Question

A proton with kinetic energy $8eV$ is moving in a uniform magnetic field. The kinetic energy of a deuteron moving in the same path in the same magnetic field will be:

• A. $2eV$
• B. $4eV$
• C. $6eV$
• D. $8eV$

Answer 1

The correct option is B $4eV$

Kinetic Energy of a charged particle in a magnetic field is $KE=\frac{P^2}{2m}=\frac{(qBr{)}^2}{2m}=\frac{q^2}{2m}{B}^2{r}^2$
$\therefore \frac{K_{\alpha }}{K_p}=(\frac{q_{\alpha }}{q_p}{)}^2\frac{m_p}{m_{\alpha }}=1\times \frac{1}{2}=\frac{1}{2}$
$\therefore K_{\alpha }=K_p\times \frac{1}{2}=8\times \frac{1}{2}=4eV$