# 3.6 Curl of a Vector Field

So far we have defined and discussed two of the three fundamental operators used in vector analysis: the gradient of a scalar and the divergence of a vector. Now we introduce the curl operator. The curl of a vector field B describes its rotational property, or circulation. The circulation of B is defined as the line integral of B around a closed contour C;

Circulation=∮C B⋅d l. (3.99)

To gain a physical understanding of this definition, we consider two examples. The first is for a uniform field B=x ˆ B 0, whose field lines are as depicted in Fig. 3-23(a). For the rectangular contour a b c d shown in the figure, we have

Circulation=∫a b x ˆ B 0⋅x ˆ d x+∫b c x ˆ B 0⋅y ˆ d y+∫c d x ˆ B 0⋅x ˆ d x+∫d a x ˆ B 0⋅y ˆ d y=B 0 Δ x−B 0 Δ x=0, (3.100)

where Δ x=b−a=c−d and, because x ˆ⋅y ˆ=0, the second and fourth integrals are zero. According to Eq. (3.100), the circulation of a uniform field is zero.

# Figure 3-23 Circulation is zero for the uniform field in (a), but it is not zero for the azimuthal field in (b).

(a) Uniform field

(b) Azimuthal field

Next, we consider the magnetic flux density B induced by an infinite wire carrying a dc current I. If the current is in free space and it is oriented along the z direction, then from Eq. (1.13),

B=ϕ ˆ μ 0 I 2 π r, (3.101)

where μ 0 is the permeability of free space and r is the radial distance from the current in the x–y plane. The direction of B is along the azimuth unit vector ϕ ˆ. The field lines of B are concentric circles around the current, as shown in Fig. 3-23(b). For a circular contour C of radius r centered at the origin in the x–y plane, the differential length vector d l=ϕ ˆ r d ϕ, and the circulation of B is

Circulation=∮C B⋅d l=∫0 2 π ϕ ˆ μ 0 I 2 π r⋅ϕ ˆ r d ϕ=μ 0 I. (3.102)

In this case, the circulation is not zero. However, had the contour C been in the x–z or y–z planes, d l would not have had a ϕ ˆ component, and the integral would have yielded a zero circulation. Clearly, the circulation of B depends on the choice of contour and the direction in which it is traversed. To describe the circulation of a tornado, for example, we would like to choose our contour such that the circulation of the wind field is maximum, and we would like the circulation to have both a magnitude and a direction with the direction being toward the tornado’s vortex. The curl operator embodies these properties. The curl of a vector field B, denoted curl B or ∇×B, is defined as

The direction of curl B is n ˆ, the unit normal of Δ s, defined according to the right-hand rule with the four fingers of the right hand following the contour direction d l and the thumb pointing along n ˆ (Fig. 3-24). When we use the notation ∇×B to denote curl B, it should not be interpreted as the cross product of ∇ and B.

# Figure 3-24 The direction of the unit vector n ˆ is along the thumb when the other four fingers of the right hand follow d l.

For a vector B specified in Cartesian coordinates as

B=x ˆ B x+y ˆ B y+z ˆ B z, (3.104)

it can be shown, through a rather lengthy derivation, that Eq. (3.103) leads to

∇×B=x ˆ ∂B z∂y−∂B y∂z+y ˆ ∂B x∂z−∂B z∂x+z ˆ ∂B y∂x−∂B x∂y=x ˆ y ˆ z ˆ∂∂x∂∂y∂∂z B x B y B z. (3.105)

Expressions for ∇×B are given in Appendix C for the three orthogonal coordinate systems considered in this chapter.

# 3.6.1 Vector Identities Involving the Curl

For any two vectors A and B and scalar V,

(1)∇×(A+B)=∇×A+∇×B, (3.106a)
(2)∇⋅(∇×A)=0, (3.106b)
(3)∇×(∇V)=0. (3.106c)

# 3.6.2 Stokes’s Theorem

For the geometry shown in Fig. 3-24, Stokes’s theorem states

Its validity follows from the definition of ∇×B given by Eq. (3.103). If ∇×B=0, the field B is said to be conservative or irrotational because its circulation, represented by the right-hand side of Eq. (3.107), is zero, irrespective of the contour chosen.

# Example 3-14: Verification of Stokes’s Theorem

For vector field B=z ˆ cos ϕ/r, verify Stokes’s theorem for a segment of a cylindrical surface defined by r=2, π/3≤ϕ≤π/2, and 0≤z≤3 (Fig. 3-25).

# Figure 3-25 Geometry of Example 3-14.

Figure 3 25 Full Alternative Text

Solution: Stokes’s theorem states that

∫S(∇×B)⋅d s=∮C B⋅d l.

Left-hand side: With B having only a component B z=cos ϕ/r, use of the expression for ∇×B in cylindrical coordinates in Appendix C gives

∇×B=r ˆ 1 r∂B z∂ϕ−∂B ϕ∂z+ϕ ˆ ∂B r∂z−∂B z∂r+z ˆ 1 r∂∂r(r B ϕ)−∂B r∂ϕ=r ˆ 1 r∂∂ϕ cos ϕ r−ϕ ˆ∂∂r cos ϕ r=−r ˆ sin ϕ r 2+ϕ ˆ cos ϕ r 2.

The integral of ∇×B over the specified surface S is

∫S(∇×B)⋅d s=∫z=0 3∫ϕ=π/3 π/2−r ˆ sin ϕ r 2+ϕ ˆ cos ϕ r 2⋅r ˆ r d ϕ d z=∫0 3∫π/3 π/2−sin ϕ r d ϕ d z=−3 2 r=−3 4.

Right-hand side: The surface S is bounded by contour C=a b c d shown in Fig. 3-25. The direction of C is chosen so that it is compatible with the surface normal r ˆ by the right-hand rule. Hence,

∮C B⋅d l=∫a b B a b⋅d l+∫b c B b c⋅d l+∫c d B c d⋅d l+∫d a B d a⋅d l,

where B a b, B b c, B c d, and B d a are the field B along segments a b, b c, c d, and d a, respectively. Over segment a b, the dot product of B a b=z ˆ cos ϕ/2 and d l=ϕ ˆ r d ϕ is zero, and the same is true for segment c d. Over segment b c, ϕ=π/2; hence, B b c=z ˆ(cos π/2)/2=0. For the last segment, B d a=z ˆ(cos π/3)/2=z ˆ/4 and d l=z ˆ d z. Hence,

∮C B⋅d l=∫d a z ˆ 1 4⋅z ˆ d z=∫3 0 1 4 d z=−3 4,

which is the same as the result obtained by evaluating the left-hand side of Stokes’s equation.