# Calculating the speed of a proton in a cyclotron?

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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.

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#2

Qu 2. Yes, if the charged particle is performing circular motion, the key is to equate its centripetal force mv

This gives you the relationship between the radius of the path, its velocity and the field strength.

^{2}/r to the the force on a moving charge in a magnetic field, BqvThis gives you the relationship between the radius of the path, its velocity and the field strength.

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(Original post by

Qu 2. Yes, if the charged particle is performing circular motion, the key is to equate its centripetal force mv

This gives you the relationship between the radius of the path, its velocity and the field strength.

**Stonebridge**)Qu 2. Yes, if the charged particle is performing circular motion, the key is to equate its centripetal force mv

^{2}/r to the the force on a moving charge in a magnetic field, BqvThis gives you the relationship between the radius of the path, its velocity and the field strength.

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#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?

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(Original post by

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?

**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?

But that doesn't seem to help much with Q.2 either. :/

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#7

(Original post by

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. :/

**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. :/

(Original post by

You calculate the speed of the protons in part a. from their kinetic energy.

**Stonebridge**)You calculate the speed of the protons in part a. from their kinetic energy.

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#8

(Original post by

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.

**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.

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**Stonebridge**)

You calculate the speed of the protons in part a. from their kinetic energy.

**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.

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. :/

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(Original post by

We can't pinpoint your possible mistake if you don't post your full working.

**Stonebridge**)We can't pinpoint your possible mistake if you don't post your full working.

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)

=1.38 x 10^7 (The answer)

However, the correct ans given in the book is 1.96 x 10^7.

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#12

(Original post by

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)

=1.38 x 10^7 (The answer)

However, the correct ans given in the book is 1.96 x 10^7.

**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)

=1.38 x 10^7 (The answer)

However, the correct ans given in the book is 1.96 x 10^7.

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.

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(Original post by

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.

**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.

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