# Urgent : Qns on Magnetic field and atoms

1. May 18, 2006

### Delzac

hi, i have a qns as follows, any help will be great apperciated.

A helium nucleus of velocity " v" enters a magnetic field and moves in a circular path of radius R. A proton having the same initial velocity that enters this magnetic field will then move in a circular path of :

a. R/4
b. R/2
c. R
d. 2R
e. 4R

by proportion(of charges and mass) i know that i either have to times by 2 or divide by 2, but which is it? R/2 or 2R? should the radius increase or decrease given that mass per charge increase?

2. May 18, 2006

### nrqed

You need a formula for this, you cannot just guess. What is the correct formula? Set the magnetic force equal to m times the centripetal acceleration $F_{magnetic} = m { v^2 \over r}$. Isolate the radius r. Notice that both the mass AND the charge plays a role in determining the radius so you need to know the charge of a helium nucleus in terms of the charge of a single proton.

PAtrick

3. May 18, 2006

### Lyuokdea

You have an equation for the force of a particle in a magnetic field right? Use that to find the acceleration of both particles, remember acceleration = force/mass.

Now do you have another equation that gives you the radius of circular motion when you have the force? Try putting both accelerations you get in there and see what you get for the different R's

~Lyuokdea

4. May 18, 2006

### Delzac

u said that there is a charge component in it, however it dun seem to be included in the equation, do i need another set of equations?

5. May 18, 2006

### siddharth

Delzac, what is the force acting on the nucleus, as it goes in a circle?

6. May 18, 2006

### Delzac

eh..... magnetic force?

7. May 18, 2006

### Delzac

anyway is the Qns done by using this equation : qvB= (mv^2)/r ??

i got my ans as 1/2 R, is that correct?

8. May 18, 2006

### siddharth

Yeah, and what is the mathematical expression for this magnetic force?

9. May 18, 2006

### Delzac

anyway is the Qns done by using this equation : qvB= (mv^2)/r ??

i got my ans as 1/2 R, is that correct?

10. May 18, 2006

### siddharth

Look, you should be more interested in knowing how you got the equation. Once you do, you'll know if your equation is right or not yourself. If you don't know what the equation means, it's pointless giving you the answer.

11. May 18, 2006

### Delzac

the equation should be derived by newton 2nd law right? ( or so it says in the book)

12. May 18, 2006

### Delzac

the 1st equation is F = qvB right?

13. May 18, 2006

### Delzac

the 2nd equation is F = (mv^2)/r ?

14. May 18, 2006

### siddharth

Yes, that is the magnetic force.

Yes, that is by newtons second law, as you (or rather, the book) said. I hope you understood why you are using these equations. Don't blindly use formulas.

The answer 1/2 R isn't right. Can you figure out what's wrong?

Last edited: May 18, 2006
15. May 18, 2006

### Delzac

but how is the 2 nd equation derived from F = ma?? ( F=mv^2/r) it is not stated in the book

16. May 18, 2006

### siddharth

This is the sort of question you should ask.

First of all, remember that you need some external force to move an object in a circle at a constant speed.

Why? This is because the velocity of the object is changing as it moves in a circle (due to the changing direction), even though the speed is constant. So, by newtons second law, you need an external force.

The rate of change in this velocity is the centripetal acceleration, and the corresponding force which is required is the centripetal force.

The centripetal acceleration a_c can be calculated as $a_c = v^2/r$, and it points towards the center of the circle. The corresponding force is given by $F_c = ma_c = m v^2/r$.

In your given question, the external force which makes the object move in a circle at a uniform speed is the magnetic force, which acts as the centripetal force. That is how you get your equation.

Last edited: May 18, 2006
17. May 18, 2006

### Delzac

the equation i am going to use is r=(mv)/(qB), am i going in the right direction?

thx for telling me about the centripetal acceleration also!

18. May 18, 2006

### siddharth

Yeah, you are going in the right direction.

19. May 18, 2006

### Delzac

R = (1 v)/(1.602 X 10^-19 X B) ?

for a proton

20. May 18, 2006

### Delzac

then for a helium nucleus, i have r= ( 4v)/(2)(1.602 X 10^-19X B)