Question

A proton, deuteron, and an $\alpha$-particle enter a magnetic field perpendicular to the field with the same velocity. What is the ratio of the radii of circular paths?

Answer 1

Step 1. Given data:

1. In the case of the proton, let mass be $m_p=m$, charge = $q$(one proton charge)
2. In the case of the deuteron, there is one proton and one neutron, so $m_d=2m$(as the mass of proton and neutron is identical), charge= $q$(one proton charge)
3. In the case of $\alpha$-particle, i.e. $H{e_2}^4$, the mass of $\alpha$-particle, $m_{\alpha }=4m$, charge = $2q$(two protons charge)
4. As a result of charges experienced by all three particles, they will feel a magnetic pull as they pass through a magnetic field zone.
5. The equation of motion is used to calculate the ratio of the trajectory.

Step 2. Formula used:

1. Force on a charged particle due to circular motion: $F=\frac{m{v}^2}{r}$, where $m,v,r$ are mass, velocity, and radius of the particle in a circular motion.
2. Force on a charged particle due to the magnetic field: $F_B=qvB.\sin \theta$, where $q,v,B$ is the charge, velocity, and magnetic field, $\theta$is the angle between the velocity vector and magnetic field vector.
3. Since all of them enter perpendicularly, the angle between the velocity vector and the field vector is $${90}^0$$. so $\sin 90°=1$
4. Thus, force on a charged particle due to the magnetic field: $F_B=qvB$

Step 3. Calculating the radius of circular path

The motion of charged particles in electric and magnetic fields. $F=qvB$.

Here, the magnetic force becomes centripetal force due to its direction towards the circular motion of the particle.

Thus, if the field and velocity are perpendicular to each other, then the particle takes a circular path.

A moving electric charge behaves like a mini-magnet as it creates its own magnetic field.

This means it experiences a force if it moves through an external magnetic field (in the same way that a mass experiences a force in a gravitational field or a charge experiences a force in an electric field.)

Since the magnetic force is perpendicular to the direction of travel, a charged particle follows a curved path in a magnetic field.

The particle continues to follow this curved path until it forms a complete circle.

Now, force on a charged particle due to circular motion = Force on a charged particle due to the magnetic field

$$\begin{gathered} \Rightarrow \frac{m{v}^2}{r}=qvB \\ \Rightarrow r=\frac{mv}{qB} \end{gathered}$$

Here, $v$ and $B$ are constant. Hence $r\propto \frac{m}{q}$, Now putting the above values in this equation, we get

$$\begin{gathered} r_p:r_d:r_{\alpha }=\frac{m}{q}:\frac{2m}{q}:\frac{4m}{2q} \\ {r}_p:r_d:r_{\alpha }=1:2:2 \end{gathered}$$

This means the proton will have the circular path of the shortest radius while the circular path of deuteron and alpha particle will be twice the radius of the circular path of the proton.

Hence, the ratio of the radii of circular paths of all the particles is $r_p:r_d:r_{\alpha }=1:2:2$.