A particle of charge +q and mass m moving under the influnce of a uniform electric field `E hati `and a uniform magnetic field `Bhatk` follows trajectory from P to Q as shown in figure. The velocities at P and Q `vhati` and `-2vhatj` respectively. Which of the following statement(s) is/are correct

Options

A. `E = 3/4 (mv^2)/(qa)`
B. Rate of work done by electric field at Pis `3/4 (mv^3)/(a)`
C. Rate of work done by electric field at P is zero
D. Rate of work done by both the fields at Q is zero

Answer

A, B, D

Explanation

Kinetic energy of the particle at point ` P = 1/2 mv^2`
K.E of the particle at point `Q = 1/2 m (2v)^2`
Increase in K.E `= 3/2 mv^2`
It comes from the work done by the electric force qE on the particle as it covers a distance 2a along the x - axis .
Thus `3/2 mv^2 = qE xx 2a rArr E = 3/4 (mv^2)/(qa)`
The rate of work done by the electric field at `P = F xx v = qE xx v = 3/4 (mv^2)/(a)`
At Q, `vecF = qvecE` is along x - axis while velocity is along negative y - axis . Hence create of work done by electric field ` = vecF.vecv=0 (because = 90^@)` . similarly , according to equation `vecF_m = q (vecv xx vec B)` force `vecF_m` is also perpendicular to velocity vector `vecv`
Hence the rate of work done by the magnetic field = 0 .