# Video: Finding the Integral of the Magnetic Field around a Closed Loop That Encloses a Current

The coil whose lengthwise cross section is shown in the accompanying figure carries a current 𝐼 and has 𝑁 evenly-spaced turns distributed along the length 𝑙. Evaluate ∮𝐁 ⋅ d𝐋 in terms of 𝜇₀ and 𝐼 for path 𝐴. Evaluate ∮𝐁 ⋅ d𝐋 in terms of 𝜇₀ and 𝐼 for path 𝐵. Evaluate ∮𝐁 ⋅ d𝐋 in terms of 𝜇₀ and 𝐼 for path 𝐶. Evaluate ∮𝐁 ⋅ d𝐋 in terms of 𝜇₀ and 𝐼 for path 𝐷.

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### Video Transcript

The coil whose lengthwise cross section is shown in the accompanying figure carries a current 𝐼 and has 𝑁 evenly-spaced turns distributed along the length 𝑙. Evaluate the closed line integral 𝐁 dot d𝐋 in terms of 𝜇 naught and 𝐼 for path 𝐴. Evaluate the closed line integral 𝐁 dot d𝐋 in terms of 𝜇 naught and 𝐼 for path 𝐵. Evaluate the closed line integral 𝐁 dot d𝐋 in terms of 𝜇 naught and 𝐼 for path 𝐶. Evaluate the closed line integral 𝐁 dot d𝐋 in terms of 𝜇 naught and 𝐼 for path 𝐷.

We’re told in the problem statement that this diagram refers to a lengthwise cross section of a coil, like a solenoid, that has a current 𝐼 running through its loops. We represent the current coming out of the page by a dot. And we represent the current going into the page by an 𝑥 or a cross. For each of the dotted line circuits that we’ve identified around this cross section — 𝐴, 𝐵, 𝐶, and 𝐷 — we want to solve for the path integral 𝐁, the magnetic field, dotted with an infinitesimal length segment, d𝐋.

Now to start off, we can recall that Ampere’s law tells us something about this path integral. It tells us that the closed line integral 𝐁 dot d𝐋 is equal to the permeability of free space, 𝜇 naught, multiplied by the current enclosed by this loop. The loops that we’ve drawn into this diagram, 𝐴, 𝐵, 𝐶, and 𝐷 are imaginary closed paths. But they’re the paths we’ll use to integrate over in Ampere’s law.

Before we begin evaluating these integrals, there’s is one more thing to cover. You may notice that each of these four paths has a direction associated with it. There are arrowheads telling us which way to move around the closed path. That path direction comes into play in our answer using something sometimes called a right-hand rule. This is a rule whose name comes from the fact that our right hand is able to help us remember it. If we look down onto our right hand so that our thumb is pointing straight up out of the page, then if we curl our fingers in the direction the path follows, either clockwise or counterclockwise, then our thumb points in the direction of positive 𝐁 dot d𝐋.

Let’s try this rule along with Ampere’s law for path 𝐴. And you’ll begin to get a feel for it. If we travel along the closed path 𝐴 and account for 𝐁 dot d𝐋 over that path, we see that there are three current segments that we crossed through. So by Ampere’s law, this integral is equal to three times 𝜇 naught times 𝐼. Now the question is, is it positive or negative three 𝜇 naught 𝐼. To figure that out, we refer to our right-hand rule. We see that 𝐴 is a path traveled counterclockwise and that the current coming through the solenoid or the coil is coming out of the page. That means that we have a positive result for our closed path integral. So around path 𝐴, Ampere’s law tells us that 𝐁 dot d𝐋 is equal to three 𝜇 naught times 𝐼.

Now we’ll consider path 𝐵. And when we consider that path on our diagram, we see that it includes two points where the current is coming out of the page and two points where it’s coming into the page. Since those currents are all equal in magnitude, the two pairs cancel one another out. And the answer to our path integral is zero.

Now we move on to consider path 𝐶. And when we look at this path on our diagram, we see that it encloses seven different multiples of 𝐼. So we write out that this closed path integral is equal to seven 𝜇 naught 𝐼. But we’re undecided so far as to the sign of this value, whether positive or negative. To figure that out, we’ll again use the right-hand rule. In this case, we curl our fingers in the clockwise direction to match the direction we walk around path 𝐶. When we do this, our thumb points into the page which matches the direction of the current in region 𝐶. Therefore, this is a positive, rather than negative, result.

Finally, we’ll consider the path integral over the region 𝐷 for 𝐁 dot d𝐋. As we look at this region, we see that it encloses two multiples of the current 𝐼. So we can write that this integral is two 𝜇 naught 𝐼 and we leave a placeholder for the sign, which we’ll determine again using our right-hand rule. If we curl our fingers counterclockwise to accord with the direction we travel through 𝐷, our thumb points out of the page. But we see that the current in the diagram goes into the page. Therefore, we have a negative sign in front of this two. So our final answer for the region over 𝐷 is negative two 𝜇 naught times 𝐼.

These are the results of Ampere’s law calculations for these four regions.