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A proton moves with a speed of 8.0 × 106 m/s at an angle of 30°...

A proton moves with a speed of 8.0 × 106 m/s at an angle of 30° to a magnetic field of 2T, shown in the figure in question 1, (the charge of a proton = 1.60 × 10-19 C). Calculate the magnitude of the magnetic force on the proton.

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Answer #1

F = qvB Sin theta

F = 1.6 × 10^-19 × 8 × 10^6 × 2 × Sin 30

F = 12.8 × 10^-13

F = 1.28 × 10^ -12 N

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