Q75 / 90 · JEE Main 2024 (30 Jan Shift 2) · Physics

A particle of charge $- q$ and mass $m$ moves in a circle of radius $r$ around an infinitely long line charge of linear density $+ λ$ . Then time period will be given as:
(Consider $k$ as Coulomb's constant)

A
$T^{2} = \frac{4 π^{2} m}{2 k λ q} r^{3}$
Correct answer
B
$T = 2 π r \sqrt{\frac{m}{2 k λ q}}$
C
$T = \frac{1}{2 π r} \sqrt{\frac{m}{2 k λ q}}$
D
$T = \frac{1}{2 π} \sqrt{\frac{2 k λ q}{m}}$

Solution

Attractive electrostatic force due to line charge on the charged particle will provide the required centripetal force. Therefore,
$q (\frac{2 k λ}{r}) = m ω^{2} r$
^{$\Rightarrow ω^{2} = \frac{2 k λ q}{m r^{2}}$}
^{$\Rightarrow {(\frac{2 π}{T})}^{2} = \frac{2 k λ q}{m r^{2}}$}
^{$\Rightarrow T = 2 π r \sqrt{\frac{m}{2 k λ q}}$}

Hence, the correct answer is option: 'B'.

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A particle of charge -q and mass m moves in a circle of radius r around an infinitely long line charge of linear density +λ. Then time period will be given as:(Consider k as Coulomb's constant)

Solution

Relevant Questions

A particle of charge $- q$ and mass $m$ moves in a circle of radius $r$ around an infinitely long line charge of linear density $+ λ$ . Then time period will be given as:
(Consider $k$ as Coulomb's constant)