Electrostatics

JEE Main level, numerical answer, medium

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Question 79 of 311

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

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Answer: 8

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\Rightarrow \left(\dfrac{2k\lambda }{r}\right)e=\dfrac{m{v}^{2}}{r}
⇒v=2kλem=2×9×109×2×10−8×1.6×10−199×10−31\Rightarrow v=\sqrt{\dfrac{2k\lambda e}{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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