An electron revolves around an infinite cylindrical wire having uniform linear charge density 2 × 10⁻⁸ 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. Given mass of electron =9 × 10⁻³¹ kg

Solution

Physics sir · 4 steps

The line charge's field falls as 1/r, so the radius cancels and the speed does not depend on orbit size.

1. Field from Gauss. Take a Gaussian cylinder of radius r and length l around the wire. Flux E(2π r l) = λ l/ε₀, so E = λ/(2πε₀ r) = 2kλ/r, pointing outward.
2. Force gives centripetal. The electron is pulled inward by eE, and this force alone keeps it on the circle. So eE = mv²/r.

   (2kλ e)/r = mv²/r ⇒ v² = (2kλ e)/m

3. Put in numbers. v² = (2 × 9 × 10⁹ × 2 × 10⁻⁸ × 1.6 × 10⁻¹⁹)/(9 × 10⁻³¹) = 6.4 × 10¹³ m² s⁻².
4. So. v = √(6.4 × 10¹³) = 8 × 10⁶ m s⁻¹, so the answer is 8.

The common slip: 5.66. Using E = kλ/r instead of 2kλ/r drops the factor 2 and gives v ≈ 5.66 × 10⁶ m s⁻¹, not 8.

The idea to keep. For a charge circling a long line charge, E ∝ 1/r and the centripetal term also has 1/r, so they cancel: v = √(2kλ q/m), the same at every radius.

Answer from the NTA final answer key. Solution by PrepWiser.