Speed of an Electron and the Magnetic field

AI Thread Summary
To find the speed of an electron with 1.60 keV of kinetic energy in a magnetic field, the correct approach involves converting keV to Joules, yielding 2.5635 e-16 J. The kinetic energy formula should be used correctly as KE = (1/2)mv^2, leading to the equation v = sqrt(2KE/m). The magnetic field can be calculated using the formula B = mv/qR, where R is the orbit radius. Clarification on the equations and their arrangement is crucial for accurate calculations.
mlsohani
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Homework Statement



An electron with kinetic energy 1.60 keV circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 24.0 cm.
a Find the speed of the electron.
b Find the Magnetic Field

Homework Equations



k=1/(2mv^2)
B=mvqr

The Attempt at a Solution


I first converted keV to Joules which came out to be 2.5635 e-16 J

I then made the equation to solve for v : v=sqrt(1/2mk)

But I came out with the velocity being e22 4.627 km/s which is not right

PLease help!
 
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Hi mlshani,

mlsohani said:

Homework Statement



An electron with kinetic energy 1.60 keV circles in a plane perpendicular to a uniform magnetic field. The orbit radius is 24.0 cm.
a Find the speed of the electron.
b Find the Magnetic Field

Homework Equations



k=1/(2mv^2)

This equation does not look right to me; do you see what it needs to be?
 
No I thought that was the equation, at least that is the one on my physics book?
Is there another equation to find the speed of an electron?
 
(mV^2)/2
 
Thank you!

I can't believe I saw that equation as having everything in the denominator!

Thank you!
 
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