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WBJEEA thin stiff insulated metal wire is bent into a circular loop with...
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
Solution:
\(\begin{aligned} & \text { Let loop makes angle } \theta \text { with vertical. } \\ & \text { in equilibrium } \tau_{\text {net }}=0 \\ & \tau_0=\mathrm{MB} \sin (90-\theta)-\mathrm{mg} . \mathrm{r} \sin \theta=0 \\ & \text { I. } \pi \mathrm{r}^2 \cdot \mathrm{B}_0 \cos \theta=\mathrm{mg} \text { r.sin } \theta \\ & \tan \theta=\frac{\pi \mathrm{rIB}_0}{\mathrm{mg}} \\ & \end{aligned}\)
\(\begin{aligned} & \text { Let loop makes angle } \theta \text { with vertical. } \\ & \text { in equilibrium } \tau_{\text {net }}=0 \\ & \tau_0=\mathrm{MB} \sin (90-\theta)-\mathrm{mg} . \mathrm{r} \sin \theta=0 \\ & \text { I. } \pi \mathrm{r}^2 \cdot \mathrm{B}_0 \cos \theta=\mathrm{mg} \text { r.sin } \theta \\ & \tan \theta=\frac{\pi \mathrm{rIB}_0}{\mathrm{mg}} \\ & \end{aligned}\)
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