Hey, guys. In this video, we're going to be discussing the general calculus form of the Biot Savart plot. Alright? Let's get to it. For any current, okay, contained by some wire, if we want to measure the magnetic field at some position r vector away. Okay, so we have this figure here above me where we have some arbitrary wire, in this case a curved one, with some arbitrary current I. We want to measure the magnetic field here. Okay, what is that going to look like? Well, Biot Savart law tells us that this is μ∶4π∙∫dl⋅rr3. Okay. We already know what r is. The only thing we need to know is what d l is. Okay? And d l is a very, very small vector, okay, in the direction of the current. So wherever this current is pointing, I can choose a really tiny infinitesimally small vector, dl in that direction. Okay? And the cross product right here is between that vector dl and the position vector r. Okay? Biot Savart's law, this integral equation reduces to our familiar magnetic field equations. For instance, we have the magnetic field due to a point charge, and we have the magnetic field due to an infinitely long current-carrying wire. That's just two examples of what the biot savart law reduces to. Okay? Let's do an example. Show that the biot savart law for a current is the same as the equation above for a point charge. Okay? So if instead of having just some large current, which is made up of a bunch of streaming electrons, we have a single charge q that is moving, and we want to use Biot-Savart law given to us, the general form, to find it for a point charge. Okay? The general form is just μ∶4π∙∫Idl⋅rvectort. Okay, what we are going to do here is we are going to substitute in the definition for current. Current, by definition, is qdt, okay? So I will plug that in. This becomes μ∶4π∙∫dqdtdl⋅rqr3. Now there's something that you can do, which you learned in calculus called implicit differentiation. Okay? If you treat these infinitesimals, dqdtdl, as implicit differentials, okay, we can reorder them and we can just say that this is μ∶4π∙∫dq dldt⋅rvectorr3. Okay? Now what dl is, is it's a very, very small amount of distance in the direction that the current is moving, or in this case the direction that the charge is moving. The rate at which that's increasing in size is just the velocity of the charge. So this becomes μ∶4π∙∫dqv⋅rr3. Okay? Now what we're integrating over is our charge dq. There's a single charge here, so the velocity doesn't change for the charge and neither do these positions. Okay? Those positions are just the position that we want to measure the magnetic field in. So we can pull all of that out. Simplifying the cross product, this becomes vrsinθ∕r3 sorry, I'm missing the 4 pi, integral of dq. And there's just a single charge, the integral of dq is just that charge q. So this becomes μ∶4π.Q,by the way,we lose the power ofrinthedenominator,qvsinθ∕r2. And that is our familiar equation for the Biot Savart law for a single moving charge. Alright? Thanks for watching, guys.

# Biot-Savart Law (Calculus) - Online Tutor, Practice Problems & Exam Prep

### Biot-Savart Law with Calculus

#### Video transcript

### Magnetic Field due to Finite Wire

#### Video transcript

Hey, guys. Let's do an example. What is the magnetic field at the position shown in the following figure due to the finite current-carrying wire. Okay? We already saw what the magnetic field due to an infinitely long current-carrying wire is. Now we want to find it due to a finite current-carrying wire. In this case, a wire of length 2a. Okay? I've centered the z-axis so that the wire extends -a to the left of the z-axis and a to the right of the z-axis. Okay? Now what Biot-Savart law tells us is that each chunk of this current-carrying wire contributes a very small amount to the magnetic field equal to μ_04πIdlsinθr2. Okay? That's the very, very small contribution of each piece dl of this current. Okay? So if I were to grab an arbitrary piece of this current, let's say right here, I'm not going to call it dl. I'm going to call it dx because it's in the x-direction. Okay? Then this is my r vector. The position from my current dl to the point at which we are interested in measuring the magnetic field. K? And what is r? R is simply whatever we call this distance, which I'll call x, and this distance, the vertical distance, which we call z. So r is the square root of x2 + z2. Okay? So let's simplify our DB right here. Okay? I, this is going to be dx, okay, instead of dl, sinθ because I'm doing the cross product between them, over r3. We're going to lose the value of r. So it's going to be μ_04πIsinθx2+z2dx. We're not quite done. The reason is, as we are integrating along x, right, as we move from -a across x all the way over to a, this angle right here, theta, changes. So we do need to know what sinθ is. Okay? We can say, though, that sinθ is just the opposite edge, which is z, over the hypotenuse r, which is the square root of x2+z2. Okay? I did make one little typo right here. This is r2. R is the square root, so the square cancels the square root. Okay? This is what sinθ looks like, though. K? So now we can plug insinθ. I'm going to minimize myself to give us some room. μ_04πIzx2+z232dx. Okay? So all we have to do is integrate this over our integration variable x, and that'll tell us what the magnetic field strength is at that point. So b is just the integral along x from -a to a of μ_0I4πzx2+z232dx. Okay? Now, you cannot use u-substitution here because the denominator, you would use sorry. U in the denominator as x2+z2, but that means you need an x in the numerator. That doesn't work. This is an integral that we've actually seen before when dealing with Coulomb's law, and it uses trigonometric substitution. I'm not going to do the integral again, I'm just going to show the result. This is something that you can find in books, in the table of integrations. It would be given to you on an exam. Okay? Unless your professor was a sadist and made you solve this from scratch on an exam. Okay. But the solution is just μ_0I4π2azz2+a2. That is the solution already plugging in the integration limits -a and a. Okay? So one thing that we could do right here is we could knock a 2 down and that would be simplified. Okay? This is the correct answer, right here, to find the magnetic field at a point z for a finite wire of length 2a. Now an interesting question is, does this result match up with the result that we know is true for an infinitely long wire? Okay? Meaning, what does the limit as a goes to infinity look like? Well, this is just μ_0I2π. None of that is involved with the limit. A over zz2+a2. Okay? The numerator and the denominator both have the same power of a. It's a in the numerator and square root of a2 in the denominator. So we can't say for sure which wins, the numerator or the denominator, because they are both growing at the same rate. I need to use a little bit of algebra to manipulate our limit. This becomes μ_0I2π. Nothing changes there.Limit. What I'm going to do is I'm going to factor out this a2. When I pull it out of the square root, it loses its power. Okay? It loses its power of 2, and it just becomes a. So this is a over zz2a2+1. Okay? That's the factored form. And now, luckily, notice we can just lose the power of a in the numerator and the denominator. So once again... oops. Wrong color. Continuing with this, this is μ_0I2π. Limit as a goes to infinity of 1 over zz2a2+1. Now, what happens when a goes to infinity? Well, this is z2a2, so that goes to 0. 0 plus 1 is just 1, so this whole thing is μ_0I2πz. And that is exactly the equation that we have for a wire that's infinitely long and straight carrying a current I if we wanted to know what was the magnetic field at this point a distance z away. Right? If we had some wire carrying some current, and I wanted to know what was the magnetic field at this point a distance z away, it would be this equation. Alright guys. Thanks for watching.

What is the magnetic field at the center of the following ring of current?

_{0}i/(2r)

_{0}i/(πr

^{2})

_{0}i/(2πr)

_{0}i / √(2πr)