A deuteron is accelerated from rest through a 10-kv, Ask an Expert

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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 ? 10-27 kg, q = 1.6 ? 10-19 C) I know the answer is 20.3 just don't know how to get it.

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