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JEE Main 2023 (10 Apr, Shift 2) · Motion of Charges in an Electric Field

Motion of Charges in an Electric Field question from JEE Main 2023 (10 Apr, Shift 2)

Q15JEE Main 2023 · 10 APR · SHIFT 2Medium

Motion of Charges in an Electric Field

An electron revolves around an infinite cylindrical wire having uniform linear charge density 2×10−8 C m−12 \times 10^{-8}\text{ C 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 ___________ ×106 m s−1\times 10^{6}\text{ m s}^{-1}. Given mass of electron =9×10−31 kg= 9 \times 10^{-31}\text{ kg}

Figure

Numerical answer type. Work it out, then open the answer.

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Answer: 8

  1. Attraction by the line charge gives the centripetal force: mv2r=2kλer\frac{mv^2}{r} = \frac{2k\lambda e}{r}.
  2. v=2kλem=2×9×109×2×10−8×1.6×10−199×10−31v = \sqrt{\frac{2k\lambda e}{m}} = \sqrt{\frac{2 \times 9 \times 10^{9} \times 2 \times 10^{-8} \times 1.6 \times 10^{-19}}{9 \times 10^{-31}}}
  3. v=6.4×1013=8×106 m/sv = \sqrt{6.4 \times 10^{13}} = 8 \times 10^{6}\text{ m/s}
  4. So the answer is 8.
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