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Image for Item 6 Four very long, current-carrying wires in the same plane intersect to form a square with side lengths o

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Answer #1

The center of the square is equidistant from the wires.

Thus, the currents wires must just add up to 0 with respect to the center of the square.

Thus,

-10 A - 8 A + 20 A + I = 0

--> I = 2 A   [ANSWER, PART A]


To cancel all fields, it must be DOWNWARD. [ANSWER, PART B]

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