question_answer

A particle of charge +q and mass m moving under the influence of a uniform electric field \[E\hat{i}\] and a uniform magnetic field \[B\hat{k}\] follows trajectory from P to Q as shown in figure. The velocities at P and Q are \[v\hat{i}\] and \[-2v\hat{j}\] respectively. Which of the following statement(s) is/are correct                                           [IIT 1991; BVP 2003]

A)            \[E=\frac{3}{4}\frac{m{{v}^{2}}}{qa}\]

B)            Rate of work done by electric field at P is \[\frac{3}{4}\frac{m{{v}^{3}}}{a}\]

C)            Rate of work done by electric field at P is zero

D)            Rate of work done by both the fields at Q is zero

Correct Answer: B

Solution :

 Kinetic energy of the particle at point \[P=\frac{1}{2}m{{v}^{2}}\]            K.E. of the particle at point \[Q=\frac{1}{2}m{{(2v)}^{2}}\] Increase in K.E. \[=\frac{3}{2}m{{v}^{2}}\] It comes from the work done by the electric force qE on the particle as it covers a distance 2a along the x-axis. Thus \[\frac{3}{2}m{{v}^{2}}=qE\times 2a\Rightarrow E=\frac{3}{4}\frac{m{{v}^{2}}}{qa}\]. The rate of work done by the electric field at P  \[=F\times v=qE\times v=3\frac{m{{v}^{3}}}{4a}\] At \[Q,\ \overrightarrow{{{F}_{e}}}=q\overrightarrow{E}\] is along x-axis while velocity is along negative y-axis. Hence rate of work done by electric field \[=\overrightarrow{{{F}_{e}}.}\,\overrightarrow{v\,}=0\ \ (\because \theta ={{90}^{o}})\] Similarly, according to equation \[\overrightarrow{{{F}_{m}}}=q(\overrightarrow{v\,}\times \overrightarrow{B})\] Force \[\overrightarrow{{{F}_{\text{m}}}}\]is also perpendicular to velocity vector \[v\]. Hence the rate of work done by the magnetic field = 0