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.


1
Expert's answer
2022-05-02T13:42:40-0400

F=qvBsin⁡α=0.21⋅10−15 N.F=qvB\sin\alpha=0.21\cdot 10^{-15}~N.


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