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Contests/Gnit Sunday Open 001 (GSO001)/Problem 13 Motion of a Charged Particle Between Two Parallel Currents
Problem 13

Motion of a Charged Particle Between Two Parallel Currents

Finished
400 ptsLv.7 IntermediateElectromagnetism
2026/03/08 19:30〜2026/03/08 21:00
Author: admin

Problem Statement

In a three-dimensional Cartesian coordinate system in vacuum, two infinitely long straight wires are placed parallel to the zzz-axis. One wire passes through (d,0,z)(d, 0, z)(d,0,z) and the other through (−d,0,z)(-d, 0, z)(−d,0,z). Both wires carry a constant current III in the +z+z+z direction.

At time ttt, a charged particle of mass mmm and charge qqq (q>0q > 0q>0) is placed at the origin (0,0,0)(0,0,0)(0,0,0) and given an initial velocity v⃗=(v1,0,v0)\vec{v} = (v_1, 0, v_0)v=(v1​,0,v0​). Here v1v_1v1​ is much smaller than v0v_0v0​ (v1≪v0v_1 \ll v_0v1​≪v0​), and the xxx-displacement of the particle is always much smaller than ddd (∣x∣≪d|x| \ll d∣x∣≪d).

The particle moves in the zzz-direction while performing small oscillations in the xxx-direction. The zzz-component of velocity may be approximated as constant at v0v_0v0​. Find the distance LLL the particle travels in the zzz-direction during one period of the xxx-oscillation.

Constraints

  • Wire position parameter: d=0.50 md = 0.50\ \mathrm{m}d=0.50 m
  • Current in wires: I=50 AI = 50\ \mathrm{A}I=50 A
  • Particle charge: q=2.0×10−4 Cq = 2.0 \times 10^{-4}\ \mathrm{C}q=2.0×10−4 C
  • Particle mass: m=4.0×10−5 kgm = 4.0 \times 10^{-5}\ \mathrm{kg}m=4.0×10−5 kg
  • Initial velocity in zzz-direction: v0=1.0×102 m/sv_0 = 1.0 \times 10^2\ \mathrm{m/s}v0​=1.0×102 m/s
  • Vacuum permeability: μ0=4π×10−7 N/A2\mu_0 = 4\pi \times 10^{-7}\ \mathrm{N/A^2}μ0​=4π×10−7 N/A2
  • Ignore gravity and electromagnetic radiation from the particle.

Input Format

The distance LLL is expressed as Aπ (m)A\pi\ (\mathrm{m})Aπ (m). Give the value of positive integer AAA.

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