question_answer

A proton of energy \[8\text{ }eV\]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

A)  \[4\text{ }eV\]         

B)  \[2\text{ }eV\]

C)  \[8\,\,eV\]         

D)  \[6\,\,eV\]

Correct Answer: C

Solution :

Let m be the mass and the charge of the particle moving in a magnetic field B, then the energy is given by           \[E=\frac{{{q}^{2}}{{B}^{2}}{{r}^{2}}}{2m}\]         where r is radius of circular path. For proton \[{{E}_{p}}=\frac{{{e}^{2}}{{B}^{2}}{{r}^{2}}}{2m}\] ?.(i) For \[\alpha \]-particle \[{{E}_{\alpha }}=\frac{{{(2e)}^{2}}{{B}^{2}}{{r}^{2}}}{2(4m)}\] ?..(ii) From Eqs. (i) and (ii), we get \[\frac{{{E}_{\alpha }}}{{{E}_{p}}}=1\] \[\Rightarrow \] s\[{{E}_{\alpha }}={{E}_{p}}=8\,eV\] (given)