JEE MainPhysicsToughMagnetics

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

Show answer and solution

(A) by work energy theorem

0 + {qE}_{0}2a = \frac{3}{2}{mv}^{2}

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

= {qE}_{0}V

= \frac{3}{4}\frac{{mv}^{3}}{a}

Q,\frac{dW}{dt} = 0

for both forces
(D)

\Delta\overset{\rightarrow}{L} = ( - m2v2a\overset{\hat{}}{k}) - ( - mva\overset{\hat{}}{k})

|\Delta\overset{\rightarrow}{L}| = 3mva

Source: Grass-G / P-22 Physcis Paper 19 Dec Eng..docx

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