Electric potential difference to accelerate electron

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SUMMARY

The discussion focuses on calculating the electric potential difference required to accelerate an electron to a specific speed for circular motion in a magnetic field. The relevant equation derived is ΔV = (qB²r²)/(2m), where q is the charge of the electron, B is the magnetic field strength (1.0 T), r is the radius of the circular path (1 cm), and m is the mass of the electron. The relationship between electric potential energy and kinetic energy is established through the equation qV = 0.5mv², confirming the connection between potential difference and the electron's acceleration.

PREREQUISITES
  • Understanding of electric potential difference and energy conversion
  • Familiarity with the Lorentz force equation: F_{B} = qvB
  • Knowledge of circular motion dynamics and centripetal force
  • Basic principles of electromagnetism, particularly in uniform magnetic fields
NEXT STEPS
  • Study the derivation of the Lorentz force and its applications in particle motion
  • Explore the relationship between electric potential and kinetic energy in charged particles
  • Learn about the behavior of electrons in magnetic fields and the concept of cyclotron motion
  • Investigate the implications of electric potential difference in various electromagnetic applications
USEFUL FOR

Physics students, educators, and professionals in electromagnetism, particularly those focusing on particle dynamics and electric-magnetic interactions.

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Homework Statement



What electric potential difference is required to accelerate and electron, initially at rest, to the speed at which it will move in a circular path with radius 1cm when the electron enters a region of uniform magnetic field of 1.0T, if the field is perpendicular to the electron's velocity? (Ignore gravity)

Homework Equations





The Attempt at a Solution



We know that,

[tex]F_{B} = qvB = \frac{mv^{2}}{r}[/tex]

So,

[tex]v = \frac{qBr}{m}[/tex]

The part I'm confused about is how to relate this knowledge the the potential difference?

Any suggestions?

EDIT: I figured it out.

[tex]\Delta V = \frac{qB^{2}r^{2}}{2m}[/tex]
 
Physics news on Phys.org
change in the potential energy is equal to gain in kinetic energy

qV = 0.5mv2
 

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