Question

In a certain region, electric and magnetic fields exist as $E\hat{i}$ and $B\hat{k}$ respectively. If a particle having mass $m$ and charge $q$ moves under the influence of these fields, it follows a trajectory from $P$ to $Q$ as indicated in the adjoining figure. If it is known that the velocities at $P$ and $Q$ are $v\hat{i}$ and $-2v\hat{j}$ respectively, then choose the correct statements out of the following.

(a) $E=\frac{3}{4}\frac{m{v}^2}{qa}$
(b) $E=\frac{3}{2}\frac{m{v}^2}{qa}$
(c) Rate of work done by magnetic field at any point is zero.
(d) Rate of work done by electric field at any point is zero.

• A. (a),(c) and (d)
• B. (b), (c)
• C. (a), (c)
• D. None of these

Answer 1

Answer: (C)

From work-energy theorem:
Work done by the electric force is equal to the change in kinetic energy.
$W_{F_E}=\Delta K.E.$
$(qE)(2a)=\frac{1}{2}m[{\left(2v\right)}^2-{\left(v\right)}^2]\left[\because W_{F_E}=F_E\times 2a\right]$
or $E=\frac{3}{4}\left(\frac{m{v}^2}{qa}\right)$

Rate of work done by magnetic field at any point
$P=\overrightarrow{F_m}\cdot \overrightarrow{v}=0$ as $\overrightarrow{(F_m}\perp \overrightarrow{v})$ always.
At P, rate of work done by electric field will be;
$P=\overrightarrow{F_E}.\overrightarrow{v}$
$\Rightarrow P=\left(qE\right)\left(v\right)cos{0}^{\circ }$
$\Rightarrow P=q\frac{3}{4}\left(\frac{m{v}^2}{qa}\right)v$
$\Rightarrow P=\frac{3}{4}\frac{m{v}^3}{a}$
At Q, rate of work done by electric field is zero because electric field is perpendicular to the velocity of the particle at point Q.