How Does Magnetic Field Influence the Path of a Doubly Charged Helium Atom?

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SUMMARY

The discussion focuses on calculating the radius of curvature for a doubly charged helium atom accelerated through a potential difference of 4.00 x 103 V in a uniform magnetic field of 0.450 T. The relevant formula for the radius of curvature is r = mV/|q|B, where m is the mass of the atom (6.68 x 10-27 kg), V is the velocity derived from kinetic energy, q is the charge (1.6 x 10-19 C), and B is the magnetic field strength. Participants clarified that the kinetic energy can be calculated using W = qΔV, leading to the determination of velocity and subsequently the radius of curvature.

PREREQUISITES
  • Understanding of kinetic energy and its relation to potential difference
  • Familiarity with the Lorentz force and its application in magnetic fields
  • Knowledge of the formula for radius of curvature in magnetic fields
  • Basic concepts of charged particle motion in magnetic fields
NEXT STEPS
  • Calculate the velocity of a charged particle using kinetic energy derived from potential difference
  • Explore the derivation of the radius of curvature formula r = mv/qB
  • Investigate the effects of varying magnetic field strengths on particle trajectories
  • Learn about the behavior of different charged particles in magnetic fields, including singly and doubly charged ions
USEFUL FOR

Physics students, researchers in atomic physics, and professionals working with particle accelerators or magnetic confinement systems will benefit from this discussion.

superjen
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A doubly charged helium atom (mass = 6.68 x 10-27 kg) is accelerated through a potential difference of 4.00x 103 V. What will be the radius of curvature of the path of the atom if it is in a uniform 0.450 T magnetic field?


the equation i was using was
r = mV/|q|B

m = 6.68 x 10^-27
V = 4.00 x 10^3
B = 0.450T
q = 1.6 x 10^-19

I don't think this is right. any help pr tips?
Thanks :)
 
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superjen said:
A doubly charged helium atom (mass = 6.68 x 10-27 kg) is accelerated through a potential difference of 4.00x 103 V. What will be the radius of curvature of the path of the atom if it is in a uniform 0.450 T magnetic field?

the equation i was using was
r = mV/|q|B

m = 6.68 x 10^-27
V = 4.00 x 10^3
B = 0.450T
q = 1.6 x 10^-19

I don't think this is right. any help pr tips?
Thanks :)

First figure the kinetic energy, from the work done on the charge.

W = q*ΔV

From ½mv² you can derive a value for v .

Then you can use your second equation derived from

F = mv² /R = qv*B

R = mv/qB
 

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