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JEE Main 202310 AprilEvening Shift
Exam
JEE Main
Year
2023
Date
10 April
Shift
Evening Shift
Subject
Physics
Chapter
Electrostatics

JEE Main 2023 · 10 April · Evening Shift · Physics · Electrostatics

An electron revolves around an infinite cylindrical wire having...

An electron revolves around an infinite cylindrical wire having uniform linear charge density 2×10−82 \times {10}^{- 8} Cm−1C{m}^{- 1} in circular path under the influence of attractive electrostatic field as shown in the figure. The velocity of electron with which it is revolving is ______________ ×106ms−1\times {10}^{6}m{s}^{- 1}. Given mass of electron =9×10−31kg= 9 \times {10}^{- 31}kg

An electron revolves around an infinite cylindrical wire having uniform linear charge density 2 × 10 ^ - 8 C m ^ - 1 in

Correct answer

0

Step-by-step solution

The electric field is given by

E=(2kλr)E = \left(\dfrac{2k\lambda}{r}\right)

The required centripetal force is

F=(mv2r)F = \left(\dfrac{m{v}^{2}}{r}\right)

The force due to the field is equal to the required centripetal force,

⇒(2kλr)e=mv2r⇒ \left(\dfrac{2k\lambda}{r}\right)e = \dfrac{m{v}^{2}}{r}

⇒v=2k\lambdaem=2×9×109×2×10−8×1.6×10−199×10−31⇒ v = \sqrt{\dfrac{2k\lambdae}{m}} = \sqrt{\dfrac{2 \times 9 \times {10}^{9} \times 2 \times {10}^{- 8} \times 1 . 6 \times {10}^{- 19}}{9 \times {10}^{- 31}}}

=1.6×4×1013=4×2×106ms−1= \sqrt{1 . 6 \times 4 \times {10}^{13}} = 4 \times 2 \times {10}^{6}m{s}^{- 1}

=8×106ms−1= 8 \times {10}^{6}m{s}^{- 1}

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