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1. *Is my solution correct?
Homework Statement
A positively charged particle of mass 7.2 X 10^-8 kg is traveling due east with a speed of 85 m/s and enters a 0.31T uniform magnetic field. The particle moves through one-quarter of a circle in 2.2 X 10^3 seconds, at which time it leaves the field heading of south. All during the motion the particle moves perpendicular to the magnetic field. a) What is the magnitude of the magnetic force acting of the particle, b) Determine the charge of the particle.
speed = distance/time
r = mv/qB
a) velocity = distance/time
= ∏r/2t
therefore r = v2t/∏
=0.119m
F= mv^2/r
= 4.37 X 10^-3
b) F = qvB
q = F/vB
= 1.66 X 10^-4C
Homework Statement
A positively charged particle of mass 7.2 X 10^-8 kg is traveling due east with a speed of 85 m/s and enters a 0.31T uniform magnetic field. The particle moves through one-quarter of a circle in 2.2 X 10^3 seconds, at which time it leaves the field heading of south. All during the motion the particle moves perpendicular to the magnetic field. a) What is the magnitude of the magnetic force acting of the particle, b) Determine the charge of the particle.
Homework Equations
speed = distance/time
r = mv/qB
The Attempt at a Solution
a) velocity = distance/time
= ∏r/2t
therefore r = v2t/∏
=0.119m
F= mv^2/r
= 4.37 X 10^-3
b) F = qvB
q = F/vB
= 1.66 X 10^-4C