Q.39 A charged particle accelerated by a potential V moves in a circular path with a velocity v in a
uniform magnetic field B that is perpendicular to the motion. Which of the following is/are correct
if the value of V is increased?
(A) Kinetic energy of the particle increases
(B) Radius of the circular path increases
(C) Time period of the motion increases
(D) Work done by the magnetic field increases

Charged particles accelerated by a potential difference V enter a uniform perpendicular magnetic field B and follow circular paths with velocity v. Increasing V boosts the particle’s speed, impacting kinetic energy, radius, period, and magnetic work in specific ways.

Correct Answers

Options (A) and (B) are correct. Kinetic energy rises due to higher speed from greater acceleration, and radius expands as it scales with velocity.

Option A: Kinetic Energy

Kinetic energy KE = (1/2)mv² increases when V rises, since v = sqrt(2qV/m) from energy conservation in acceleration. Higher V means higher v, thus higher KE.

Option B: Radius Increases

Radius r = mv/(qB) grows linearly with v, and since v increases with sqrt(V), r also increases for fixed B and q.

Option C: Time Period

Time period T = 2πm/(qB) stays constant, independent of v or V, as it depends only on m, q, and B. Increasing V does not affect T.

Option D: Work by Magnetic Field

Magnetic force is always perpendicular to velocity, so work done W = F⃗ · dr⃗ = 0. No change occurs; work remains zero.

Effects Summary