Question

A beam of a mixture of $\alpha$ particles and protons are accelerated through the same potential difference before entering into the magnetic field of strength $B$ . If ${r_1} = 5cm$ , then ${r_2}$ is

(A) $5\;cm$
(B) $5\sqrt 2 cm$
(C) $10\sqrt 2 cm$
(D) $20\;cm$

Answer 1

We know that both alpha particles and protons are charged particles. If a charged particle perpendicular to the uniform magnetic field. The magnetic Lorentz force is acting perpendicular to the velocity. This supplies the necessary centripetal force required for circular motion.

Formula used:
$v = \dfrac{{qBr}}{m}$
Where $v$ is the velocity of the particle, $q$ stands for the charge of the particle, $B$ stands for the magnetic field, $r$ is the radius of the circular path, and $m$ is the mass of the particle.

Complete Step by step solution:
The velocity of a charged particle moving in a circular path in a uniform magnetic field is given by,
$v = \dfrac{{qBr}}{m}$
From this equation, we can write the expression for radius as,
$r = \dfrac{{mv}}{{qB}}$
We know that the momentum of the particle,
$P = mv$
Therefore, the radius can be written as
$r = \dfrac{P}{{qB}}$
We know that momentum can be written as,
$P = \sqrt {2mE}$
Substituting in the above equation, we get
$r = \dfrac{{\sqrt {2mE} }}{{qB}}$
We can write the energy of a charged particle in a potential $V$ as,
$E = qV$
Substituting in the above equation, we get
$r = \dfrac{{\sqrt {2mqV} }}{{qB}}$
This equation can be rearranged as,
$r = \sqrt {\dfrac{m}{q}} \dfrac{{\sqrt {2V} }}{B}$
The radius ${r_1}$ can be written as,
${r_1} = \sqrt {\dfrac{{{m_1}}}{{{q_1}}}} \dfrac{{\sqrt {2V} }}{B}$
The radius ${r_2}$ can be written as,
${r_2} = \sqrt {\dfrac{{{m_2}}}{{{q_2}}}} \dfrac{{\sqrt {2V} }}{B}$
The ratio of the two radii can be written as,
$\dfrac{{{r_2}}}{{{r_1}}} = \dfrac{{\sqrt {\dfrac{{{m_2}}}{{{q_2}}}} \dfrac{{\sqrt {2V} }}{B}}}{{\sqrt {\dfrac{{{m_1}}}{{{q_1}}}} \dfrac{{\sqrt {2V} }}{B}}}$
Cancelling the common terms and rearranging, we get
$\dfrac{{{r_2}}}{{{r_1}}} = \sqrt {\dfrac{{{m_2}}}{{{q_2}}}} \times \sqrt {\dfrac{{{q_1}}}{{{m_1}}}}$
This can be written as,
$\dfrac{{{r_2}}}{{{r_1}}} = \sqrt {\dfrac{{{m_2}}}{{{m_1}}}} \times \sqrt {\dfrac{{{q_1}}}{{{q_2}}}}$
We know that the mass of an alpha particle is four times that of the mass of protons. i.e.
${m_\alpha } = 4{m_p}$
Here ${m_2}$ is the mass of the alpha particle and ${m_p}$ is the mass of the proton.
Therefore we can write,
$\sqrt {\dfrac{{{m_2}}}{{{m_1}}}} = \sqrt {\dfrac{4}{1}}$
Also, the charge of alpha particles are two times that of the charge of protons, i.e.
${q_\alpha } = 2{q_p}$
Here ${q_2}$ is the charge of the alpha particle and ${q_1}$ is the charge of the proton, then we can write
${q_2} = 2{q_p}$
Then the ratio will be
$\sqrt {\dfrac{{{q_1}}}{{{q_2}}}} = \sqrt {\dfrac{1}{2}}$
Putting these values in the expression
$\dfrac{{{r_2}}}{{{r_1}}} = \sqrt {\dfrac{{{m_2}}}{{{m_1}}}} \times \sqrt {\dfrac{{{q_1}}}{{{q_2}}}}$
We get,
$\dfrac{{{r_2}}}{{{r_1}}} = \sqrt {\dfrac{4}{1}} \times \sqrt {\dfrac{1}{2}} = \sqrt 2$
From this ${r_2} = {r_1}\sqrt 2$
It is given that, the value of ${r_1} = 5cm$
Then,
${r_2} = 5\sqrt 2 cm$
The answer is: Option (B): $5\sqrt 2 cm$ .

Note:
The radius $r = \dfrac{{mv}}{{qB}}$ is known as the cyclotron radius. A cyclotron is a device employed to accelerate charged particles to high energies. It works on the principle that a charged particle moving normal to magnetic flux experiences magnetic Lorentz force. Because of this force, the particle moves in a circular path.