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Electromagnetism (Phys. 12)

 
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May27-06, 09:21 PM   #1
 

Electromagnetism (Phys. 12)


An ion with a change of 1.6*10^-19 enters a 0.075 T magnetic field. If the ion follows a circular path of radius 0.083m, what is the momentum of the ion?

I tried manipulating Fc = Fmagnet into Momentum = QBr, but that didn't work... any ideas?

Thanks!
 
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May27-06, 09:30 PM   #2
 
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What do you mean it didn't work?
 
May27-06, 09:36 PM   #3
 
Quote by Doc Al
What do you mean it didn't work?
Well, it's a multiple choice question, and the answer I got wasn't any of them.
 
May27-06, 10:40 PM   #4
 

Electromagnetism (Phys. 12)


Quote by Drevin
Well, it's a multiple choice question, and the answer I got wasn't any of them.
[tex]F = qvB[/tex]

You obviously know this. But what must the magnetic force do to keep the charge moving in a circle?
 
May27-06, 10:46 PM   #5
 
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Quote by Drevin
I tried manipulating Fc = Fmagnet into Momentum = QBr, but that didn't work... any ideas?
Write down the complete equation and the subsequent steps as you would have, for your homework. What you've written above is not an equation.
 
May28-06, 07:55 AM   #6
 
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Quote by Drevin
Well, it's a multiple choice question, and the answer I got wasn't any of them.
Your solution, "Momentum = QBr", is certainly correct. What were the other choices? If the choices were numerical, perhaps you made an arithmetic error.
 
May28-06, 02:02 PM   #7
 
Quote by Doc Al
Your solution, "Momentum = QBr", is certainly correct. What were the other choices? If the choices were numerical, perhaps you made an arithmetic error.
The choices are:
A) 2.7 * 10^-26 kgm/s
B) 1.0 * 10^-21 kgm/s
C) 1.7 * 10^-12 kgm/s
D) 5.2 * 10^-4 kgm/s

The answer I got was 9.96 * 10^-22 using Momentum = QBR
mv = QBr
P = (1.6*10^-19)(0.075)(0.083)
P = 9.96 * 10^-22

That's what I did.
 
May28-06, 02:18 PM   #8
 
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Quote by Drevin
The choices are:
A) 2.7 * 10^-26 kgm/s
B) 1.0 * 10^-21 kgm/s
C) 1.7 * 10^-12 kgm/s
D) 5.2 * 10^-4 kgm/s

The answer I got was 9.96 * 10^-22 using Momentum = QBR
mv = QBr
P = (1.6*10^-19)(0.075)(0.083)
P = 9.96 * 10^-22

That's what I did.
Your answer is correct, except it is written to 3 sig figs, when it should only be written to 2. Do this, and you'll find that one the the given choices is correct.

I misunderstood what you'd written in the OP in my earlier post. After re-reading, it is clear what you did. My bad, sorry.
 
May28-06, 03:18 PM   #9
 
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Quote by Drevin
The choices are:
A) 2.7 * 10^-26 kgm/s
B) 1.0 * 10^-21 kgm/s
C) 1.7 * 10^-12 kgm/s
D) 5.2 * 10^-4 kgm/s

The answer I got was 9.96 * 10^-22 using Momentum = QBR
mv = QBr
P = (1.6*10^-19)(0.075)(0.083)
P = 9.96 * 10^-22
You should realize that your answer is extremely close to one of those choices, so close that if you had to choose (and you do!) there should be no question as to which one to pick. (The other choices are off by many orders of magnitude--they aren't even close!) And, as Gokul says, if you had worked out your answer to 2 significant figures--as you should have, given the data--you would have found a perfect match. (Since the data you were given only had two significant figures, your answer must have no more than two significant figures.)
 
May28-06, 03:36 PM   #10
 
Quote by Doc Al
You should realize that your answer is extremely close to one of those choices, so close that if you had to choose (and you do!) there should be no question as to which one to pick. (The other choices are off by many orders of magnitude--they aren't even close!) And, as Gokul says, if you had worked out your answer to 2 significant figures--as you should have, given the data--you would have found a perfect match. (Since the data you were given only had two significant figures, your answer must have no more than two significant figures.)
Ah, yes.. you're correct! (Our class doesn't tend to use sig figs very often, so I think that's where I got confused)

Thanks so much for the help!

I'll write down my steps a little more clearly next time I post a question, too.
 
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