Question

The charge on a particle Y is double the charge on particle X. These two particles X and Y after being accelerated through the same potential difference enter a region of uniform magnetic field and describe circular paths of radii \[{R_1}\] and \[{R_2}\] respectively. The ratio of the mass of X to that of Y is
A. \[{\left( {\dfrac{{2{R_1}}}{{{R_2}}}} \right)^2}\]
B. \[{\left( {\dfrac{{{R_1}}}{{2{R_2}}}} \right)^2}\]
C. \[\dfrac{{R_1^2}}{{2R_2^2}}\]
D. \[\dfrac{{2{R_1}}}{{{R_2}}}\]

Answer 1

Hint:When a charged particle is moving in a region with a magnetic field which is perpendicular to the motion of the electron. The magnetic force acts inward which makes the path of the motion circular. The velocity is imparted in the charged particle by application of potential difference.

Formula used:
\[{F_m} = qVB\]
Here \[{F_m}\] is the magnetic field acting on the charge particle q moving with speed v in a magnetic field strength B where the motion is perpendicular to the magnetic field.
\[{F_c} = \dfrac{{m{v^2}}}{r}\]
Here \[{F_c}\] is the centrifugal force acting on a body of mass m moving with speed v in a circular path of radius r.
\[\dfrac{{m{v^2}}}{2} = qV\]
Here kinetic energy is equal to the work done by the potential difference.

Complete step by step solution:
It is given that the charge of particle Y is double the charge on the particle X,
\[{q_Y} = 2{q_X}\]
The two of the particles are accelerated through the same potential difference, let the potential difference is V, then the speeds will be as,
For the particle X,
\[\dfrac{{{m_X}v_X^2}}{2} = {q_X}V \\ \]
\[\Rightarrow {v_x} = \sqrt {\dfrac{{2{q_X}V}}{{{m_X}}}} \\ \]
For the particle Y,
\[\dfrac{{{m_y}v_y^2}}{2} = {q_y}V \\ \]
\[\Rightarrow {v_Y} = \sqrt {\dfrac{{2{q_Y}V}}{{{m_Y}}}} \]

Let the magnetic field strength of the region is B, then in circular motion the outward force will be balanced by the inward magnetic force,
\[\dfrac{{m{v^2}}}{r} = qvB \\ \]
\[\Rightarrow r = \dfrac{{mv}}{{Bq}} \\ \]
For the particle X,
\[{r_X} = \dfrac{{{m_X}{v_X}}}{{B{q_X}}} \\ \]
\[\Rightarrow {r_X} = \dfrac{{{m_X}\sqrt {\dfrac{{2{q_x}V}}{{{m_X}}}} }}{{B{q_X}}} \\ \]
\[\Rightarrow {r_X} = \dfrac{1}{B}\sqrt {\dfrac{{2V{m_X}}}{{{q_x}}}} \\ \]
Similarly for particle Y.
\[{r_Y} = \dfrac{1}{B}\sqrt {\dfrac{{2V{m_Y}}}{{{q_Y}}}} \]

On dividing the expression of the radius of the particles, we get
\[\dfrac{{{R_1}}}{{{R_2}}} = \dfrac{{\dfrac{1}{B}\sqrt {\dfrac{{2V{m_X}}}{{{q_X}}}} }}{{\dfrac{1}{B}\sqrt {\dfrac{{2V{m_Y}}}{{{q_Y}}}} }} \\ \]
\[\Rightarrow \dfrac{{{R_1}}}{{{R_2}}} = \sqrt {\dfrac{{{m_X}}}{{{m_Y}}} \times \dfrac{{{q_y}}}{{{q_X}}}} \\ \]
\[\Rightarrow \dfrac{{{m_X}}}{{{m_Y}}} \times \dfrac{{2{q_X}}}{{{q_X}}} = {\left( {\dfrac{{{R_1}}}{{{R_2}}}} \right)^2} \\ \]
\[\therefore \dfrac{{{m_X}}}{{{m_Y}}} = \dfrac{{R_1^2}}{{2R_2^2}}\]
So, the ratio of the mass of X to the mass of Y is \[\dfrac{{R_1^2}}{{2R_2^2}}\].

Therefore, the correct option is C.

Note: As both the particles are in the same region of the magnetic field, so the ratio of the masses is independent of the magnetic field strength as well the nature of the charge on the particle; whether positive or negative.