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 the 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

Answer: (B)

When the wire is brought in a uniform field E, then the torque is given by
$\tau =qELsin\theta$
$=qEL\theta$ $\left[\because \theta isverysmall\right]$
The moment of inertia of rod AB about O is
$I=m{\left(\frac{L}{2}\right)}^2+m{\left(\frac{L}{2}\right)}^2=\frac{m{L}^2}{2}$
As $\tau =I\alpha$

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