Radius of motion of particle through magnetic field.

  • #1
KingNothing
869
4
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.
 
Last edited:

Answers and Replies

  • #2
Mentz114
5,432
292
'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
20,601
4,559
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.
 

Suggested for: Radius of motion of particle through magnetic field.

Replies
1
Views
7K
  • Last Post
Replies
17
Views
2K
Replies
3
Views
762
Replies
2
Views
19K
Replies
0
Views
3K
Replies
0
Views
2K
Replies
2
Views
2K
Replies
4
Views
2K
  • Last Post
Replies
6
Views
3K
Replies
1
Views
671
Top