University:
Santa Fe College
Course:
PHY 2004 | Applied Physics 1
Academic year:

2024
Views:

311

Pages:

5
Author:

Jabin F.

Equations for uniformly accelerated motion: a = constant s = ut + 1/2 at^2 v = u + at v^2 = u^2 + 2as Motion under gravity near earth's surface is an example of uniformly accelerated motion. Q: A particle of mass m and charge q is thrown from the origin at t = 0 with initial velocity v₀ into a uniform magnetic field B(-k). Find F_x(t), v_y(t), and the equation of the path of the particle. Solution: Initial force: F_m = q(v × B) = q(v₀i × B(-k)) = qv₀Bk EOM at any instant of time: F_net = ma q(v_xi + v_yi) × B(-k) = m(a_xi + a_yi) qv_xBk - qv_yBi = ma_xi + ma_yi qBvy = max -qBvy = max -qB(d²y/dt²) = max d²y/dt² = -(qB/m)y This is a 2nd order differential equation. Solution of this equation: y = C₁cos(ωt) + C₂sin(ωt) At t = 0, y = 0 ⇒ C₁ = 0 y = C₂sin(ωt) By differentiating: dy/dt = C₂ωcos(ωt) At t = 0, dy/dt = v₀ ⇒ C₂ω = v₀ ⇒ C₂ = v₀/ω y = (v₀/ω)sin(ωt) ω = qB/m Equation of path: x = v₀t y = (v₀/ω)sin(ωt) x = (m/2B)(v₀/ω)sin(2ωt) y = (m/2B)(v₀/ω)(1 - cos(2ωt)) Squaring and adding equations (1) and (2): x² + y² = (m/2B)(v₀/ω)²(sin²(2ωt) + (1 - cos(2ωt))^2) x² + y² = (m/2B)(v₀/ω)²(1 - cos(2ωt)) Conclusion: When a charged particle is thrown into a uniform magnetic field such that the initial velocity is perpendicular to the magnetic field, then the particle moves in a circle of radius r = mv₀/(2qB) with time of revolution T = 2πm/(qB)