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Radius of ion's path in field

  • Thread starter rinarez7
  • Start date
1. A singly charged positive ion has a mass = 1.08e-26 kg
After being accelerated through a potential difference of 240 V the ion enters a magnetic field of 0.595 T in a direction perpendicular to the field.
The charge on the ion is 1.602 e-19 C.
Find the radius of the ion's path in the field.
Answer in units of cm.



2. m= 1.08e-26 kg
potential difference= 240 V
B= 0.595 T
q= 1.602e-19
sin theta= 1 (enters in direction perpendicular to the field)
radius= mv/ qB




3. I am confused on exactly how to calculate velocity and the relevance of the potential difference. But, I tried:
F=qvBsin theta= ma
F=qvB= ma
1.602e-19(v) (0.595T)= 1.08e-26(-9.81m/s^2)
v= 1.112e-6 m/s
I then plugged this into r= mv/ qB
r= 1.08e-26kg (1.112e-6 m/s )/ 1.602e-19 (0.595T) and calculated r= 1.8997 e-11 cm, but this incorrect. What am I missing?
 

rock.freak667

Homework Helper
6,230
31
Try using the fact that the potential difference*charge of the ion gives the kinetic energy of it. (you can now find v)
 
Thank you! I just needed that missing piece!
 

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