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VITEEE 2015 Paper

Section: Physics

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Question : 12 of 120

Marks: +1, -0

Two points 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 i. The minimum time needed by rod to align itself along the field is

2π $\sqrt{\frac{mL}{qE}}$
$\frac{\pi}{2}$ $\sqrt{\frac{mL}{2qE}}$
π $\sqrt{\frac{2mL}{qE}}$
2π $\sqrt{\frac{3mL}{qE}}$

Solution:

Torque when the wire is brought in a uniform field E.

τ = qEL sin θ

= qELθ [Since θ 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{mL^2}{2}

∴ α =

\frac{\tau}{l}

=

\frac{qEL\theta}{\frac{mL^2}{2}}

⇒

\omega^2 \theta

=

\frac{2qEL\theta}{mL^2}

[Since θ =

\omega^2 \theta

]

⇒

\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}

∴ t =

\frac{T}{4}

=

\frac{\pi}{2}

\sqrt{\frac{mL}{2qE}}

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