Question

An electron, a proton, a deuteron, and an alpha particle, each having speed are in a region of constant magnetic field perpendicular to the direction of the velocities of the particles. The radius of the circular orbits of these particles is respectively $R_{e,}{R}_p,R_d$and $R_{\alpha }$. It follows that
 A. $R_e=R_p$
B.$R_p=R_d$
C.$R_d=R_{\alpha }$
D.$R_p=R_{\alpha }$

Answer 1

Hint: When a charged particle moves with definite velocity and enters a uniform magnetic field $B$, then it experiences a magnetic force perpendicular to the direction of motion and it travels a circular path. Then by equating magnetic force with centripetal force, we can derive the equation of the radius of a circular path.


Formula used:
The radius,$R$ of the circular path in a magnetic field,$B$can be expressed in the following way:
$R={\displaystyle \frac{mv}{qB}}$
Here $m\mathrm{\&}v$are the mass and velocity of the particle with charge $q$.


Complete answer:
When a particle carrying charge$q$, moving with velocity $\overrightarrow{v}$enters into a magnetic field $\overrightarrow{B}$, it experiences a magnetic force, $\overrightarrow{F}=q(\overrightarrow{B}\times \overrightarrow{v})$

Or,$F=q(Bv\sin {90}^o)$ [Since $\overrightarrow{B}$is perpendicular to $\overrightarrow{v}$]
Or,$F=qBv$ ……..(i)
As a particle moves in a circular path, then magnetic force becomes a centripetal force ${\displaystyle \frac{m{v}^2}{R}}$.
Hence by equating magnetic force with centripetal force,
$qBv={\displaystyle \frac{m{v}^2}{R}}$
Or,$R={\displaystyle \frac{mv}{qB}}$
Here we have four charged particles: an electron($e$), a proton ($p$), a deuteron ($d$), and an alpha particle $(\alpha )$. They all have equal speed,$v$and move in a region of the constant magnetic field,$B$.
Therefore the radius of the circular path mainly depends on ${\displaystyle \frac{mass(m)}{ch\arg e(q)}}$ratio.
Or,$R$ $\alpha$ ${\displaystyle \frac{m}{q}}$
Let us check ${\displaystyle \frac{m}{q}}$ratio of each charged particle in the following table,
Let the Mass of a proton be $m$and charge $q$.

	Proton	Electron	deuteron	deuteron
Mass($m$)	$m$	${\displaystyle \frac{m}{1836}}$	$2m$	$4m$
Charge($q$)	$q$	$q$	$q$	$2q$
${\displaystyle \frac{mass}{ch\arg e}}\left({\displaystyle \frac{m}{q}}\right)$	${\displaystyle \frac{m}{q}}$	${\displaystyle \frac{m}{q\times 1836}}$	${\displaystyle \frac{2m}{q}}$	${\displaystyle \frac{4m}{2q}}={\displaystyle \frac{2m}{q}}$

As we know the mass of an electron, deuterium and an alpha particle are ${\displaystyle \frac{1}{1836}}$, $2$ and $4$ times the mass of the proton.
Therefore ${\displaystyle \frac{m}{q}}$ ratio for deuterium and an alpha particle are equal, hence their radius of the circular orbit would be equal i.e,$R_d=R_{\alpha }$.

Thus, option (C) is correct.

Note:Neutron does not feel any magnetic force while other charged particles experience that force. A neutron is a neutral particle, having no charge. But for charged particle trajectory curvature is proportional to the mass by charge ratio for a definite velocity.