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QJEEAdv24Ph5dd6: JEE Advanced 2024 Physics - Class 12 JEEAdv24Ph5dd6 Solution: A thin stiff insulated metal wire is bent into a circular loop…

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

JEE Advanced 2024 Paper 2 Online Physics - Magnetism Question 2 English

Multiple Choice Options

  • A tan⁡θ=πrIB0/(mg)\tan \theta=\pi r I B_0 /(m g)
  • B tan⁡θ=2πrIB0/(mg)\tan \theta=2 \pi r I B_0 /(m g)
  • C tan⁡θ=πrIB0/(2mg)\tan \theta=\pi r I B_0 /(2 m g)
  • D tan⁡θ=mg/(πrIB0)\tan \theta=m g /\left(\pi r I B_0\right)

Step-by-Step Verified Solution

JEE Advanced 2024 Paper 2 Online Physics - Magnetism Question 2 English Explanation

Let loop makes angle θ\theta with vertical.

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

τ0=MBsin⁡(90−θ)−mg.rsin⁡θ=0\tau_0=\mathrm{MB} \sin (90-\theta)-\mathrm{mg} . \mathrm{r} \sin \theta=0

I\mathrm{I} . πr2⋅B0cos⁡θ=mgr⋅sin⁡θ\pi \mathrm{r}^2 \cdot \mathrm{B}_0 \cos \theta=\mathrm{mg} \mathrm{r} \cdot \sin \theta

tan⁡θ=πrIB0mg\tan \theta=\frac{\pi \mathrm{rIB}_0}{\mathrm{mg}}

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