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

2024
Views:

410

Pages:

6
Author:

Jabin F.

# Magnetic force on a charge particle in a magnetic field F = q(V x B) Force is perpendicular to both V and B. # Power of magnetic force is zero. # Work done by magnetic force is zero. # In magnetic field, speed of particle doesn't change. # A charge particle follows a circular path in a magnetic field. Q2) A particle of mass m is thrown vertically upward from earth's surface with initial speed (GM/2R)^1/2 where M = mass of earth, R = radius of earth. What is maximum height reached by particle? Note: When radius of earth is mentioned in question then we can't use mgh as potential energy or force = mg. We use Solution: Method 1: Using conservation of energy: (KE + PE)i = (KE + PE)f 1/2 m(GM/2R) - GMm/R = 1/2 m(0)^2 - GMm/(R+h) GMm[1/4R - 1/R] = - GMm/(R+h) 1/4R - 1/R = - 1/(R+h) 4R = R+h h = 3R Ans: h = 3R Method (3) using EOM: F = ma -GMm/(R+x)^2 = m dv/dt -GMm/(R+x)^2 dx = v dv -GMm ∫ dx/(R+x)^2 = ∫ v dv -GMm [1/(R+x)] = v^2/2 -GMm [1/(R+h) - 1/R] = 1/2 [v^2 - GM/2R] -GMm [1/4R - 1/R] = 1/2 v^2 v^2 = 3GM/2R v = √(3GM/2R) Ans Q2) A particle of mass m is dropped from a height 3R above earth's surface, where R is radius of earth. If M is mass of earth then calculate speed of particle near earth's surface. Solution: using conservation law of energy (KE + PE)i = (KE + PE)f 1/2mv^2 + GMm/4R = 1/2mv^2 - GMm/R 0 = -GMm/4R + 1/2mv^2 - GMm/R -GMm/4R = 1/2mv^2 - GMm/R 1/2mv^2 = GMm [1 - 1/4] v^2 = 3GM/2R v = √(3GM/2R) Ans Q. A particle with time-varying mass m(t) = m₀(1-t/τ), where m₀ and τ are positive constants, moves along the x-axis under the action of a constant positive force F for 0 ≤ t ≤ τ. If the particle is at rest at time t = 0, then at time t = t, its velocity will be: W_total = KE_f - KE_i as W_total = W_nc + W_c W_nc + W_c = KE_f - KE_i W_c = ∫ F dt = ∫ F dx W_nc = U_i - U_f W_nc = (KE + U)_i - (KE + U)_f