Finding the change in velocity through a potential difference

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
6 replies · 2K views
Patches1532
Messages
4
Reaction score
0
Homework Statement
Question states: A singly charged ion of unknown mass moves in a circle of radius 12.5 cm in a magnetic field of 1.2 T. The ion was accelerated through a potential difference of 7 kV before it entered the magnetic field. What is the mass of the ion?
Relevant Equations
qB=mv/r
We're given the equation qB=mv/r which is simple enough. I just don't know how to find the velocity given the acceleration through a potential difference. I tried using the radial acceleration equation given to me but I end up with the square root of a negative...and that breaks math... I assume you're first supposed to convert to volts 7kV=7000V, but what equation do I use after that? Any help would be appreciated. Thanks.
 
Physics news on Phys.org
I ended up putting a negative in the radial acceleration equation in my notes...I missed these classes and copied notes from a friend. In our book it has it as a=v^2/r which will allow me to solve for V. Thank you for the help.
 
Patches1532 said:
I ended up putting a negative in the radial acceleration equation in my notes...I missed these classes and copied notes from a friend. In our book it has it as a=v^2/r which will allow me to solve for V. Thank you for the help.
I thought you had difficulty finding the velocity of the particle when it enters the magnetic field. That is calculated using the 7000 V potential difference and is not related to ##v^2/r.##
 
Then maybe I'm misunderstanding the notes. So in the question, we're given B, r, and p.d. and we are asked to solve for mass (m). In our notes, the only equations it gives us are for radial acceleration, F of the magnetic field=qvBCos(theta), F=ma, and qB=mv/r. We're not told how to plug in p.d. into any equation.
 
Yes that rings a bell. I'll look at the link and try to figure it out. Thank you for your time.