Question

A charged particle having charge Q1 accelerated through a potential difference V enters a perpendicular magnetic field in which it experiences force F. If V is increased to 5v the particle will experience a force​

Answer 1

A charged particle having charge Q1 accelerated through a potential difference V.

Let velocity of particle (of mass m) becomes u.

so, kinetic energy = 1/2 mu²

electric potential energy = Q1V

so, kinetic energy = electric potential energy

or, 1/2 mu² = Q1V

or, u = √{2Q1.V/m}

we know, force exerted on moving charge particle is given by, F = q(v × B)

where v is velocity of charged particle and B is magnetic field.

here, a magnetic force F is acting perpendicularly on particle. also, v = u = √{2Q1V/m}, q = Q1

so, F = Q1 × √{2Q1V/m}B

= √{2Q1²V/m}B

here it is clear that force is directly proportional to square root of V(i.e., potential difference ).

so, if V is increased to 5V

then, force, F is increased √5F

hence, the particle will experience a force √5F