One end of a straight copper rod, carrying a current of 2A, is at the origin and the other end is at the point A whose co ordinates are (1m, 1m, 1m). A uniform magnetic field B→=1.5i^ T exists in that region, Direction of current in the rod is from the origin to point A. Then magnitude of force experienced by the rod is

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3 N

22 N

32 N

43 N

answer is C.

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Detailed Solution

F→=i l→×B→. Here l→=(i^+j^+k^)m∴ F→=2(i^+j^+k^)×1.5 i^ N=3(−k^+j)N∴ F=312+12 N=32 N

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