Question

A proton, a deuteron and an $\alpha $- particle with the same kinetic energy enter a region of uniform magnetic field moving at right angles to B. What is the ratio of the radii of their circular paths?
A. \[1:\sqrt{2}:\sqrt{2}\]
B. \[1:\sqrt{2}:1\]
C. \[\sqrt{2}:1:1\]
D. \[\sqrt{2}:\sqrt{2}:1\]

Answer 1

Hint: When a charged particle enters a uniform magnetic field with some speed it experiences magnetic force. Now if the particle has entered at a right angle to the field, the magnetic force provides necessary centripetal force and the particle performs a circular motion with constant speed. The radius of that circular path depends on the mass and the charge of the particle.

Complete step by step solution:
Let there be a charged particle with charge $q$ and mass $m$ enter perpendicular to a constant magnetic field $B$ with a constant speed $v$ , the charge would experience a magnetic force $F$ and this force will provide a centripetal force to the particle and particle takes a circular path of radius $r$.
We can write:
 $ \Rightarrow F=q(\vec{v}\times \vec{B})=\dfrac{m{{v}^{2}}}{r} \\
  \Rightarrow qvB=\dfrac{m{{v}^{2}}}{r} \\
  \Rightarrow r=\dfrac{mv}{qB} $
Also, kinetic energy K and momentum$P=mv$ can be related as:
$ K=\dfrac{1}{2}m{{v}^{2}}=\dfrac{{{P}^{2}}}{2m} \\
  \Rightarrow P=\sqrt{2mK}=mv $

Now from equation of radius and momentum: $r=\dfrac{\sqrt{2mK}}{Bq}\Rightarrow r\propto \dfrac{\sqrt{m}}{q}$.
Let ${{m}_{\alpha }},{{m}_{p}},{{m}_{d}}\And {{q}_{\alpha }},{{q}_{p}},{{q}_{d}}\And {{r}_{\alpha }},{{r}_{p}},{{r}_{d}}$ be the mass, charge and radius by the alpha-particle, proton and deuteron respectively. These are related as:
$ \Rightarrow {{q}_{\alpha }}=2{{q}_{p}}\And {{q}_{d}}=2{{q}_{p}}
 {{m}_{\alpha }}=4{{m}_{p}}\And {{m}_{d}}=2{{m}_{p}} $
$
\Rightarrow {{r}_{p}}:{{r}_{d}}:{{r}_{\alpha }}=\dfrac{\sqrt{{{m}_{p}}}}{{{q}_{p}}}:\dfrac{\sqrt{{{m}_{d}}}}{{{q}_{d}}}:\dfrac{\sqrt{{{m}_{\alpha }}}}{{{q}_{\alpha }}} $
$ \Rightarrow {{r}_{p}}:{{r}_{d}}:{{r}_{\alpha }}=\dfrac{\sqrt{{{m}_{p}}}}{{{q}_{p}}}:\dfrac{\sqrt{2{{m}_{p}}}}{{{q}_{p}}}:\dfrac{\sqrt{4{{m}_{p}}}}{2{{q}_{p}}} $
$ \Rightarrow {{r}_{p}}:{{r}_{d}}:{{r}_{\alpha }}=1:\sqrt{2}:1$

Hence the correct option is (B).

Additional information: Magnetic force can alter the direction of a particle’s motion but cannot alter its energy. Since the force is also perpendicular to the magnetic field, it cannot change a particle’s velocity component parallel to the field. Also, the electric force does work in displacing a charged particle, whereas the magnetic force associated with a constant magnetic field does not work when a charged particle is displaced.

Note: If a charged particle enters a magnetic field at an angle (not perpendicular), the velocity of the particle is resolved into two mutually perpendicular components, one along the field and the other perpendicular to the field. The component perpendicular component causes a circular motion and the one along the field makes the particle move along the field and as a result we get a spiral motion.