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Charged particle in magentic field problem

by tomanator
Tags: charged, field, magentic, particle
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tomanator
#1
May27-10, 05:02 AM
P: 10
1. The problem statement, all variables and given/known data

A deuteron (the nucleus of an isotope of hydrogen) has a mass 3.34 10−27 kg
and a charge of +1.60 10−19 C. The deuteron travels in a circular path with
a radius of 7.32mm in a magnetic field with magnitude 2.00T.
(i) What is the angle between the velocity and magnetic field vectors?
(ii) Calculate the speed of the deuteron.
(iii) Calculate the time required for it to make half a revolution.

2. Relevant equations

F=qVBsin(theta)

R=(mv)/(qB)


3. The attempt at a solution

I think I know how to solve parts ii and iii using the second equation listed, rearranged for V, then the time period would be ((pi)R)/V (half revolution)

However with part i it seems to me there is no way of calculating the angle without knowing V? Am I missing something here? It seems stupid that they would ask for a calcualtion including V without working it out first, which comes after
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ehild
#2
May27-10, 07:44 AM
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What is the direction of the force the magnetic field exerts on the particle?

ehild
tomanator
#3
May27-10, 08:01 AM
P: 10
Quote Quote by ehild View Post
What is the direction of the force the magnetic field exerts on the particle?

ehild
Is that a question you know the answer to and are asking me to think about or is it something you want to know haha sorry i'm a bit confused. The force will be perpendicular to the field and the velocity using the right hand screw rule

ehild
#4
May27-10, 08:36 AM
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Charged particle in magentic field problem

Perfect. The force is the same as the centripetal force for the circular orbit the particle travels. This plane of the circle is perpendicular to the magnetic field. The velocity vector is in the same plane as the circle, is not it?

If the velocity of the particle were not perpendicular to the magnetic field, that is, it had both a parallel and normal component, the magnetic field would not influence the parallel component, the particle would move along a helix.

ehild
tomanator
#5
May27-10, 10:18 AM
P: 10
Quote Quote by ehild View Post
Perfect. The force is the same as the centripetal force for the circular orbit the particle travels. This plane of the circle is perpendicular to the magnetic field. The velocity vector is in the same plane as the circle, is not it?

If the velocity of the particle were not perpendicular to the magnetic field, that is, it had both a parallel and normal component, the magnetic field would not influence the parallel component, the particle would move along a helix.

ehild
Still not sure how to solve the angle for part (i) though!
ehild
#6
May27-10, 10:59 AM
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What do you think? The particle moves in a plane that is perpendicular to the magnetic field. Can the velocity vector point out of this plane? What is the angle between any vector in-plane and the magnetic field?

ehild
tomanator
#7
May27-10, 11:16 AM
P: 10
Quote Quote by ehild View Post
What do you think? The particle moves in a plane that is perpendicular to the magnetic field. Can the velocity vector point out of this plane? What is the angle between any vector in-plane and the magnetic field?

ehild
No the magentic force is perpendicular to both the vector and magnetic field vectors. This means the field and the velocity vectors are in the same plane. Thus they are not always perpendicular.
ehild
#8
May27-10, 01:42 PM
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The particle moves along a circle with uniform speed. The circle defines the plane in which both the velocity and the force lie. The force is radial, and rotates around while the particle is moving. The magnetic field is perpendicular to the force at every instant: it is normal to the plane of the circle. The velocity also lies in the plane of the circle. If a vector is normal to a plane, it is perpendicular to every vector lying in the plane.

ehild
tomanator
#9
May27-10, 02:35 PM
P: 10
Quote Quote by ehild View Post
The particle moves along a circle with uniform speed. The circle defines the plane in which both the velocity and the force lie. The force is radial, and rotates around while the particle is moving. The magnetic field is perpendicular to the force at every instant: it is normal to the plane of the circle. The velocity also lies in the plane of the circle. If a vector is normal to a plane, it is perpendicular to every vector lying in the plane.

ehild
sorry it seems I have become very confused. You are correct, i apologise for my questioning of your knowledge :)
ehild
#10
May27-10, 05:43 PM
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Well, one never knows who is there at the other end.
I suppose you know the answer to part i now?

ehild
tomanator
#11
May27-10, 06:23 PM
P: 10
Quote Quote by ehild View Post
Well, one never knows who is there at the other end.
I suppose you know the answer to part i now?

ehild
Very true... If I'm assuming all my understanding is now correct, there is no calculation to part i, just the fact that they are perpendicular ie 90 degrees
ehild
#12
May27-10, 11:22 PM
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The circular orbit is crucial. If the particle had a component parallel with B, it would move along a helix. But it travels along a circle, so there is no parallel component of v.

ehild
cartonn30gel
#13
May28-10, 04:57 AM
P: 68
Drawing a picture might help a lot. I would highly recommend doing that if you haven't alreadly done it.


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