² + Y²³/ 2.
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I will put that in a 3D box because we are done.
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Hopefully that gets you a good start on Biot-Savart Law.
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It is a tricky law and very tough to implement.
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The concept is simple, actually using it definitely takes some experience and practice.
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Do that and in the next lesson we are going to come up to Amperes law and
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talk about other ways to find the magnetic field when you have a certain symmetry considerations you can use.
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Thank you so much for watching www.educator.com.
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We will see you soon, make it a great day everybody.