Answer
We can rank the circuits according to the magnitude of the magnetic field produced at the center of curvature:
$(a) \gt (c) \gt (b)$
Work Step by Step
We can write the expression for the magnetic field produced by an arc of current:
$B = \frac{\mu_0~i~\phi}{4~pi~r}$
The semi-circle at the top of each circuit contributes the same magnetic field for each circuit.
Also, each straight section contributes zero magnetic field at the center.
In circuit (b), the small semi-circle is above the center. The current in this section produces a magnetic field in the opposite direction, so circuit (b) has the smallest net magnetic field.
Circuit (a) has a larger net magnetic field at the center than circuit (c), since an arc with a smaller radius produces a larger magnetic field at the center.
We can rank the circuits according to the magnitude of the magnetic field produced at the center of curvature:
$(a) \gt (c) \gt (b)$
