Force Direction of Alpha Particle in Magnetic Field

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SUMMARY

The discussion centers on calculating the magnetic force acting on an alpha particle moving east at 2.0 x 105 m/s in a magnetic field of 0.20 T directed upwards. Using the equation F = qvBsinø, the force is determined to be 1.28 x 10-14 N directed south. The right-hand rule confirms that the force is perpendicular to both the velocity and magnetic field vectors, leading to the conclusion that the force cannot point in the same direction as either vector. Instead, it points south, as the charge of the alpha particle is positive.

PREREQUISITES
  • Understanding of magnetic fields and forces
  • Familiarity with the right-hand rule for cross products
  • Knowledge of the equation F = qvBsinø
  • Basic concepts of particle physics, specifically alpha particles
NEXT STEPS
  • Study the right-hand rule in detail for vector cross products
  • Explore the implications of magnetic forces on charged particles
  • Learn about the behavior of different charged particles in magnetic fields
  • Investigate applications of magnetic fields in particle accelerators
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of charged particles in magnetic fields.

qswdefrg
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Homework Statement


An alpha particle (two protons and two neutrons) traveling east at 2.0 x 105 m/s enters a magnetic field of 0.20 T pointing straight up. What is the force acting on the alpha particle?

Homework Equations


F = qvBsinø

The Attempt at a Solution


With the fingers of the right hand pointing straight up, and the thumb pointing east, the palm points south.
F = qvBsinø = (2 x 1.6 x 10-19 C)(2.0 x 105 m/s)(0.20 T)sin90º = 1.28 x 10-14 N

Need a little help on the direction. This was the answer found when I checked it, but that can't be right. Isn't the correct answer [out]?
 
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qswdefrg said:
With the fingers of the right hand pointing straight up, and the thumb pointing east, the palm points south.
F = qvBsinø = (2 x 1.6 x 10-19 C)(2.0 x 105 m/s)(0.20 T)sin90º = 1.28 x 10-14 N

'Looks okay to me. :approve:
Need a little help on the direction. This was the answer found when I checked it, but that can't be right. Isn't the correct answer [out]?
Out of what? :-p

The magnetic force, [tex]\vec F = q \vec v \times \vec B[/tex] involves the cross product between two vectors: the velocity vector and the magnetic field vector.

The cross product means that the result will be perpendicular to both vectors. In other words, the force is perpendicular to both the magnetic field and also perpendicular to the velocity.

Still in other words, if the magnetic field points completely in an up/down direction, the force does not point in the up/down direction at all. If the velocity points completely in the east/west direction, the force does not point in the east/west direction at all. So in this problem, the only direction left that's perpendicular to both is the north/south direction. The right hand rule let's you know that the force points south instead of north (and because the charge is positive -- reverse direction for negative charges).
 
So imagine your cardinal directions are in the plane of the computer screen. That is, East is ------> that way. So the magnetic field pointing "up" would be OUT of the screen, rather than North.

Otherwise, since the problem is written as if you're working a piece of paper, the cardinal directions are in the plane of your keyboard, with East pointing toward your Enter/Return button. "Up" is then up toward the ceiling if you're keyboard is flat.

Right hand rule for your cross product: put your fingers along v (east), curl them into B (out of the screen toward yourself, or if you're using the keyboard way then up to the ceiling) and your thumb points in the direction of the force (which, note, is in the same plane as your cardinal directions).
 

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