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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Quizplus · Accepted Answer

First, we can use the potential difference to find the initial kinetic energy of the deuteron:
ΔV = KE/q
KE = qΔV = (1.6 x 10^-19 C)(10,000 V) = 1.6 x 10^-15 J
Using the conservation of mechanical energy, we can equate the initial kinetic energy to the kinetic energy in the circular motion:
KE = (1/2)mv^2 = (1/2)(3.3 x 10^-27 kg)(v^2)
Solving for v gives:
v = √(2KE/m) = √(2(1.6 x 10^-15 J)/(3.3 x 10^-27 kg)) = 1.2 x 10^7 m/s
The radius of the circular path can be found using the formula for the centripetal force:
F = mv^2/r = qvB
Solving for r gives:
r = mv/qB = (3.3 x 10^-27 kg)(1.2 x 10^7 m/s)/(1.6 x 10^-19 C)(1.6 T) = 1.3 x 10^-2 m = 13 mm, which is closest to choice B.

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