Answer to Question #335861 in Electricity and Magnetism for Mariaa

Question #335861

A proton moving horizontally at 3.5 x 10^4m/s enters a uniform magnetic field of 0.075 T directed upward 30° with the horizontal. Find the magnitude of the magnetic force acting on the proton.

Expert's answer

"F=qvB\\sin\\alpha=0.21\\cdot 10^{-15}~N."

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