Hey, guys. In this video, we're going to be talking about Ampere's law and how to use it. Alright? Let's get to it. Any magnetic field, b, must satisfy the following equation. This is what's known as a line integral. Okay? I am integrating b along some line that creates an arbitrary shape, which I called s. Okay? And this line integral always equals μ₀ I_{enclosed}. Okay? Notice that this integral has this little circle in the integral sign. All that means is that it's a closed line. Okay? We saw this before for Gauss's law, where we use the circle for the surface integral. Okay? And we said that that was a closed surface. So this has to be a closed loop or a closed line, which we call an amperean loop. Okay? And what Ampere's law says is that this integral depends only on the current enclosed by that amperean loop. And a consequence of that is that the magnetic field is only going to depend upon the current enclosed by that amperean loop as well, just like we saw in Gauss's law. Okay? How to apply this is pretty straightforward. If we take this arbitrary curve s in the figure above me, which surrounds a current I, everywhere along this loop, I can measure the magnetic field, and I can do the dot product with dl. All dl is is it's a very small it's an infinitesimal segment that goes along the curve. Right? So at this point, dl is going to be like this. This point, dl is going to be like this. It just points tangential to the curve. Right? Tangential everywhere. And the magnetic field is going to depend it's going to point in something like this. It would look something like this. It would look something like this. Something like this. Something like this, etcetera. And I could do the dot products, add all those up, integrate it along the curve s, and that would equate to a measurement times the charge enclosed. Okay? Let's apply Ampere's law for a scenario that we've already seen a bunch of times. Using Ampere's law, find the magnetic field due to an infinitely long current carrying wire. So I'm just going to draw the current as going into the page. Okay. Exactly like the figure above. Now, what does that magnetic field look like? Alright. Well, I'm going to stick my thumb in the direction of that current into the page, and the magnetic field is going to curl around it exactly like that. Okay? Which means if I were to draw an empyrean loop that was a circle of some radius r, the magnetic field is always going to be parallel to that line segment dl. Remember, that line segment always points tangential to the curve. Right? So Ampere's law says that this integral, b⋅dl, is going to just be bdl. That dot product is always going to be 1 because they always point parallel. And this equals μ₀ I_{enclosed}. Okay? Now just like we did for Gauss's law, we cheated a little bit because we knew what the magnetic field looked like. We had to cheat with Gauss's law as well by knowing what the electric field looks like. Furthermore, along this circle, because it's of a constant radius, the magnetic field is constant. So as I go along this circle, as I change my dl, the magnetic field doesn't change. So I can pull that field out. Okay? And lastly, I just need to solve this integral. Well, if I go around the circle one revolution, I've covered a distance equal to the circumference. So that magnetic field is just 2 pi r, and I enclosed is just I. I over 2 pi r. Exactly what we would expect it to be, and what we've seen it to be over and over again. Okay? So Ampere's law tells us what the magnetic field due to an infinitely long wire is much, much more quickly than Biot Savart law does, for instance. Alright. Thanks for watching, guys.

# Ampere's Law (Calculus) - Online Tutor, Practice Problems & Exam Prep

### Ampere's Law with Calculus

#### Video transcript

### Magnetic Field Inside a Solenoid

#### Video transcript

Hey guys, let's do an example. What is the magnetic field along the axis of a solenoid? Okay, so let's take some sort of solenoid like this. It's got some current going through it. It's got n number of loops spread out over a distance, l. Now, what is the magnetic field going to look like straight down the center of this solenoid? We want to use Ampere's law to find what that magnetic field is. Ampere's law tells us that the line integral b·dl across an Amperean loop has to equal μ_0 times the charge enclosed. Now, the particular Amperian loop I'm going to choose looks like this. I'm going to go down the axis of the solenoid, only for the length of the solenoid, where I know that the magnetic field is constant and straight. Like for Gauss's law, we are going to cheat a bit because we already know things about what the magnetic field should be before using Ampere's law. Then I'm going to go straight up. So it's a 90 degree angle straight up. Up. Infinitely high. So there is a gap here. And then I'm going to come back across. This little gap here takes us all the way up to infinity. So, this Amperian loop has four steps. It has step 1, which is along the axis, step 2, which is straight up perpendicular to the axis all the way up to infinity, step 3, which is parallel to the axis coming back but infinitely far away. And step 4, which is perpendicular to the axis coming back down. So this integral becomes the first path, b·dl, plus the second path, b·dl, plus the third path, b·dl, plus the fourth path, b·dl. Now the thing about paths 2 and paths 4 is that above the solenoid, the magnetic field line points parallel to the axis. So you're going to get a magnetic field line that is perpendicular to dl. Here's dl. Well, that magnetic field line actually points in the opposite direction. Because they are perpendicular, the dot product is always 0. So right away, for 2 and for 4, it's 0. Now what about 3? The thing is, 3 is parallel to the axis, so there would be a component of the magnetic field along it. The problem is that it's infinitely far away. Things in physics always drop to 0 when you go infinitely far away. Otherwise, two objects that are infinitely far apart could still interact with one another. Gravity goes to 0, gravitational potential energy goes to 0, electric force, electric potential energy, electric potential, electric field, etcetera, they all become 0 infinitely far away so that two things can't interact when they are infinitely far away. So the magnetic field line along path 3 is 0. This only leaves path 1. And they're parallel along path 1. So this integral becomes bdl from 0 to l, just the length of that axis, which runs the length of the solenoid, capital L. The magnetic field is going to be constant along the axis, so I can pull it out. And this just becomes bL. Now, this is the left half of Ampere's law. What about the right half of Ampere's law? That tells us it's μ_0 times the enclosed current. If there was a single loop here, just a single loop, we would have a single current I. For each loop, we have an additional current I. How many loops are there? There are n loops. So this is times n. Each of them carries I. So if we get out of the way, so we see the equation right here, this whole thing becomes bL equals μ_0nI or b is μ_0n over lI. This is the exact equation for a solenoid that we're expecting. Sometimes written like this, where little n is the number of turns per unit length. Okay guys? Thanks for watching.

A solid, cylindrical conductor carries a uniform current density, J. If the radius of the cylindrical conductor is R, what is the magnetic field at a distance ? from the center of the conductor when r < R? What about when r > R?

## Do you want more practice?

More sets### Your Physics tutor

- A solid conductor with radius a is supported by insulating disks on the axis of a conducting tube with inner r...
- As a new electrical technician, you are designing a large solenoid to produce a uniform 0.150-T magnetic field...
- A closed curve encircles several conductors. The line integral ∲B .dl around this curve is 3.83 * 10^-4 T m. (...
- The value of the line integral of →B ⋅ ds around the closed path in FIGURE EX29.21 is 1.38 x 10⁻⁵ T m. What ar...
- (I) A 2.5-mm-diameter copper wire carries a 28-A dc current (uniform across its cross section). Determine the ...
- You want to get an idea of the magnitude of magnetic fields produced by overhead power lines. You estimate tha...
- (III) Suppose the current in the coaxial cable of Problem 34, Fig. 28–45, is not uniformly distributed, but in...