ch29-p075 - 75. Let the length of each side of the square...

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42 82 1 ππ 2 aR R => , as one can check numerically (that 82 1 π 2 > ). 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 () 0 center 2 2 22 4. 42 ii a B aa µ π ⎛⎞ == ⎜⎟ ⎝⎠ + 0 π On the other hand, the magnetic field at the center of a circular wire of radius R is 0 /2 iR (e.g., Eq. 29-10). Thus, the problem is equivalent to showing that
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