question_answer

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 :

A)  4 eV                     

B)  2 eV

C)  8 eV                     

D)  6 eV

Correct Answer: C

Solution :

                Let m be the mass and q 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 a-particle                 \[{{E}_{\alpha }}=\frac{{{(2e)}^{2}}{{B}^{2}}{{r}^{2}}}{2(4m)}\]   ???.. (ii) From Eqs. (i) and (ii), we,get                 \[\frac{{{E}_{\alpha }}}{{{E}_{p}}}=1\] \[\Rightarrow \]               \[{{E}_{\alpha }}={{E}_{p}}=8\,eV\]                                (given)