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Asked by Roy

A proton with speed v perpendicular to a magnetic field B is experiencing a force F.If the speed of the proton is doubled,the new force is (a)F/2(b)F(c)2F(d)4F
12 years ago

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Answered by Steve
since F = qv x B,
2v -> 2F
12 years ago
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A proton with speed v perpendicular to a magnetic field B is experiencing a force F.If the speed of the proton is doubled,the new force is (a)F/2(b)F(c)2F(d)4F

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