Answer to Question #89707 in Electricity and Magnetism for Khadeeja

Question

A proton a deuteron (nucleus of a deuterium atom) and an alpha particle having the same kinetic energy are moving in circular trajectories constant magnetic field. If our R1, R2 and R3 denote respectively the radius of trajectories of these particles then show the relationship between the three.

Answer 1

A circular trajectory occurs when the magnetic field is perpendicular to the velocity direction. Writing down the 2nd Newton's law for an arbitrary charged particle, we obtain:

$$q v B = m a = \frac{m v^2}{R}$$

Hence,

$$R = \frac{m v}{q B}$$

Kinetic energy of a particle is defined as follows:

$$E_k=\frac{m v^2}{2} = \frac{(mv)^2}{2m} \, \Rightarrow \, mv = \sqrt{2m E_k}$$

As a result,

$$R = \frac{\sqrt{2 E_k}}{B} \frac{\sqrt{m}}{q}$$

Hence, the ratio of radii is

$$R_1 : R_2 : R_3 = \frac{\sqrt{m_1}}{q_1} : \frac{\sqrt{m_2}}{q_2} : \frac{\sqrt{m_3}}{q_3}$$

For a proton, a deuteron and alpha particle we have:

$$proton: \quad m_1 = m_0, \, q_1 = q_0 \\ deuteron: \quad m_2 = 2m_0, \, q_2 = q_0\\ \alpha-particle: \quad m_3 = 4m_0, \, q_3 = 2q_0$$

Hence,

$$R_1 : R_2 : R_3 = 1 : \sqrt{2} : 1$$