A proton is entering magnetic field B = 0.2i - 0.5j + 0.4k. If the proton moves...?
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A proton is entering magnetic field B = 0.2i - 0.5j + 0.4k. If the proton moves at v = 4000j - 5000k in meters per second, find the force vector acting on the proton. Proton charge is +1.6 x 10^-19 C I'm really lost, but I know it should be simple.
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Answer:
The force vector F = q* v cross B So take the cross product of v & B and multiply by the charge So v X B = -900i - 1000j - 800k Therefore F = 1.6x10^-19*( -900i - 1000j - 800k) = -1.44x10^-16 i - 1.60x10^-16j - 1.28x10^-16k
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