Notice that the closed loop does not actually enclose the wire. Thus the line integral over this loop must be zero.

Though we stated this general fact in the text, we did not prove it. This exercise completes the proof. Notice from our figure from the last problem that the closed loop consists of a circle that almost encloses the wire, and a randomly shaped loop that almost encloses the wire. We thus break up the loop into two sections. We can approximate the line integral of the first section, the circle, using what we already know about line integrals of circles around a wire. The line integral over the circle is thus approximately . We also know that the line integral of the complete closed loop (both sections) is zero, implying that the line integral over the second section (the odd-shaped curve) must be - . Since the second segment is oriented in the opposite direction as the right hand rule would dictate for our wire, the negative sign is attached to the expression. No matter the shape of that second segment, it will have the same value for its line integral. Thus we have shown that this property applies to all closed loops, not just circular ones.

Though this problem looks quite complex, the property that div B = 0 greatly simplifies our work. Gauss' Law states that

Because the divergence of any magnetic field must be zero, then the surface integral of the magnetic field over a closed surface must also be zero. Since the sphere is a closed surface, the surface integral over the sphere is necessarily zero.