Rankers Physics

Force Acting on Moving Charges: Practice Problem & Solution

A deuteron of kinetic energy $50\text{ keV}$ is describing a circular orbit of radius $0.5\text{ metre}$ in a plane perpendicular to magnetic field $B$. The kinetic energy of the proton that describes a circular, orbit of radius $0.5\text{ metre}$ in the same plane with the same $B$ is (1991)
$25\text{ keV}$
$50\text{ keV}$
$200\text{ keV}$
$100\text{ keV}$

Solution Explained:

To solve this problem, we apply the core principles of Force Acting on Moving Charges. Understanding the underlying formula is key to arriving at the correct answer below:

Using $r = \frac{\sqrt{2mK}}{qB}$, equating radii for deuteron and proton where $m_d = 2m_p$ yields $K_p = 2K_d = 100\text{ keV}$.

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