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JEE Advanced 2024Paper 2
Exam
JEE Advanced
Year
2024
Paper
Paper 2
Subject
Physics
Chapter
Magnetic Effects of Current

JEE Advanced 2024 · Paper 2 · Physics · Magnetic Effects of Current

A 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 mm and radius rr and it is in a uniform vertical magnetic field B0B_0, as shown in the figure. Initially, it hangs vertically downwards, because of acceleration due to gravity gg, on two conducting supports at P\mathrm{P} and Q\mathrm{Q}. When a current II is passed through the loop, the loop turns about the line PQ\mathrm{PQ} by an angle θ\theta given by A thin stiff insulated metal wire is bent into a circular loop with its two ends extending tangentially from the same

Options

  1. Atan⁡θ=πrIB0/(mg)\tan \theta=\pi r I B_0 /(\mathrm{mg})
  2. Btan⁡θ=2πrIB0/(mg)\tan \theta=2 \pi r I B_0 /(\mathrm{mg})
  3. Ctan⁡θ=πrIB0/(2mg)\tan \theta=\pi r I B_0 /(2 m g)
  4. Dtan⁡θ=mg/(πrIB0)\tan \theta=m g /\left(\pi r I B_0\right)

Correct answer

A. tan⁡θ=πrIB0/(mg)\tan \theta=\pi r I B_0 /(\mathrm{mg})

Step-by-step solution

 Let loop makes angle θ with vertical.  in equilibrium τnet =0τ0=MBsin⁡(90−θ)−mg.rsin⁡θ=0 I. πr2⋅B0cos⁡θ=mg r.sin θtan⁡θ=πrIB0mg\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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