Question

A particle of charge $+q$ and y mass m moving under the influence of a uniform electric field $E\hat{i}$ and uniform magnetic field $E\hat{i}$ follows a trajectory from $P$ to $Q$ as shown in figure. The velocities at $P$ and $Q$ are $v\hat{i}$ and $-2\hat{j}$ Which of the following statement $(s)$ is/are correct ?

• A. $E=\frac{3}{4}\left[\frac{m{v}^2}{qa}\right]$
• B. Rate of work done by the electric field at P is $=\frac{3}{4}\left[\frac{m{v}^2}{qa}\right]$
• C. Rate of work done by the electric field at P is zero
• D. Rate of work done by both the fields at Q is zero

Answer 1

Answer: (A), (B), (D)

Magnetic force does not do work. From work-energy theorem :
$W_{F_e}=\Delta KE$ or $(qE)(2a)=\frac{1}{2}m[4v^2-v^2]$
or $E=\frac{3}{4}\left(\frac{m{v}^2}{qa}\right)$
$\therefore$ Correct option is (a).
At $P$, rate of work done by electric field ,br>
$=F_e.v=(qE)(v)\cos 0^{\circ }$
$=q\left(\frac{3}{4}\frac{m{v}^2}{qa}\right)v=\frac{3}{4}\left(\frac{m{v}^3}{a}\right)$
Therefore, option (b) is also correct.
Rate of work done at $Q$ :
of electric field = $F_e\cdot v=(qE)(2v)\cos 90^{\circ }=0$
and of magnetic field is always zero. Therefore, option (d) is also correct.
Note that $F_e=qE\hat{i}$