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,…

Question

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 $\mathrm{P}$ and $\mathrm{Q}$. When a current $I$ is passed through the loop, the loop turns about the line $\mathrm{PQ}$ by an angle $\theta$ given by

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Let loop makes angle } 	heta 	ext { with vertical. } \ & 	ext { in equilibrium } 	au_{	ext {net }}=0 \ & 	au_0=\mathrm{MB} \sin (90-	heta)-\mathrm{mg} . \mathrm{r} \sin 	heta=0 \ & 	ext { I. } \pi \mathrm{r}^2 \cdot \mathrm{B}_0 \cos 	heta=\mathrm{mg} 	ext { r.sin } 	heta \ & 	an 	heta=\frac{\pi \mathrm{rIB}_0}{\mathrm{mg}} \ & \end{aligned}$

A thin stiff insulated metal wire is bent into a circular loop with its two ends extending tangentially from…

Solution