A charge particle having charge q is accelerated through a potential difference

Question

A charge particle having charge q is accelerated through a potential difference V and then it enters a perpendicular magnetic field in which it experiences a force F. If V is increased to 5V, the particle will experience a force

Options:

A . F

B . 5F

C . F5

D . √5F

Answer: Option D
:
D

\frac{1}{2} m v^{2} = q V \Rightarrow v = \sqrt{\frac{2 q V}{m}}

. Also

F = q v B

\Rightarrow F = q B \sqrt{\frac{2 q V}{m}}

hence

F \propto \sqrt{V}

which gives

F^{'} = \sqrt{5} F

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