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Urgent : Qns on Magnetic field and atoms

  • Thread starter Delzac
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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?
 

nrqed

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Delzac said:
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?
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 [itex] F_{magnetic} = m { v^2 \over r} [/itex]. 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
 
147
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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
 
389
0
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?
 

siddharth

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Delzac, what is the force acting on the nucleus, as it goes in a circle?
 
389
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eh..... magnetic force?
 
389
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anyway is the Qns done by using this equation : qvB= (mv^2)/r ??

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

siddharth

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Delzac said:
eh..... magnetic force?
Yeah, and what is the mathematical expression for this magnetic force?
 
389
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anyway is the Qns done by using this equation : qvB= (mv^2)/r ??

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

siddharth

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

i got my ans as 1/2 R, is that correct?
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.
 
389
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the equation should be derived by newton 2nd law right? ( or so it says in the book)
 
389
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the 1st equation is F = qvB right?
 
389
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the 2nd equation is F = (mv^2)/r ?
 

siddharth

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Delzac said:
the 1st equation is F = qvB right?
Yes, that is the magnetic force.

the 2nd equation is F = (mv^2)/r ?
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:
389
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but how is the 2 nd equation derived from F = ma?? ( F=mv^2/r) it is not stated in the book
 

siddharth

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Delzac said:
but how is the 2 nd equation derived from F = ma?? ( F=mv^2/r) it is not stated in the book
:smile: 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 [itex] a_c = v^2/r [/itex], and it points towards the center of the circle. The corresponding force is given by [itex] F_c = ma_c = m v^2/r [/itex].

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:
389
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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!
 

siddharth

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Delzac said:
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!
Yeah, you are going in the right direction.
 
389
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R = (1 v)/(1.602 X 10^-19 X B) ?

for a proton
 
389
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then for a helium nucleus, i have r= ( 4v)/(2)(1.602 X 10^-19X B)
 
389
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then from those 2 equations, u equate and get r=2R?
 
389
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then the circular path of a helium nucleus is 2R?
 

siddharth

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Yes, that answer looks right.
 
389
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Thx for all the help : )
 

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