Q. A proton of energy $8\,eV $ is moving in a circular path in a uniform magnetic field. The energy of an alpha particle moving in the same magnetic field and along the same path will be
J & K CETJ & K CET 2004
Solution:
Let $m$ be the mass and the charge of the particle moving in a magnetic field $B$, then the energy is given by
$E=\frac{q^{2} B^{2} r^{2}}{2 m}$ where $r$ is radius of circular path.
For proton $E_{p}=\frac{e^{2} B^{2} r^{2}}{2 m}$ ...(i)
For $\alpha$ -particle $E_{\alpha}=\frac{(2 e)^{2} B^{2} r^{2}}{2(4 m)}$ ...(ii)
From Eqs. (i) and (ii),
we get $\frac{E_{\alpha}}{E_{p}}=1$
$\Rightarrow E_{\alpha} =E_{p}=8\, eV$ (given)
