A deuteron is accelerated from rest through a 5-kV potential difference and then moves perpendicularly to a uniform magnetic field with B = 1.2 T. What is the radius of the resulting circular path? (deuteron: m = 3.3 x 10-27kg, q = 1.6 x 10-19C)

Principles of Physics: A Calculus-Based Text
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Author:Raymond A. Serway, John W. Jewett
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Chapter30: Nuclear Physics
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A deuteron is accelerated from rest through a 5-kV potential difference and then moves perpendicularly to a uniform magnetic field with B = 1.2 T. What is the radius of the resulting circular path? (deuteron: m = 3.3 x 10-27kg, q = 1.6 x 10-19C)

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