A charged particle of mass 'm' and charge 'q' moving under the influence of uniform electric field $E \vec{i}$ and a uniform magnetic field $\vec{Bk}$ follows a trajectory from point P to Q as shown in figure. The velocities at P and Q are respectively, $v \vec{i} and - 2 v \vec{j}$ . Then which of the following statements (A, B, C, D) are not correct ? (Trajectory shown is schematic and not to scale)

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$E = \frac{3}{4} (\frac{m v^{2}}{q a})$

$Rate of work done by the electric field at P is \frac{3}{4} (\frac{m v^{3}}{a})$

Rate of work done by both the fields at Q is zero

The difference between the magnitude of angular momentum of the particle at P and Q is 2mav.

answer is A.

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Detailed Solution

(A) by work energy theorem

$W_{mag} + W_{e l e} = \frac{1}{2} m (2 v)^{2} - \frac{1}{2} m (v)^{2}$

$\begin{aligned} 0 + q E_{0} 2 a = \frac{3}{2} m v^{2} \\ E_{0} = \frac{3}{4} \frac{mv}{q} \end{aligned}$

(B) Rate of work done at A = power of electric force

$= {qE}_{0} V = \frac{3}{4} \frac{{mv}^{3}}{a}$

$(C) at Q, \frac{d w}{d t} = 0 for both forces$

(D) $Δ \vec{L} = (- m 2 v 2 a \hat{k}) - (- m v a \hat{k}) \Rightarrow | Δ \vec{L} | = 3 mva$

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