Particle accelerators: A deuteron has the same charge as a proton but approximately twice the proton mass. Suppose deuterons are being accelerated in a cyclotron in an orbit of radius 75 cm at a frequency of 8.0 MHz. What magnetic field magnitude is needed? (e = 1.60 × 10^-19 C, m_proton = 1.67 × 10^-27 kg) A) 0.011 T B) 0.11 T C) 1.1 T D) 11 T E) 110 T

Question

Quizplus · Accepted Answer

The frequency of the cyclotron is related to the magnetic field and the mass of the particle being accelerated by the following equation:

f = Bq/2πm

where f is the frequency, B is the magnetic field magnitude, q is the charge of the particle, and m is the mass of the particle.

For deuterons, q = 1.60 × 10^-19 C and m = 2(1.67 × 10^-27 kg) = 3.34 × 10^-27 kg. Plugging in these values along with the given frequency of 8.0 MHz (or 8.0 × 10^6 Hz) and solving for B, we get:

B = 2πmf/q = 2π(3.34 × 10^-27 kg)(8.0 × 10^6 Hz)/(1.60 × 10^-19 C) ≈ 1.1 T

Therefore, the best choice is C, with a magnetic field magnitude of 1.1 T.

Particle Accelerators: a Deuteron Has the Same Charge as a Proton