Question

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


$\left(A\right)$ $E=\frac{3}{4}\left(\frac{m{v}^2}{qa}\right)$
$\left(B\right)$ Rate of work done by the electric field at $P$ is $\frac{3}{4}\left(\frac{m{v}^2}{a}\right)$.
$\left(C\right)$ 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 $2mav$.

• A. $A,C,D$
• B. $B,C,D$
• C. $A,B,C$
• D. $A,B,C,D$

Answer 1

The correct option is C $A,B,C$

$\left(A\right)$ By work energy theorem:

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

$0+q{E}_{0}2a=\frac{3}{2}m{v}^2$

$\Rightarrow E_0=\frac{3}{4}\frac{m{v}^2}{qa}$

$\left(B\right)$ Rate of work done at $P$ = power of electric force
$w=q{E}_{0}v=\frac{3}{4}\frac{m{v}^3}{a}$.

$\left(C\right)$ At $Q$, $\frac{dw}{dt}=0$ for both the fields

$\left(D\right)$ The difference between magnitudes of angular momentum of the particle at $P$ and $Q$.

$\Delta \overrightarrow{L}=(-m(2v\left)\right(2a\left)\hat{k}\right)-(-mva\hat{k})$

$|\Delta \overrightarrow{L}|=3mva$

Hence, $\left(C\right)$ is the correct answer.