What is the Radius of an Ion's Path in a Magnetic Field?

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SUMMARY

The radius of a singly charged positive ion's path in a magnetic field can be calculated using the formula r = mv / (qB). Given a mass of 1.08e-26 kg, a charge of 1.602e-19 C, and a magnetic field strength of 0.595 T, the velocity must first be determined from the potential difference of 240 V. The correct approach involves equating the kinetic energy gained from the potential difference to the ion's motion in the magnetic field, leading to the correct calculation of the radius.

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  • Understanding of kinetic energy and potential difference in physics
  • Familiarity with the formula for the radius of a charged particle in a magnetic field
  • Basic knowledge of circular motion and magnetic forces
  • Ability to manipulate scientific notation and perform unit conversions
NEXT STEPS
  • Learn how to derive velocity from kinetic energy using the equation KE = qV
  • Study the relationship between magnetic fields and charged particle motion
  • Explore advanced topics in electromagnetism, such as Lorentz force
  • Practice problems involving charged particles in magnetic fields to reinforce concepts
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Students and professionals in physics, particularly those studying electromagnetism, as well as engineers working with charged particle systems in magnetic fields.

rinarez7
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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?
 
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Hi rinarez7

rinarez7 said:
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)

the acceleration to use in this problem is not gravity! in fact you can likely neglect gravity... check at the end that your calculated accerlation is much larger than g.

the ion will undergo circular motion due to the magnetic field, and i think you have used this to find r, however v is unknown as you don't yet know a...

rinarez7 said:
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?

to find v, think about the energy of the ion due to being accelerated by the potential difference and how it relates to velocity
 

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