Question

An α particle and a proton travel with same velocity in a magnetic field perpendicular to the direction of their velocities, find the ratio of the radii of their circular path
A . 4:1
B . 1:4
C . 2:1
D . 1:2

Answer 1

Hint: In this question we have two charged particles that are both moving at the same speed—one is a proton and the other is an alpha particle. Since they are both charged, they will be affected by the magnetic force and deviate from their path as they enter an area of magnetic field perpendicularly, thus we need to calculate the ratio of the radii of their circular paths.



Formula used:
Magnetic force on the particle:
${F_m}$= q(v×B); here, q denotes the charge, v is the velocity of the particle, and B the magnetic field.
The centripetal force is given by,
${{F}_{c}}=\dfrac{m{{v}^{2}}}{r}$
Here m is the mass of the particle, v is the velocity of the particle, and r is the radius.



Complete answer:
F=q(v×B) is the formula for the magnetic force acting on a charged particle in a magnetic field.

The magnetic force will have a magnitude of F=qvB since the angle between the velocity vector and the magnetic field is 90 degrees.

For both the particles the force will be in the same direction since both are positively charged particles.

The force acting on the alpha particle will be two times as strong as the force acting on the proton because both have the same velocity, the magnetic field value is the same for both, and force is directly proportional to the magnitude of charge. The charge on the alpha particle is also twice the charge on the proton.

If the proton has mass ${m_p}$, the alpha particle, let's say ${m_α}$, will have
${m_α}$ = 4 ${m_p}$.
The charge on the alpha particle will be 2e if the proton's charge is e.
The answer indicates that both particles are traveling in the same magnetic field at the same speed. Therefore, in order to achieve a circular trajectory when they are introduced to a magnetic field, the lorentz force and centripetal force must be equal. As a result,
$qvB=\dfrac{m{{v}^{2}}}{r}$
Here, m is the particle's mass, v is its velocity, q is its charge, B is the magnetic field they are entering, and r is the radius of the circular path.

After solving the relation above, we get;
$r=\dfrac{mv}{qB}$
The radius of the circular path taken by the two particles will only be influenced by their mass and charge because they both moved at the same speed and entered the same magnetic field.
The radius of proton, ${{r}_{p}}=\dfrac{{{m}_{p}}}{q}=\dfrac{{{m}_{p}}}{e}$
The radius of alpha particle, ${{r}_{\alpha }}=\dfrac{{{m}_{\alpha }}}{q}=\dfrac{4{{m}_{p}}}{2e}$
Now, dividing the above equations we get
$\dfrac{{{r}_{p}}}{{{r}_{\alpha }}}=\dfrac{\dfrac{{{m}_{p}}}{e}}{\dfrac{4{{m}_{p}}}{2e}}$
$\dfrac{{{r}_{p}}}{{{r}_{\alpha }}}=\dfrac{1}{2}$
Therefore, the ratio of the radii of the circular pathways proton and alpha particle is $\dfrac{1}{2}$
The correct answer is D.


Note: Two protons and two neutrons from the atom's nucleus combine to form the positively charged alpha particles. Alpha particles and protons both are positively charged and gets deflected in a magnetic field because of experiencing magnetic force. But if the angle between the velocity vector and the magnetic field was zero, then both of them would have passed undeflected.