Question

Consider a long thin conducting wire carrying a uniform current I. A particle having mass " M " and charge "q" is released at a distance " a " from the wire with a speed $v_0$ along the direction of current in the wire. The particle gets attracted to the wire due to magnetic force. The particle turns round when it is at distance x from the wire. The value of x is [$\mu_0$ is vacuum permeability]

• A. $ae^{-\frac{4\pi mv_0}{q\mu_0I}}$
• B. $a\left[1-\frac{mv_0}{2q\mu_0I}\right]$
• C. $a\left[1-\frac{mv}{q\mu_0I}\right]$
• D. $\frac{a}{2}$

Answer 1

Answer: (A)

A $ae^{-\frac{4\pi mv_0}{q\mu_0I}}$

Answer 2

We work in cylindrical coordinates. Choose the vector potential in the gauge
  A₋z = –(μ₀I)/(2π) ln r
so that B = (μ₀I)/(2πr) φ̂. Then the Lagrangian is

  L = ½ m (ṙ² + ż²) – q (μ₀I)/(2π) ln(r) ż.

Since z is a cyclic coordinate, its canonical momentum

  p_z = m ż – q (μ₀I)/(2π) ln r              (1)

is constant. At t = 0 the particle is at r = a with ż = v₀ so that

  p_z = m v₀ – q (μ₀I)/(2π) ln(a).

When the particle “turns round” (i.e. when its z–velocity reverses) we have ż = –v₀ and r = x so that

  p_z = – m v₀ – q (μ₀I)/(2π) ln(x).

Equate the two expressions for p_z from (1):

  m v₀ – q (μ₀I)/(2π) ln(a) = – m v₀ – q (μ₀I)/(2π) ln(x).

Solving for ln(x):

  m v₀ + m v₀ = – q (μ₀I)/(2π)[ln(x) – ln(a)]   ⇒ 2m v₀ = – q (μ₀I)/(2π) ln(x/a)   ⇒ ln(x/a) = – (4π m v₀)/(q μ₀I).

Thus,
  x = a exp[ –(4π m v₀)/(q μ₀I) ].

This matches option (A).