Question

Two point masses, m each carrying charges -q and +q are attached to the ends of a massless rigid non-conducting wire of length 'L'. When this arrangement is placed in a uniform electric field, then it deflects through an angle $\theta$. The minimum time needed by rod to align itself along the field is :

• A. $2\pi \sqrt{\frac{mL}{qE}}$
• B. $\frac{\pi }{2}\sqrt{\frac{mL}{2qE}}$
• C. $\pi \sqrt{\frac{2mL}{qE}}$
• D. $2\pi \sqrt{\frac{3mL}{qE}}$

Answer 1

The correct option is B $\frac{\pi }{2}\sqrt{\frac{mL}{2qE}}$

Torque when the wire is brought in a uniform field E.
$\tau =qELsin\theta$
$=qEL\theta$ [$\because \theta$ is very small]
Moment of inertia of rod AB about O
$I=m{\left(\frac{L}{2}\right)}^2+m{\left(\frac{L}{2}\right)}^2=\frac{m{L}^2}{2}$
$\tau =I\alpha$
$\therefore \alpha =\frac{\tau }{I}=\frac{qEL\theta }{\frac{m{L}^2}{2}}$
$\Rightarrow {\omega }^2\theta =\frac{2qEL\theta }{m{L}^2}[\because \alpha ={\omega }^2\theta ]$
$\Rightarrow {\omega }^2=\frac{2qE}{mL}$
Time period of the wire
$T=\frac{2\pi }{\omega }=2\pi \sqrt{\frac{mL}{2qE}}$
The rod will become parallel to the field in time $\frac{T}{4}$.
$\therefore t=\frac{T}{4}=\frac{\pi }{2}\sqrt{\frac{mL}{2qE}}$