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. $E=\frac{3}{4}\left[\frac{mv^2}{qa}\right]$
• B. Rate of work done by the electric field at P is $=\frac{3}{4}\left[\frac{mv^2}{qa}\right]$
• 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 $E=\frac{3}{4}\Big(\frac{mv^2}{qa}\Big)$
$\therefore $ Correct option is (a).
At $P$, rate of work done by electric field ,br>
$= F_e . v = (qE) (v)\cos 0^{\circ}$
$=q\left(\frac{3}{4}\frac{mv^2}{qa}\right)v=\frac{3}{4}\left(\frac{mv^3}{a}\right)$
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 $F_e = q E \hat{i}$