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A tricky problem that I can't figure out

  • Thread starter Clay
  • Start date

Clay

Hi everybody. I'm taking non-calculus physics at NCSU. There is a tricky homework question that I'm unable to solve. Here it is. My main problem is that I don't know where to start. Any advice would be greatly appreciated. Also, if anybody knows of some web links that explain RC and complex DC currents and how to calculate voltage, current, etc... on them, I would appreciate that link. Here we go:

---------------
A singly charged heavy ion following a circular path in a uniform magnetic field of magnitude 0.053 T is observed to complete five revolutions in 1.58 ms. Calculate the mass of the ion.
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Shouldn't the centripetal force equal the magnetic force, so:
qvB = ((m)(v)(v))/r

and then I could solve for m?

My problem is that I don't know what "singly charged heavy ion" means in terms of the charge. Does that give me the charge? There's no explanation of this term in the book.

Thanks for any help!
 

Tom Mattson

Staff Emeritus
Science Advisor
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Hi, and welcome to PF.

Originally posted by Clay
Shouldn't the centripetal force equal the magnetic force, so:
qvB = ((m)(v)(v))/r

and then I could solve for m?
Yes.

My problem is that I don't know what "singly charged heavy ion" means in terms of the charge. Does that give me the charge?
There is some ambiguity there because it does not tell you whether the ion is positive or negative. In other words, it does not tell you whether the ion has lost an electron or gained an extra one. Nevertheless, you can still get the mass because only the magnitude of the charge is important for this question. That magnitude is the same as the magnitude of the charge on an electron:
1.6×10-19C.
 

Clay

Great! Thanks. I got it correct. This is what I did.

started with qvB = ((m)(v)(v))/r

simplified to qB = (mv)/r

solved for m = (qBr)/v

Figured that 5 revolutions in 1.58 ms was equal to 1 rev in 3.17e-4 sec

v, then was equal to (2(pi)r)/3.17e-4

substituted for v in the simplified equation, and did some algebra:
m = (qB(3.17e-4))/2(pi)

solved for m = 4.28e-25 kg

voila! Thanks for the help! I just needed a push in the right direction.
 

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