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Contests/Gnit Sunday Open 003 (GSO003)/Problem 7 Thomson Experiment with Exact Trajectory Geometry
Problem 7

Thomson Experiment with Exact Trajectory Geometry

Finished
400 ptsLv.7 IntermediateElectromagnetism
2026/04/05 19:30〜2026/04/05 21:00
Author: admin

Problem Statement

In a three-dimensional Cartesian coordinate system (x,y,z)(x, y, z)(x,y,z), an electron (mass mmm, charge −e-e−e, e>0e > 0e>0) is accelerated from rest through a potential difference VVV and injected from the origin (0,0,0)(0, 0, 0)(0,0,0) in the +x+x+x direction with initial speed v0v_0v0​. Gravity and relativistic effects are negligible.

Step 1: Velocity Selector In the region 0≤x≤L0 \le x \le L0≤x≤L, a uniform electric field E⃗=(0,E0,0)\vec{E} = (0, E_0, 0)E=(0,E0​,0) and a uniform magnetic field B⃗=(0,0,B0)\vec{B} = (0, 0, B_0)B=(0,0,B0​) are applied simultaneously. The injected electron travels straight along the xxx-axis without any deflection.

Step 2: Magnetic Deflection The electric field is turned off (E⃗=0⃗\vec{E} = \vec{0}E=0), while the magnetic field B⃗\vec{B}B remains. The same electron with initial speed v0v_0v0​ is injected again from the origin. Inside 0≤x≤L0 \le x \le L0≤x≤L, the electron is deflected by the Lorentz force and follows a curved path in the xyxyxy-plane, exiting the magnetic field region through the plane x=Lx = Lx=L.

Step 3: Drift Space The region x>Lx > Lx>L is a field-free drift space. The electron travels in a straight line and hits a fluorescent screen (parallel to the yzyzyz-plane) placed at x=L+Dx = L + Dx=L+D. The yyy-coordinate of the impact point on the screen is recorded as displacement YYY.

The instrument is precisely set up with drift length D=3LD = \sqrt{3}LD=3​L. The observed displacement on the screen is exactly Y=(3−3)LY = (3 - \sqrt{3})LY=(3−3​)L.

From these results, find the specific charge e/me/me/m of the electron.

Important: Do not use any small-angle approximation (such as sin⁡θ≈θ\sin\theta \approx \thetasinθ≈θ or y≈L2/(2R)y \approx L^2/(2R)y≈L2/(2R)). Derive the result using exact geometric relations.

Constraints

  • Uniform electric field strength: E0=1.76×104 V/mE_0 = 1.76 \times 10^4\,\mathrm{V/m}E0​=1.76×104V/m
  • Uniform magnetic flux density: B0=1.0×10−3 TB_0 = 1.0 \times 10^{-3}\,\mathrm{T}B0​=1.0×10−3T
  • Length of magnetic field region: L=5.0×10−2 mL = 5.0 \times 10^{-2}\,\mathrm{m}L=5.0×10−2m

Input Format

The specific charge e/me/me/m can be written as A×1011 C/kgA \times 10^{11}\,\mathrm{C/kg}A×1011C/kg for some real number AAA.

Compute 100A100A100A and give the answer as a positive integer.

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