Final answer:
The line integral of the magnetic field around the closed path in question can be calculated using Ampère's Law, relating the current density to the enclosed current within the path. Only one segment of the rectangular path contributes to the integral as it's the only part normal to the current density direction. The enclosed current is found by multiplying the current density by the area enclosed by the path, and the line integral is this current times the permeability of free space.
Step-by-step explanation:
The question concerns the calculation of the line integral of the magnetic field vector (B) around a closed path in a region with a uniform current density (J) using Ampère's Law. Given a uniform current density of 16 A/m² in the positive z direction, the line integral of B · ds along a closed path can be related to this current density through the enclosed current, utilizing the symmetry of the problem. According to Ampère's Law, the line integral of the magnetic field around any closed path is equal to the total enclosed current times the permeability of free space (μ0). The enclosed current is the current density times the area (A) through which the path encloses current.
For the specific path described, composed of three straight-line segments forming a rectangle in the xy-plane, and since J is in the z direction, only the segment from (4d, 0, 0) to (4d, 3d, 0) will have current passing through it normally, which will contribute to the integral. We can find the enclosed current by integrating the current density over the area of the rectangle formed by the closed path. The area A is the width times the height of the rectangle (3d × 4d). The enclosed current I is then J times A, and the line integral ∫ B · ds would be this enclosed current times μ0, for a rectangular loop in a uniform magnetic field.