Question

A charged particle of mass $'m'$ and charge $'q'$ moving under the influence of uniform electric field $E\overrightarrow{i}$ and a uniform magnetic field $B\overrightarrow{k}$ follows a trajectory from point $P$ to $Q$ as shown in figure. The velocities at $P$ and $Q$ are respectively, $v\overrightarrow{i}$ and $-2v\overrightarrow{i}.$ Then which of the following statements (A, B, C D) are the correct ? (Trajectory shown is schematic and not to scale)
$\left(A\right)\quad E = \frac{3}{4}\left(\frac{m\upsilon^{2}}{qa}\right)$
$\left(B\right) \quad$ Rate of work done by the electric field at $P$ is $\frac{3}{4} \left(\frac{m\upsilon^{3}}{a}\right)$
$\left(C\right) \quad$ Rate of work done by both the fields at $Q$ is zero
$\left(D\right) \quad$ The difference between the magnitude of angular momentum of the particle at $P$ and $Q$ is $2\, ma \upsilon.$

• A. (A) , (C) , (D)
• B. (A), (B), (C)
• C. (A), (B), (C), (D)
• D. (B), (C) , (D)

Answer 1

Answer: (B)

Option $(A)$
$W = k_f - k_i$
$qE\left(2a - 0\right) =\frac{1}{2}m\left(2V\right)^{2}-\frac{1}{2}mV^{2}$
$qE2a=\frac{3}{2}mV^{2}$
$E=\frac{3}{4} \frac{mv^{2}}{qa}$
Option $\left(B\right)$
Rate of work done $P=\vec{F}. \vec{V}=FV\,cos\,\theta=FV$
Powe$r = qEV$
Power $=q\left(\frac{3}{4} \frac{mV^{2}}{qa}\right)V$
Power $=q \frac{3}{4} \frac{mV^{3}}{qa}$
Power $=\frac{3}{4} \frac{mV^{3}}{a}$
Option $(C)$
Angle between electric force and velocity is $90º$, hence rate of work done will be zero at $Q.$
Option $(D)$ Initial angular momentum $L_i=m\,V\,$
Final angular momentum $L_f=m(2V) (2a)$
Change in angular momentum $Lf - Li = 3m\,Va$
(Note : angular momentum is calculated about $O$)