Magnetism and Centripetal force question

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SUMMARY

The discussion focuses on calculating the mass of a singly charged ion that completes 7.0 revolutions in 1.3 x 10^-3 seconds within a magnetic field of 4.5 x 10^-2 T. The correct formula to use is derived from equating centripetal force and magnetic force, leading to the equation m = qBT/2π. By substituting the values, users can determine the mass accurately. The final mass calculation confirms that the formula is indeed applicable for this scenario.

PREREQUISITES
  • Understanding of centripetal force and magnetic force equations
  • Familiarity with the concepts of orbital period and frequency
  • Knowledge of the relationship between charge, magnetic field, and velocity
  • Basic algebra for manipulating equations
NEXT STEPS
  • Study the derivation of the centripetal force equation in magnetic fields
  • Learn about the properties of heavy ions and their behavior in magnetic fields
  • Explore the application of the Lorentz force in charged particle motion
  • Investigate the significance of magnetic field strength in ion mass determination
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Students in physics, particularly those studying electromagnetism and ion dynamics, as well as researchers working with heavy ions in magnetic fields.

jumpfreak
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Homework Statement



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

Homework Equations



I'm not sure of which equation to use. Maybe...
Fc = Fb
m4(pi^2)r/T^2 = QvB



The Attempt at a Solution



m4(pi^2)r/(1.3x10^-3s/7)^2 = Qv(4.5x10^-2)

This might be the wrong formula.
 
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jumpfreak said:

Homework Statement



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

Homework Equations



I'm not sure of which equation to use. Maybe...
Fc = Fb
m4(pi^2)r/T^2 = QvB



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
 
One caution:

In the previously cited equation m = qBT/2\pi, T is the orbital period = (1.3 x 10^-3)/7 sec
 
Last edited:

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