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

Correct option is 27.4×10−12 N

Questions

A uniform magnetic field B is acting from south to north and is of magnitude 1.5 Wb/m². If a proton having mass $= 1.7 \times 10^{- 27} kg and charge = 1.6 \times 10^{- 19} C$ 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

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7.4×10−19 N

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detailed solution

Correct option is B

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