Verified Solution • 2025–26 Session moderate

QJEEM24Ph325a: JEE Main 2024 Physics - Class 12 JEEM24Ph325a Solution: A particle of charge ' ' and mass ' ' moves in a circle of…

A particle of charge ' −q-q ' and mass ' mm ' moves in a circle of radius ' rr ' around an infinitely long line charge of linear charge density ' +λ+\lambda '. Then time period will be given as :

(Consider kk as Coulomb's constant)

Multiple Choice Options

  • A T2=4π2m2kλqr3T^2=\frac{4 \pi^2 m}{2 k \lambda q} r^3
  • B T=12πrm2kλqT=\frac{1}{2 \pi r} \sqrt{\frac{m}{2 k \lambda q}}
  • C T=12π2kλqmT=\frac{1}{2 \pi} \sqrt{\frac{2 k \lambda q}{m}}
  • D T=2πrm2kλqT=2 \pi r \sqrt{\frac{m}{2 k \lambda q}}

Step-by-Step Verified Solution

amp;2kλqr=mω2ramp;ω2=2kλqmr2amp;(2πT)2=2kλqmr2amp; T=2πrm2kλq\begin{aligned}& \frac{2 \mathrm{k} \lambda \mathrm{q}}{\mathrm{r}}=\mathrm{m} \omega^2 \mathrm{r} \\& \omega^2=\frac{2 \mathrm{k} \lambda \mathrm{q}}{\mathrm{mr}^2} \\& \left(\frac{2 \pi}{\mathrm{T}}\right)^2=\frac{2 \mathrm{k} \lambda \mathrm{q}}{\mathrm{mr}^2} \\& \mathrm{~T}=2 \pi \mathrm{r} \sqrt{\frac{\mathrm{m}}{2 \mathrm{k} \lambda \mathrm{q}}}\end{aligned}

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