Charged particle in magentic field problem

In summary: What is the point you are trying to make?In summary,A deuteron has a mass 3.34× 10−27 kg and a charge of +1.60 × 10−19 C. It travels in a circular path in a magnetic field with a magnitude of 2.00T.
  • #1
tomanator
10
0

Homework Statement



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.

Homework Equations



F=qVBsin(theta)

R=(mv)/(qB)

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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  • #2
What is the direction of the force the magnetic field exerts on the particle?

ehild
 
  • #3
ehild said:
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
 
  • #4
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
 
  • #5
ehild said:
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!
 
  • #6
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
 
  • #7
ehild said:
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.
 
  • #8
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
 
  • #9
ehild said:
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 :)
 
  • #10
Well, one never knows who is there at the other end.
I suppose you know the answer to part i now?

ehild
 
  • #11
ehild said:
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
 
  • #12
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
 
Last edited:
  • #13
Drawing a picture might help a lot. I would highly recommend doing that if you haven't alreadly done it.
 

1. What is a charged particle in a magnetic field problem?

A charged particle in a magnetic field problem is a type of physics problem that involves calculating the motion of a charged particle in a magnetic field. The particle can be either a positively or negatively charged particle, and the magnetic field can be created by a permanent magnet or an electric current.

2. How does a magnetic field affect a charged particle?

A magnetic field exerts a force on a charged particle, causing it to move in a circular or helical path. The direction of the force is perpendicular to both the direction of the particle's motion and the direction of the magnetic field. The strength of the force depends on the charge and velocity of the particle, as well as the strength and direction of the magnetic field.

3. What is the equation for the force on a charged particle in a magnetic field?

The equation for the force on a charged particle in a magnetic field is F = qvB, where F is the force (in Newtons), q is the charge of the particle (in Coulombs), v is the velocity of the particle (in meters per second), and B is the strength of the magnetic field (in Tesla). This equation is known as the Lorentz force equation.

4. How is the motion of a charged particle in a magnetic field calculated?

To calculate the motion of a charged particle in a magnetic field, you can use the Lorentz force equation to determine the force acting on the particle. Then, you can use Newton's second law of motion (F=ma) to determine the acceleration of the particle. Finally, you can use equations of motion (such as s = ut + 0.5at^2) to calculate the position, velocity, and acceleration of the particle at different points in time.

5. What are some real-world applications of charged particles in magnetic fields?

Charged particles in magnetic fields have many practical applications, including particle accelerators, mass spectrometers, and magnetic resonance imaging (MRI) machines. They are also used in everyday devices such as electric motors, generators, and speakers. In addition, charged particles in Earth's magnetic field play a crucial role in the formation of auroras (also known as the northern and southern lights).

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