29.6.1 Line Integrals

We’ve flirted with the idea of a line integral ever since introducing the concept of work in Chapter 9, but now we need to take a more serious look at what a line integral represents and how it is used. FIGURE 29.22a shows a curved line that goes from an initial point i to a final point f.

FIGURE 29.22 Integrating along a line from i to f.

The figure labeled ("a") illustrates a curved line that connects two points.
The figure labeled (b) illustrates a curved line divided into small and equal segments.
Full Alternative Text

Suppose, as shown in FIGURE 29.22b, we divide the line into many small segments of length Δs. The first segment is Δs1, the second is Δs2, and so on. The sum of all the Δs’s is the length l of the line between i and f. We can write this mathematically as

l=∑kΔsk→∫ifds (29.11)

where, in the last step, we let Δs→ds and the sum become an integral.

This integral is called a line integral. All we’ve done is to subdivide a line into infinitely many infinitesimal pieces, then add them up. This is exactly what you do in calculus when you evaluate an integral such as ∫x dx. In fact, an integration along the x-axis is a line integral, one that happens to be along a straight line. Figure 29.22 differs only in that the line is curved. The underlying idea in both cases is that an integral is just a fancy way of doing a sum.

The line integral of Equation 29.11 is not terribly exciting. FIGURE 29.23a makes things more interesting by allowing the line to pass through a magnetic field. FIGURE 29.23b again divides the line into small segments, but this time Δ s⃗k is the displacement vector of segment k. The magnetic field at this point in space is B⃗k.

FIGURE 29.23 Integrating B⃗⋅ds⃗ along a line from i to f.

The figure labeled (“a”) illustrates a line in a magnetic field.
The figure labeled (b) illustrates the magnetic field at a segment of a line.
Full Alternative Text

Suppose we were to evaluate the dot product B⃗k⋅Δs⃗k at each segment, then add the values of B⃗k⋅Δs⃗k due to every segment. Doing so, and again letting the sum become an integral, we have

∑kB⃗k⋅Δs⃗k→∫ifB⃗⋅ds⃗=the line integral of B⃗⋅ds⃗ from i to f

Once again, the integral is just a shorthand way to say: Divide the line into lots of little pieces, evaluate B⃗k⋅Δs⃗k for each piece, then add them up.

Although this process of evaluating the integral could be difficult, the only line integrals we’ll need to deal with fall into two simple cases. If the magnetic field is everywhere perpendicular to the line, then B⃗ ⋅ds⃗=0 at every point along the line and the integral is zero. If the magnetic field is everywhere tangent to the line and has the same magnitude B at every point, then B⃗ ⋅ds⃗=B ds at every point and

∫ifB⃗⋅ds⃗=∫ifB ds=B∫ifds=Bl (29.12)

We used Equation 29.11 in the last step to integrate ds along the line.

Tactics Box 29.3 summarizes these two situations.

TACTICS BOX 29.3 Evaluating line integrals

1 If B⃗ is everywhere perpendicular to a line, the line integral of B⃗⋅ds⃗ is

∫ifB⃗⋅ds⃗=0
The figure illustrates the evaluation of line integral when the magnetic field is perpendicular to a line.

2 If B⃗ is everywhere tangent to a line of length l and has the same magnitude B at every point, then

∫ifB⃗⋅ds⃗=Bl
The figure illustrates the evaluation of line integral when the magnetic field is parallel to a line.

Workbook Exercises 23–24