Hint: Here we have two charged particles, one is a proton and the other is an alpha particle and they both are moving in a straight line towards the positive x-axis. They enter a region of magnetic field perpendicularly and we need to fight the trajectory since they both are charged they will be acted upon by the magnetic force
Complete step by step answer: The magnetic force on a charged particle placed in a magnetic field is given by \[\overrightarrow{F}=q(\overrightarrow{v}\times \overrightarrow{B})\]
The angle between the velocity vector and the magnetic field is \[90{}^\circ \], so the magnitude of the magnetic force will be,
$F=qvB$
The direction is given as \[=v\widehat{i}\times B(-\widehat{k})\].
Since we are interested in finding the direction, taking only unit vectors we get,
$ =\widehat{i}\times (-\widehat{k}) $
$ =\widehat{j} $
So the force acts towards the positive y-axis that is in an upward direction for both the particles. Now let us find the magnitude for both of them. Since both have the same velocity and the value of the magnetic field is the same for both and force is directly proportional to the magnitude of charge and the charge on the alpha particle is twice the value of charge on the proton, so the force acting on the alpha particle will be two times in magnitude the force acting on the proton. So the trajectory looks like,
The ratio of the radii of the circular paths of these two particles can be found as;
Let the mass of the proton be $m_p$ then the mass of the alpha particle say ${m_{\alpha}}$ will be $4{m_p}$. Similarly if the charge on proton is $e$ then the charge on alpha particle will be $2e$. It is given in the question that both the particle are moving with same velocity in the same magnetic field. So, when they are entered in the magnetic field then lorentz force must be balanced by centripetal force inorder to obtain the circular trajectory. So we get,
$qvB=m\dfrac{v^2}{r}$
Here $m$= mass of the particle, $v$=velocity of the particle, $q$= Charge on the particle, $B$= magnetic field in which they are entering and $r$= radius of the circular path
After solving the above relation we get;
$r=\dfrac{mv}{qB}$
Since both the particles travelled with the same speed and entered in the same magnetic field, the radius of the circular path travelled by them will depend only on the mass and charge of the particle.
$r_p=\dfrac{m_p}{q}= \dfrac{m_p}{e}$...............(Here $r_p$ is the radii for proton and $q=e$)
$\Rightarrow {r_{\alpha}}=\dfrac{m_{\alpha}}{q}=\dfrac{4{m_p}}{2e}$..........(Here ${m_{\alpha}}=4{m_p}$ and $q=4e$)
Now dividing the above equations so that we can obtain the ratio.
$\dfrac{r_p}{r_{\alpha}}=\dfrac{\dfrac{m_p}{e}}{\dfrac{4{m_p}}{2e}}\\$
After cancellation of $m_p$ and $e$ we get,
$\therefore \dfrac{r_p}{r_{\alpha}}=\dfrac{1}{2}$
Hence, the ratio of the radii of the circular paths of proton and alpha particles is $\dfrac{1}{2}$.
Proton has a charge $q$ and the charge on the alpha particle which is a doubly ionized helium atom is $2q$. Both are positive and when they enter the magnetic field, they experience the force. If the angle between the velocity vector and the magnetic field was zero, then both of them would have passed undeflected.
Verified
