MCQ
A thin stiff insulated metal wire is bent into a circular loop with its two ends extending tangentially from the same point of the loop. The wire loop has mass $m$ and radius $r$ and it is in a uniform vertical magnetic field $B_0$, as shown in the figure. Initially, it hangs vertically downwards, because of acceleration due to gravity $g$, on two conducting supports at $P$ and $Q$. When a current $/$ is passed through the loop, the loop turns about the line $P Q$ by an angle $\theta$ given by
  • $\tan \theta=\pi r I B_0 /(m g)$
  • B
    $\tan \theta=2 \pi r I B_0 /(m g)$
  • C
    $\tan \theta=\pi r I B_0 /(2 m g)$
  • D
     $\tan \theta=m g /\left(\pi r / B_0\right)$

Answer

Correct option: A.
$\tan \theta=\pi r I B_0 /(m g)$
a
Let loop makes angle $\theta$ with vertical.

$\text { in equilibrium } \tau_{\text {net }}=0$

$\tau_0= MB \sin (90-\theta)- mg \cdot r \sin \theta=0$

$\text { I. } \pi r ^2 \cdot B _0 \cos \theta= mg r \cdot \sin \theta$

$\tan \theta=\frac{\pi rIB }{ mg }$

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