# Magnetism and Centripetal force question

by jumpfreak
Tags: centripetal, force, magnetism
 P: 1 1. The problem statement, all variables and given/known data One method for determing masses of heavy ions involves timing their orbital period in a known magnetic field. What is the mass of a singly charged ion that makes 7.0 revolutions in 1.3 x 10^-3 seconds in a 4.5 x 10^-2 T field. a) 2.1x10^-25 kg b) 1.3x10^-24 kg c) 6.5x10^-23 kg a) 5.0x10^-20 kg 2. Relevant equations I'm not sure of which equation to use. Maybe.... Fc = Fb m4(pi^2)r/T^2 = QvB 3. The attempt at a solution m4(pi^2)r/(1.3x10^-3s/7)^2 = Qv(4.5x10^-2) This might be the wrong formula.
HW Helper
P: 6,653
 Quote by jumpfreak 1. The problem statement, all variables and given/known data One method for determing masses of heavy ions involves timing their orbital period in a known magnetic field. What is the mass of a singly charged ion that makes 7.0 revolutions in 1.3 x 10^-3 seconds in a 4.5 x 10^-2 T field. a) 2.1x10^-25 kg b) 1.3x10^-24 kg c) 6.5x10^-23 kg a) 5.0x10^-20 kg 2. Relevant equations I'm not sure of which equation to use. Maybe.... Fc = Fb m4(pi^2)r/T^2 = QvB 3. The attempt at a solution m4(pi^2)r/(1.3x10^-3s/7)^2 = Qv(4.5x10^-2) This might be the wrong formula.
It is the right formula.

If $mv^2/r = qvB$ where $v = 2\pi r/T$ then:

$$qB = m2\pi/T$$

$$m = qBT/2\pi$$

Just plug in the numbers.

AM
 P: 137 One caution: In the previously cited equation $$m = qBT/2\pi$$, T is the orbital period = (1.3 x 10^-3)/7 sec

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