An electron is accelerated from rest through a potential difference of V. This electron experiences force F in a uniform magnetic field. On increasing the potential difference to V’ , the force experienced by the electron in the same magnetic field becomes 2F. Then, the ratio $\dfrac{{{\text{V'}}}}{{\text{V}}}{\text{ }}$ ​ is equal to
A. $\dfrac{4}{1}$
B. $\dfrac{2}{1}$
C. $\dfrac{1}{2}$
D. $\dfrac{1}{4}$

HINT: Proceed the solution of this question first by considering Kinetic energy as work done which is also equal to charge multiplied by potential difference and write velocity in formula potential difference. Hence write equations for both cases and divide them so you will get the required ratio.

Complete step-by-step answer:
In this question it is given

Potential difference V
Electron experiences of force F
A uniform magnetic field B

We know that \[{\text{K}}{\text{.E}}{\text{. = }}\dfrac{1}{2}{\text{m}}{{\text{v}}^2}\] which is equal to work done
And if a charge moves in potential difference V which is equal to work done that is equal to eV
Hence on equalising K.E. equal to eV

\[ \Rightarrow {\text{K}}{\text{.E}}{\text{. = }}\dfrac{1}{2}{\text{m}}{{\text{v}}^2}\] = eV

Hence velocity v can be written as

∴${\text{v = }}\sqrt {\dfrac{{{\text{2 eV}}}}{{\text{m}}}} $​​

We know that force F = qvB here electron charge q=e so
Force F can be written as


$ \Rightarrow {\text{F = evB}}$
On putting value of v from above expression
$ \Rightarrow {\text{F = e}}\sqrt {\dfrac{{{\text{2 ev}}}}{{\text{m}}}} {\text{B}}$ ………………. (1)
Hence in 2nd case
$ \Rightarrow {\text{F' = e}}\sqrt {\dfrac{{{\text{2 eV'}}}}{{\text{m}}}} {\text{B}}$…………………(2)


It is given that on increasing the potential difference to V’ , the force experienced by the electron in the same magnetic field becomes 2F
$ \Rightarrow {\text{F' = 2F}}$
On putting from equation (1 ) and (2)
$ \Rightarrow {\text{ e}}\sqrt {\dfrac{{{\text{2 eV'}}}}{{\text{m}}}} {\text{B = 2e}}\sqrt {\dfrac{{{\text{2 eV}}}}{{\text{m}}}} {\text{B}}$
On cancelling
$ \Rightarrow {\text{ }}\sqrt {\dfrac{{{\text{2 eV'}}}}{{\text{m}}}} {\text{ = 2}}\sqrt {\dfrac{{{\text{2 eV}}}}{{\text{m}}}} $
On squaring both side
$ \Rightarrow {\text{ }}\dfrac{{{\text{2 eV'}}}}{{\text{m}}}{\text{ = 4}} \times \dfrac{{{\text{2 eV}}}}{{\text{m}}}$
$ \Rightarrow {\text{ V' = 4V}}$
$ \Rightarrow {\text{ }}\dfrac{{{\text{V'}}}}{{\text{V}}}{\text{ = 4}}$

Hence option A is correct.

Note: It should be noted here that 1 eV is the kinetic energy produced by an electron or proton acting on a potential difference of 1 volt.
The energy formula with respect to charge and potential difference is ${\text{E = Q}}{\text{.V}}$.