Question

A proton of energy $8eV$ 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

Answer 1

Answer: (C)

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 $\frac{E_{\alpha }}{E_p}=1$
$\Rightarrow E_{\alpha }=E_p=8eV$ (given)