Question

A particle of charge +q and mass m moving under the influence of a uniform electric field $E\hat{i}$ and uniform magnetic field $B\hat{k}$ follows a trajectory from P to Q as shown in fig. The velocities at P and Q are $v\hat{i}$ and $-2v\hat{j}$ 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 zero
• C. Rate or work done by both the fields at Q is zero
• D. Rate of work done by the electric field at P is $\frac{3}{4}\left[\frac{m{v}^3}{a}\right]$

Answer 1

Answer: (A), (C), (D)

From the work- energy theorem
$W=△K.E.$
$\Rightarrow qE.2a=\frac{1}{2}m[4{\upsilon }^2-{\upsilon }^2]$
$\Rightarrow E=\frac{3}{4}\left(\frac{m{\upsilon }^2}{qa}\right)$
At P rate of work done by E field will be
$=\overrightarrow{F}.\overrightarrow{\upsilon }=\left(qE\right)\left(\upsilon \right)\cos 0°$
$=\frac{3}{4}\left(\frac{m{\upsilon }^3}{a}\right)$
At Q rate of work done
$=\overrightarrow{F}.\overrightarrow{\upsilon }=\left(qE\right)\left(2\upsilon \right)\cos 90°=0$
And work done by magnetic field is always zero.