A charged particle of mass 'm' and charge 'q' moving under the influence of uniform electric field Ei⃗ and a uniform magnetic field Bk⃗ follows a trajectory from point P to Q as shown in figure. The velocities at P and Q are respectively, vi⃗ and - 2vj⃗ . Then which of the following statements (A, B, C, D) are the correct ?
(Trajectory shown is schematic and not to scale) :
(A) E = 3/4( mv²/qa )
(B) Rate of work done by the electric field at P is 3/4( mv³/a )
(C) Rate of work done by both the fields at Q is zero
(D) The difference between the magnitude of angular momentum of the particle at P and Q is 2mav.

Solution

Physics sir · 4 steps

The magnetic force never does work, so the whole kinetic energy gain from P to Q comes from E.

1. Work-energy theorem. The magnetic force is always perpendicular to the velocity, so it does no work. The electric force qE acts along x, and from P to Q the particle moves 2a along x. That work equals the gain in kinetic energy.

   qE(2a) = 1/2m(2v)² - 1/2mv² = 3/2mv² ⇒ E = 3/4(mv²/qa)

2. Rate of work. At P the velocity vî is along the electric force, so the power is qEv = 3/4(mv³/a). At Q the velocity -2vĵ is perpendicular to Eî, so the electric power is zero, and the magnetic power is always zero. So (A), (B) and (C) are all correct.
3. Angular momentum about O. At P, r⃗ = aĵ and v⃗ = vî, so |L⃗_P| = mav. At Q, r⃗ = 2aî and v⃗ = -2vĵ, so |L⃗_Q| = m(2a)(2v) = 4mav. The difference is 3mav, not 2mav, so (D) is wrong.
4. So. Statements (A), (B) and (C) are correct, which is option (b).

The idea to keep. In combined E and B fields, the B field only bends the path. Speed changes come entirely from the electric field, so use work-energy with only the displacement along E.

Answer from the NTA final answer key. Solution by PrepWiser.