Magnetics 1 Question 3
3. A proton and an $\alpha$-particle (with their masses in the ratio of $1: 4$ and charges in the ratio of $1: 2$ ) are accelerated from rest through a potential difference $V$. If a uniform magnetic field $B$ is set up perpendicular to their velocities, the ratio of the $\operatorname{radii} r _p: r _{\alpha}$ of the circular paths described by them will be
(a) $1: \sqrt{2}$
(b) $1: \sqrt{3}$
(c) $1: 3$
(d) $1: 2$
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Answer:
Correct Answer: 3. (a)
Solution:
- Radius of path of charged particle $q$ in a uniform magnetic field $B$ of mass ’ $m$ ’ moving with velocity $v$ is
$$ \begin{aligned} r & =\frac{m v}{B q}=\frac{m \sqrt{(2 q V / m)}}{B q} \\ \Rightarrow \quad r & \propto \frac{\sqrt{m}}{\sqrt{q}} \text { so, required ratio is } \\ \Rightarrow \quad \frac{r _p}{r _{\alpha}} & =\sqrt{\frac{m _p}{m _{\alpha}}} \times \sqrt{\frac{q _{\alpha}}{q _p}} \\ & =\sqrt{\frac{1}{4}} \times \sqrt{\frac{2}{1}}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}} \end{aligned} $$
