The discussion revolves around the relationship between ion momentum, magnetic fields, and circular motion, specifically focusing on a problem involving an ion's momentum in a magnetic field of 0.075 T and a circular path of radius 0.083 m.
Participants are actively engaging with the problem, with some providing guidance on significant figures and the importance of clarity in presenting calculations. There is recognition of the original poster's calculations, with suggestions for improvement rather than a definitive resolution.
There is mention of a multiple-choice format for the problem, which adds a layer of complexity regarding the correctness of the calculated answer in relation to the provided options. The discussion also highlights the importance of significant figures based on the data given.
Well, it's a multiple choice question, and the answer I got wasn't any of them.Doc Al said:What do you mean it didn't work?
Drevin said:Well, it's a multiple choice question, and the answer I got wasn't any of them.
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.Drevin said:I tried manipulating Fc = Fmagnet into Momentum = QBr, but that didn't work... any ideas?
Your solution, "Momentum = QBr", is certainly correct. What were the other choices? If the choices were numerical, perhaps you made an arithmetic error.Drevin said:Well, it's a multiple choice question, and the answer I got wasn't any of them.
The choices are:Doc Al said:Your solution, "Momentum = QBr", is certainly correct. What were the other choices? If the choices were numerical, perhaps you made an arithmetic error.
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.Drevin said: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.
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.)Drevin said: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
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)Doc Al said: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.)