Question

A charged particle of mass ${}^{\prime }{m}^{\prime }$ and charge ${}^{\prime }{q}^{\prime }$ 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)
$\left(A\right)E=\frac{3}{4}\left(\frac{m{\upsilon }^2}{qa}\right)$
$\left(B\right)$ Rate of work done by the electric field at $P$ is $\frac{3}{4}\left(\frac{m{\upsilon }^3}{a}\right)$
$\left(C\right)$ Rate of work done by both the fields at $Q$ is zero
$\left(D\right)$ The difference between the magnitude of angular momentum of the particle at $P$ and $Q$ is $2ma\upsilon .$

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

Answer 1

Answer: (B)

Option $(A)$
$W=k_f-k_i$
$qE\left(2a-0\right)=\frac{1}{2}m{\left(2V\right)}^2-\frac{1}{2}m{V}^2$
$qE2a=\frac{3}{2}m{V}^2$
$E=\frac{3}{4}\frac{m{v}^2}{qa}$
Option $\left(B\right)$
Rate of work done $P=\vec{F}.\vec{V}=FVcos\theta =FV$
Powe$r=qEV$
Power $=q\left(\frac{3}{4}\frac{m{V}^2}{qa}\right)V$
Power $=q\frac{3}{4}\frac{m{V}^3}{qa}$
Power $=\frac{3}{4}\frac{m{V}^3}{a}$
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 $L_i=mV$
Final angular momentum $L_f=m(2V)(2a)$
Change in angular momentum $Lf-Li=3mVa$
(Note : angular momentum is calculated about $O$)