42 82 1ππ2aRR=>, as one can check numerically (that 821π2>). 75. Let the length of each side of the square be a. The center of a square is a distance a/2 from the nearest side. There are four sides contributing to the field at the center. The result is ()0center22224.42iiaBaaµπ⎛⎞==⎜⎟⎝⎠+0πOn the other hand, the magnetic field at the center of a circular wire of radius Ris 0/2iR(e.g., Eq. 29-10). Thus, the problem is equivalent to showing that
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