A uniform magnetic field B is acting from south to north and is of magnitude 1.5 Wb/m2. If a proton having mass =1.7×10−27 kg and charge =1.6×10−19C moves in this field vertically downwards with energy 5 Me V, then the force acting on it will be

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7.4×1012 N

7.4×10−12 N

7.4×1019 N

7.4×10−19 N

answer is B.

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Detailed Solution

F =qvB and K=12mv2⇒F=qB2Km=1.6×10−19×1.52×5×106×1.6×10−191.7×10−27=7.344×10−12 N

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