A deuteron is accelerated from rest through a 10-kv, Physics

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A deuteron is accelerated from rest through a 10-kv

A deuteron is accelerated from rest through a 10-kV potential difference and then moves perpendicularly to a uniform magnetic field with B = 1.6 T. What is the radius of the resulting circular path? (deuteron: m = 3.3 x10-27 kg, q = 1.6 x10-19 C) the answer is 20.3mm

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Physics: A deuteron is accelerated from rest through a 10-kv

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