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Radius of motion of particle through magnetic field.

  • #1
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A singly charged particle of Li-7, an isotope of Lithium, is accelerated through a potential difference of 229V and then enters a magnetic with magnitude 0.723T perpendicular to the path of motion. What is the radius of the ion's path in the magnetic field?
I know that [tex]\[
R = \frac{{mv}}{{\left| q \right|B}}
\][/tex], but I am unsure how to calculate the velocity, and I certainly don't know how to calculate the charge. My educated guess is that velocity and charge are directly proportional over a potential difference so that they don't matter in this problem, but I don't have any equations relating the three.
 
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Answers and Replies

  • #2
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'Singly charged' means having the charge of one electron. So you do know the charge and now can calculate the velocity generated by the voltage drop.
 
  • #3
Astronuc
Staff Emeritus
Science Advisor
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Well, in the case of an atom, or more precisely, ion, singly charged usually means a single + charge (i.e. it has one less electron than the neutral atom).

The velocity can be determined from the kinetic energy, which is equal to the energy received from being accelerated (assume from rest) across a potential difference (229 V). One unit charge 'q' or 'e' is receives 1 eV of energy from a 1 V potential.
 

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