In a mass spectrometer used for measuring the masses of ions, the ions are initially accelerated by an electric potential V and then made to describe semicircular paths of radius R using a magnetic field B. If V and B are kept constant, the ratio (charge on the ion/mass of the ion) will be proportional to-
A. $\dfrac{1}{R}$
B. $\dfrac{1}{{{R^2}}}$
C. ${R^2}$
D. $R$

Hint: Since we know that if there is a charged particle q which is moving with a velocity v entering in a magnetic field with direction 90 degrees, the magnetic field will make the charged particle move in a circular motion. We will equate the centripetal force and the magnetic force in order to find the value of v (velocity). We will put this value of v into the formula for kinetic energy. This will be equal to the electrical energy.

Formula used: ${F_c} = \dfrac{{m{v^2}}}{r}$, $K.E = \dfrac{1}{2}m{v^2}$

Step-By-Step answer:
The value of magnetic force will be-
$
   \Rightarrow {F_m} = q\left( {\vec v \times \vec B} \right) \\
    \\
   \Rightarrow {F_m} = q\sin 90^\circ vB \\
    \\
   \Rightarrow {F_m} = qvB \\
$
The figure of the above situation will be-


Since the particle makes a semi-circular path, the magnetic force will provide a necessary centripetal force. This is only possible if the angle between v and q is 90 degrees.
We know the formula for centripetal force is ${F_c} = \dfrac{{m{v^2}}}{r}$. Therefore-
 $
   \Rightarrow \dfrac{{m{v^2}}}{r} = qvB \\
    \\
   \Rightarrow r = \dfrac{{mv}}{{qB}} \\
$
Since the value of radius given in the question is R. So-
$
   \Rightarrow R = \dfrac{{mv}}{{qB}} \\
    \\
   \Rightarrow v = \dfrac{{RqB}}{m} \\
$
As we already know the formula for kinetic energy is $K.E = \dfrac{1}{2}m{v^2}$. Applying this formula, we get-
$ \Rightarrow K.E = \dfrac{1}{2}m{v^2}$
Putting the value of v from the above calculations, we will have-
$
   \Rightarrow K.E = \dfrac{1}{2}m{\left( {\dfrac{{qRB}}{m}} \right)^2} \\
    \\
   \Rightarrow K.E = \dfrac{1}{2}m\dfrac{{{q^2}{R^2}{B^2}}}{{{m^2}}} \\
    \\
   \Rightarrow K.E = \dfrac{{{q^2}{R^2}{B^2}}}{{2m}} \\
$
An electric potential V was applied in order to get the charge particle accelerated. The particle applied some velocity based on which its kinetic energy is $K.E = \dfrac{{{q^2}{R^2}{B^2}}}{{2m}}$.
Now, this kinetic energy should be equal to electrical energy qV where V is the electric potential.
$
   \Rightarrow K.E = qV \\
    \\
   \Rightarrow \dfrac{{{q^2}{R^2}{B^2}}}{{2m}} = qV \\
    \\
   \Rightarrow \dfrac{{q{R^2}{B^2}}}{{2m}} = V \\
$
Since the ratio asked for by the question is (charge on the ion/mass of the ion), we will arrange the above equation accordingly. We will get-
$ \Rightarrow \dfrac{q}{m} = \dfrac{{2V}}{{{B^2}{R^2}}}$
Since it was given in the question that V and B are kept constant, we can say that-
$ \Rightarrow \dfrac{q}{m} \propto \dfrac{1}{{{R^2}}}$
Hence, option B is the correct option.

Note: Magnetic energy, attraction or repulsion that occurs because of their travel through electrically charged particles. The magnetic force between two moving loads is defined as the impact of a magnetic field created by the other filled. The electrical potential, the work required to transfer the unit charge from a point of reference toward an electrical field to a specific point. The reference point usually is the Moon, although some position outside its reach may be considered.