Question

A deuteron (the nucleus of an isotope of hydrogen) has a mass of 3.34×10−27 kg and a charge of 1.60×10−19 C . The deuteron travels in a circular path with a radius of 7.30 mm in a magnetic field with a magnitude of 2.50 T

Find the time required for it to make 12 of a revolution.

Through what potential difference would the deuteron have to be accelerated to acquire this speed?

Answer 1

https://www.dspace.espol.edu.ec/bitstream/123456789/5200/1/8539.pdf

Answer 2

in terms of process, for the circular motion you need a centripetal force \(F = m \omega^2r = qvB\)

that second part is from the Lorentz Force Law. (from Wiki: " If a particle of charge q moves with velocity v in the presence of an electric field E and a magnetic field B, then it will experience a force \(\large \mathbf{F} = q\left[\mathbf{E} + (\mathbf{v} \times \mathbf{B})\right]\)"


for motion in a circle of radius r at speed v, we can say \(v = \omega r\) so \(m \omega^2r = q\omega rB\) or \( \omega = \dfrac{qB}{m}\)

we also know that \(\omega = \dfrac{2\pi}{T}\) where T is the period or time of 1 full revolution....

we can also calculate the kinetic energy of the rotating particle as \( \dfrac{1}{2} m v^2\) ...and use that to find out the voltage drop required to produce that amount of energy in the particle...for the second part of your question.