You are given two vector fields. Vector field one magnitude decreases toward the east and is toward the top of the page. Vector field two is cylindrically symmetric, toward positive $phi$ direction and is decreasing with increasing $r$ distance from $z$..
a) Is the curl of field one non-zero? Explain. What is this field line integral on the dashed path?
b) Is the curl of the field two zero? If is zero, where can its curl be non-zero? What is this field line counterclockwise integral around the dashed path shown? Explain. What is the physical explanation of the curl of the field two inside the dashed path? Was this your prediction?
The curl vector shows the density of angular momentum of a given vector. Therefore if the vector field rotates about a center it has nonzero curl. The first field in the figure has zero curl (does not rotate). Stokes theorem says:
$oint F*dR=iint (nabla times F) dS$
where the integral surface $Sigma$ “stands” on the countur $C$ on the left integral. Since $(nabla times F) =0$ the the line integral on the contour $C$ is zero.
This field rotates about a center. It has a nonzero curl in all points of space except the center of rotation. The line integral is nonzero on lines along the field rotation (1 and 2). Since along line 1 the field is parallel to the displacement $dr$, then the integral is positive. Also because along line 2 the field is opposing the displacement the integral is negative. Elsewhere (lines 3 and 4), the line integral is zero since the displacement is perpendicular to the field.
Since path 1 is longer than 2 the total line integral along the closed contour is positive.
The field rotates counterclockwise so its curl is positive (out of page). Also the normal to the surface (if the contour rotation is counterclockwise as in the figure) is out of page. Thus the integral (because of the Stokes theorem) is positive (the surface integral of the curl is positive).