A particle of charge + q and mass m moving under the influence of a uniform electric field E i and a uniform magnetic field B k follows trajectory from P to Q as shown in fig. The velocities at P and Q are v i and 

- 2 v j respectively. Which of the following statements (s) is/are correct

detailed solution

Correct option is A

(i) W=(K.E.)Q−(K.E.)P∴ qE×2a=12m(2v)2−12mv2or qE×2a=12m×3v2E=34mv2qa(ii) The rate of work done by electric field at P=Fv=qEv=q34mv2qav=34mv3a(iii) At Q, the force due to electric field is q E along X-axis. Here velocity v is along negative y-axis. Hence rate of doing work Fe .v= 0            ∵θ=90∘The force due to magnetic fieldFm = q (v x B)Here Fm- is also perpendicular to velocity vector v,∴ Rate of doing work in magnetic fieldFm. v = 0