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The coil whose lengthwise cross section is shown in the accompanying figure carries a current ๐ผ and has ๐ evenly-spaced turns distributed along the length ๐. Evaluate the closed line integral ๐ dot d๐ in terms of ๐ naught and ๐ผ for path ๐ด. Evaluate the closed line integral ๐ dot d๐ in terms of ๐ naught and ๐ผ for path ๐ต. Evaluate the closed line integral ๐ dot d๐ in terms of ๐ naught and ๐ผ for path ๐ถ. Evaluate the closed line integral ๐ dot d๐ in terms of ๐ naught and ๐ผ for path ๐ท.

Weโre told in the problem statement that this diagram refers to a lengthwise cross section of a coil, like a solenoid, that has a current ๐ผ 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 ๐ฅ or a cross. For each of the dotted line circuits that weโve identified around this cross section โ ๐ด, ๐ต, ๐ถ, and ๐ท โ we want to solve for the path integral ๐, the magnetic field, dotted with an infinitesimal length segment, d๐.

Now to start off, we can recall that Ampereโs law tells us something about this path integral. It tells us that the closed line integral ๐ dot d๐ is equal to the permeability of free space, ๐ naught, multiplied by the current enclosed by this loop. The loops that weโve drawn into this diagram, ๐ด, ๐ต, ๐ถ, and ๐ท are imaginary closed paths. But theyโre the paths weโll use to integrate over in Ampereโs law.

Before we begin evaluating these integrals, thereโs 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 ๐ dot d๐.

Letโs try this rule along with Ampereโs law for path ๐ด. And youโll begin to get a feel for it. If we travel along the closed path ๐ด and account for ๐ dot d๐ over that path, we see that there are three current segments that we crossed through. So by Ampereโs law, this integral is equal to three times ๐ naught times ๐ผ. Now the question is, is it positive or negative three ๐ naught ๐ผ. To figure that out, we refer to our right-hand rule. We see that ๐ด 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 ๐ด, Ampereโs law tells us that ๐ dot d๐ is equal to three ๐ naught times ๐ผ.

Now weโll consider path ๐ต. 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 itโs 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 ๐ถ. And when we look at this path on our diagram, we see that it encloses seven different multiples of ๐ผ. So we write out that this closed path integral is equal to seven ๐ naught ๐ผ. But weโre undecided so far as to the sign of this value, whether positive or negative. To figure that out, weโll 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 ๐ถ. When we do this, our thumb points into the page which matches the direction of the current in region ๐ถ. Therefore, this is a positive, rather than negative, result.

Finally, weโll consider the path integral over the region ๐ท for ๐ dot d๐. As we look at this region, we see that it encloses two multiples of the current ๐ผ. So we can write that this integral is two ๐ naught ๐ผ and we leave a placeholder for the sign, which weโll determine again using our right-hand rule. If we curl our fingers counterclockwise to accord with the direction we travel through ๐ท, 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 ๐ท is negative two ๐ naught times ๐ผ.

These are the results of Ampereโs law calculations for these four regions.