How Do Electric and Magnetic Fields Affect an Electron's Path?

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SUMMARY

The discussion focuses on calculating the electric field (E) required to keep an 828-eV electron undeflected in a velocity selector with a magnetic field (B) of 14.5 mT. Using the Lorentz force law, the relationship qE = qvB is established, leading to the conclusion that the electron's speed (v) can be derived from its kinetic energy. The necessary calculations involve converting the electron's energy from electronvolts to joules and applying the formula v = E/B to find the appropriate value of E.

PREREQUISITES
  • Understanding of the Lorentz force law
  • Knowledge of kinetic energy calculations in electronvolts
  • Familiarity with electromagnetic fields and their interactions
  • Basic algebra for solving equations
NEXT STEPS
  • Calculate electron speed from kinetic energy using the formula v = sqrt(2 * KE / m_e)
  • Learn about the conversion of electronvolts to joules (1 eV = 1.602 x 10^-19 J)
  • Explore the implications of undeflected motion in electric and magnetic fields
  • Study the application of velocity selectors in particle physics experiments
USEFUL FOR

Students and professionals in physics, particularly those studying electromagnetism, particle physics, or engineering applications involving electric and magnetic fields.

nemzy
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i have no idea how to do this problem:

a velocity selector consists of electric and magnetic fields described by the expressions E=E(k hat) and B=B(j hat), with B=14.5 mT. Find the vale of E such that a 828-eV electron moving along the positive x-axis is undeflected.



hmm..since it is undeflected we can say that qE=qvB, right? and v=E/B...

but how could u solve for E? and how does the 828 eV electron fit into this problem?
 
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nemzy said:
i have no idea how to do this problem:

a velocity selector consists of electric and magnetic fields described by the expressions E=E(k hat) and B=B(j hat), with B=14.5 mT. Find the vale of E such that a 828-eV electron moving along the positive x-axis is undeflected.



hmm..since it is undeflected we can say that qE=qvB, right? and v=E/B...

but how could u solve for E? and how does the 828 eV electron fit into this problem?
You have correctly stated the Lorentz force law:
[tex]\vec{F} = q \vec{E} + q \vec{v} \times \vec{B}[/tex]

Determine electron speed from the electron kinetic energy of 828 eV. Since the net force = 0,
[tex]\vec{v} = \frac{-\vec{E}}{\vec{B}}[/tex]

AM
 
[tex]1eV=1.602 \times 10^{-19} J[/tex]
[tex]m_e=9.11 \times 10^{-31} Kg[/tex]

M B
 

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