Question

Two points masses, $m$ each carrying charges $-q$ and $+q$ are attached to the ends of a mass less rigid non-conducting wire of length $'L'$. When this arrangement is placed in a uniform electric field, then it deflects through an angle $i$. 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}{2 qE}}$
• C. $ \pi \sqrt{\frac{2 mL}{qE}}$
• D. $2 \pi \sqrt{\frac{3 mL}{qE}}$

Answer 1

The Correct Option is B

Torque when the wire is brought in a uniform field $E$.
$\tau=q E L \sin \theta$
$=q E L \theta[\because \theta$ is very small $]$
Moment of inertia of rod AB about $O$
$I=m\left(\frac{L}{2}\right)^{2}+m\left(\frac{L}{2}\right)^{2}=\frac{m L^{2}}{2}$
As $\tau=I \alpha$.

So $\alpha=\frac{\tau}{I}=\frac{q E L \theta}{\frac{m L^{2}}{2}}$
$\Rightarrow \omega^{2} \theta=\frac{2 q E L \theta}{m L^{2}}\left[\because \theta=\omega^{2} \theta\right]$
$\Rightarrow \omega^{2}=\frac{2 q E}{m L}$
The time period of the wire is
$T=\frac{2 \pi}{\omega}=2 \pi \sqrt{\frac{m L}{2 q E}}$
The rod will become parallel to the field in time $\frac{T}{4}$
So $t=\frac{T}{4}=\frac{\pi}{2} \sqrt{\frac{m L}{2 q E}}$