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

AI Thread Summary
To find the electric field (E) that allows an 828-eV electron to remain undeflected in a velocity selector with a magnetic field (B) of 14.5 mT, the Lorentz force law is applied. The electron's speed can be calculated from its kinetic energy, using the conversion of 1 eV to joules. The relationship qE = qvB simplifies to v = E/B, indicating that the electric field must balance the magnetic force. The discussion emphasizes the need to determine the electron's speed to solve for E accurately. Understanding these principles is crucial for solving the problem effectively.
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:
\vec{F} = q \vec{E} + q \vec{v} \times \vec{B}

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

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

M B
 
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