# Calculating the speed of a proton in a cyclotron?

Watch
Announcements
This discussion is closed.
#1
This is straight from the Edexcel A2 Student's Book. I would normally use K.E. = eV for questions like this, but that won't work here. I don't understand how to solve this. It's a cyclotron, so perhaps I could use r = mv/Bq, but I don't know B. 0
5 years ago
#2
Qu 2. Yes, if the charged particle is performing circular motion, the key is to equate its centripetal force mv2/r to the the force on a moving charge in a magnetic field, Bqv
This gives you the relationship between the radius of the path, its velocity and the field strength.
0
#3
(Original post by Stonebridge)
Qu 2. Yes, if the charged particle is performing circular motion, the key is to equate its centripetal force mv2/r to the the force on a moving charge in a magnetic field, Bqv
This gives you the relationship between the radius of the path, its velocity and the field strength.
Yes, but um, the field strength is not given in the question. How would I calculate the speed without it?
0
5 years ago
#4
27km distance (27,000m) 11,000 times a second. 1/11,000s for 27km, speed = distance/time, speed = 27,000/(1/11,000) and I don't have a calculator to hand haha, but that might be the speed?
0
#5
(Original post by Brandon_97)
27km distance (27,000m) 11,000 times a second. 1/11,000s for 27km, speed = distance/time, speed = 27,000/(1/11,000) and I don't have a calculator to hand haha, but that might be the speed?
Okay. OKAY. REALLY embarrassed right now, but okay, yes, that's exactly it, that gives the answer. xD Um. Wow. Okay. xD Thank you so much!

But that doesn't seem to help much with Q.2 either. :/
0
5 years ago
#6
You calculate the speed of the protons in part a. from their kinetic energy.
0
5 years ago
#7
(Original post by You-know-who)
Okay. OKAY. REALLY embarrassed right now, but okay, yes, that's exactly it, that gives the answer. xD Um. Wow. Okay. xD Thank you so much!

But that doesn't seem to help much with Q.2 either. :/
I'm not sure where you are stuck, you have the mass, the velocity, and the equation....?

(Original post by Stonebridge)
You calculate the speed of the protons in part a. from their kinetic energy.
I think the point in the question is to compare the two kinetic energies, due to the fact they are largely different, as the protons travel at 0.99c.
0
5 years ago
#8
(Original post by Phichi)
I'm not sure where you are stuck, you have the mass, the velocity, and the equation....?

I think the point in the question is to compare the two kinetic energies, due to the fact they are largely different, as the protons travel at 0.99c.
I was discussing 2a.
0
5 years ago
#9
(Original post by Stonebridge)
I was discussing 2a.
0
#10
(Original post by Stonebridge)
You calculate the speed of the protons in part a. from their kinetic energy.

(Original post by Phichi)
I'm not sure where you are stuck, you have the mass, the velocity, and the equation....?

I think the point in the question is to compare the two kinetic energies, due to the fact they are largely different, as the protons travel at 0.99c.
I tried doing that for Q.3, didn't work. Brandon_97's method seemed to work though, thank you for that.

For question 2, the first thing I did was equate eV =1/2mv^2.

1MeV = 1 x 10^6 x 1.6 x 10^-19
1 x 10^6 x 1.6 x 10^-19 = 0.5 x 1.67 x 10^-27 x v^2.

From there, my answer comes out to be 1.38 x 10^7. The correct ans is 1.96 x 10^7. :/
0
#11
(Original post by Stonebridge)
We can't pinpoint your possible mistake if you don't post your full working.
Um, sorry, but that was my full working for question 2. eV =1/2mv^2 (the formula)

1MeV = 1 x 10^6 x 1.6 x 10^-19 (Finding the energy in Joules)

1 x 10^6 x 1.6 x 10^-19 = 0.5 x 1.67 x 10^-27 x v^2 (Putting the values in the formula)

However, the correct ans given in the book is 1.96 x 10^7.
0
5 years ago
#12
(Original post by You-know-who)
Um, sorry, but that was my full working for question 2. eV =1/2mv^2 (the formula)

1MeV = 1 x 10^6 x 1.6 x 10^-19 (Finding the energy in Joules)

1 x 10^6 x 1.6 x 10^-19 = 0.5 x 1.67 x 10^-27 x v^2 (Putting the values in the formula)

However, the correct ans given in the book is 1.96 x 10^7.
I realised that and deleted my original post. As you can see. I thought initially you were giving the answer to the final part of the question, the bit you originally asked about.
Unfortunately you have replied to something I deleted.

Yes. I get the same answer. Sometimes mark schemes or book answers can be wrong or misprinted.
What value does that give for the magnetic field?
Does it agree with the book answer?
You can check by trying both your answer and the given answer to find the value to the other part.
0
#13
(Original post by Stonebridge)
I realised that and deleted my original post. As you can see. I thought initially you were giving the answer to the final part of the question, the bit you originally asked about.
Unfortunately you have replied to something I deleted.

Yes. I get the same answer. Sometimes mark schemes or book answers can be wrong or misprinted.
What value does that give for the magnetic field?
Does it agree with the book answer?
You can check by trying both your answer and the given answer to find the value to the other part.
All the other answers correspond to the first answer given in the book, none of them work with my value. But okay, atleast now I know I didn't go completely wrong somewhere. Thank you so much for your time. 0
1 month ago
#14
No you are actually right. It is 1.38x10^7.
0
1 month ago
#15
This thread is 5 years old.
0
X
new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### Should there be a new university admissions system that ditches predicted grades?

No, I think predicted grades should still be used to make offers (707)
33.91%
Yes, I like the idea of applying to uni after I received my grades (PQA) (891)
42.73%
Yes, I like the idea of receiving offers only after I receive my grades (PQO) (394)
18.9%
I think there is a better option than the ones suggested (let us know in the thread!) (93)
4.46%