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) A) 19 mm B) 13 mm C) 20 mm D) 10 mm E) 9.0 mm

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The Answer of 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) A) 19 mm B) 13 mm C) 20 mm D) 10 mm E) 9.0 mm

A Deuteron Is Accelerated from Rest Through a 10-KV Potential