(A) by work energy theorem
Wmag+Wele=21m(2v)2−21m(v)2
0+qE02a=23mv2
E0=43qamv2
(B) Rate of work done at A= power of electric force
=qE0V
=43amv3
(C) at Q,dtdw=0 for both forces
(D) ΔL=(−m2 v 2ak^)−(− mvak )
∣ΔL∣=3mva
A charged particle of mass ‘ m ’ and charge ‘ q ’ moving under the influence of uniform electric field E i and a uniform magnetic field B k follows a…
JEE Main 2020 — Physics Electromagnetism
2020mcqmedium
A charged particle of mass ‘ m ’ and charge ‘ q ’ moving under the influence of uniform electric field Ei^ and a uniform magnetic field Bk^ follows a trajectory from point P to Q as shown in figure. The velocities at P and Q are respectively, vi and −2vj . Then which of the following statements (A, B, C, D) are the correct? (Trajectory shown is schematic and not to scale)
(A) E=23(qamv2)
(B) Rate of work done by the electric field at P is 23(amv3)
(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 .
Official previous-year question
Held on 9 Jan 2020 · Verified 6 Jul 2026.
Options
- A
(A),(C),(D)
- B
(B),(C),(D)
- C
(A),(B),(C)
- D
(A),(B),(C),(D)
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