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)

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Quizplus · Accepted Answer

The kinetic energy gained by the deuteron is equal to the work done by the electric field, which is qV. This energy is converted into kinetic energy, (1/2)mv^2. Solving for v gives v = sqrt(2qV/m). The radius of the circular path in a magnetic field is given by r = mv/(qB). Substituting the expression for v, we get r = sqrt(2mV/q)/B. Plugging in the given values, we find r ≈ 13 mm.

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