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JEE Main 20244 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Morning Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2024 · 4 April · Morning Shift · Physics · Electrostatics

An infinitely long positively charged straight thread has a linear...

An infinitely long positively charged straight thread has a linear charge density λCm−1\lambda \mathrm{Cm}^{-1}. An electron revolves along a circular path having axis along the length of the wire. The graph that correctly represents the variation of the kinetic energy of electron as a function of radius of circular path from the wire is :An infinitely long positively charged straight thread has a linear charge density λ Cm^ -1 . An electron revolves alongAn infinitely long positively charged straight thread has a linear charge density λ Cm^ -1 . An electron revolves alongAn infinitely long positively charged straight thread has a linear charge density λ Cm^ -1 . An electron revolves alongAn infinitely long positively charged straight thread has a linear charge density λ Cm^ -1 . An electron revolves alongAn infinitely long positively charged straight thread has a linear charge density λ Cm^ -1 . An electron revolves along

Options

  1. AThe net current flowing in the given circuit is_______ A.
  2. BThe net current flowing in the given circuit is_______ A.
  3. CThe net current flowing in the given circuit is_______ A.
  4. DThe net current flowing in the given circuit is_______ A.

Correct answer

C. The net current flowing in the given circuit is_______ A.

Step-by-step solution

An infinitely long positively charged straight thread has a linear charge density λ Cm^ -1 . An electron revolves along
Electric field E\mathrm{E} at a distance r\mathrm{r} due to infinite long wire is E=2kλrE=\frac{2 k \lambda}{r} Force of electron ⇒F=eE\Rightarrow F=e EF=e(2kλr)F=2kλer\begin{aligned} & F=e\left(\frac{2 k \lambda}{r}\right) \\ & F=\frac{2 k \lambda e}{r} \end{aligned}<br/>Thisforcewillproviderequiredcentripetalforce<br />This force will provide required centripetal force\begin{aligned} \mathrm{F} & =\frac{\mathrm{m}^2}{\mathrm{r}}=\frac{2 \mathrm{k} \lambda \mathrm{e}}{\mathrm{r}} \\ \mathrm{v} & =\sqrt{\frac{2 \mathrm{k} \lambda \mathrm{e}}{\mathrm{m}}} \\ \mathrm{KE} & =\frac{1}{2} \mathrm{mv}^2=\frac{1}{2} \mathrm{~m}\left(\frac{2 \mathrm{k} \lambda \mathrm{e}}{\mathrm{m}}\right) \\ & =\mathrm{k} \lambda \mathrm{e} \end{aligned}$$
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