A particle of charge ` +q` and mass `m` moving under the influence of a uniform electric field ` E hat(i)` and uniform magnetic field ` B hat(k)` follows a trajectory from ` P to Q` as shown in fig. The velocities at `P and Q` are ` v hat(i)` and ` - 2v hat(j)` . 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 `P` is `(3)/(4)(mv^(3))/(a)`
C. Rate of work done by electric field at `P` is zero
D. Rate of word down by both the fields at `Q` is zero

Answer

1,2,3

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)=qExx2aimplies E=(3)/(4)(mv^(2))/(qa)`
The rate of work done by the electric field at `P=Fxxv=qExxv=(3)/(4)(mv^(2))/(a)`
At `A, vec(F)_(e)=qvec(E)` is along `x-` axis while velocity is along negative `y-` axis. Hence rate of work done by electric field `=vec(F)_(e).vec(v)=0 ( :. theta =90^(@))`. Similarly, according to equation `vec(F)_(m)=q(vec(v)xxvec(B))` force `vec(F)_(m)` is also perpendicular to velocity vector `vec(v)`.
Hence the rate of work done by the magnetic field `=0`.