What Is the Correct Period and Frequency for an Electron in a Magnetic Field?

AI Thread Summary
The discussion focuses on calculating the period and frequency of an electron with 34 KeV of kinetic energy moving in a magnetic field of 0.270 T. The user successfully computed the radius of the electron's circular orbit as 2.2 mm using classical kinetic energy formulas. However, they encountered discrepancies with the frequency and period calculations, which were deemed incorrect by their homework platform, WebAssign. Despite the values appearing reasonable, the user seeks clarification on potential mistakes, possibly related to rounding. The conversation highlights the challenges of applying classical physics principles to particle motion in magnetic fields.
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Homework Statement


Find the period and frequency of the motion of an electron of kinetic energy, E, 34 KeV as is moves in a circular orbit perpendicular to a magnetic field, B, of 0.270 T.

Homework Equations


For speed, v:
upload_2015-10-3_16-1-3.png


For charge, q:
upload_2015-10-3_16-6-15.png


For mass, m:
upload_2015-10-3_15-59-20.png


For radius, R:
upload_2015-10-3_15-33-33.png


For frequency, f:
upload_2015-10-3_15-42-32.png


For period, T:
upload_2015-10-3_15-52-33.png


The Attempt at a Solution



We were told to treat this classically, so I computed the radius by solving for the speed with the classical kinetic energy formula,
upload_2015-10-3_16-1-33.png
, from which I found
upload_2015-10-3_15-33-33.png
, which gave me R = 2.2 mm, which is the correct radius. I then went on to use the relation
upload_2015-10-3_15-42-32.png
and this is where things went horribly wrong. I got the answer
upload_2015-10-3_15-47-35.png
, which seems very reasonable to me. I obtained
upload_2015-10-3_15-50-29.png
for the frequency, which also seems very reasonable to me. The only problem is, the answers are wrong or at least our homework website (WebAssign) says that those are the wrong answer. Anyone see a mistake I may be making?
 

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Just a rounding issue? 7.56 GHz according to WolframAlpha.
Looks correct.
 
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