GREPhysics.NET: GR0177 #62

GR0177 #62

Problem

GREPhysics.NET Official Solution

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Electromagnetism $\Rightarrow$ }Cyclotron Frequency

The cyclotron frequency can be easily derived by equating centripetal force with the Lorentz force. It applies to this problem since the magnetic field is perpendicular to the velocity of the particle. $qvB = mv^2/r = mv \omega \Rightarrow \omega = qB/m$ , where one recalls the angular frequency $\omega = v/r$ .

Solving for m in the cyclotron frequency expression above, one has $m=qB/(2\pi f)$ , where one recalls that the angular frequency is related to the frequency by $\omega = 2\pi f$ .

Plug in the numbers to get choice (A).

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Comments

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booyah
2019-04-02 05:25:52

I have always thought the \"cyclotron frequency\" referred to the angular frequency, so I left out the 2pi. I feel like ETS should have made which type of frequency they were referring to in the question statement more clear

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TigerTed8
2007-09-15 11:59:45

I get the correct formula, but I don't know how to plug in the numbers correctly. Can anyone advise?

Blake7
2007-09-21 20:37:24

The real trick to remember here is that this B-field here is in mks Tesla and happily that is what ETS provides us with; with a Pi in it, to boot! The Pi's will cancel into the angular frquency and the 1600 will cancel out the 1.6 in the electron charge so you'll get something like

m = [2(1.6E-19Coulomb)(Pi/4Tesla)]/[2Pi(1600Hz)] =1/4 E -22 =2.5E-23 kg=> (A)

Drill every day with ETS problems and PERTINENT formulas, get plenty of sleep, eat right, keep a good attitude and good luck!

yummyhat
2017-10-26 19:56:59

lmao dont think thats the kind of advice he wanted, but yeah plant your corn early

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