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Newton's 2nd Law and Orbital Motion

  • Thread starter runevxii
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  • #1
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Here's the problem...unfortunately I don't remember much about orbital motion. I'm a bit stuck on where to begin. If somebody could give me a little advice on how to tackle this problem I would appreciate it.

Recall that the magnetic force on a charge q moving with velocity v in a magnetic field B is equal to qvXB. If a charged particle moves in a circular orbit with a fixed speed v in the presence of a constant magnetic field, use the relativistic form of Newton's 2nd law to show that the frequency of its orbital motion is

f=((qB)/(2pim))(1-(v^2/c^2))^(1/2)
 

Answers and Replies

  • #2
Tide
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If the speed is constant then

[tex]\frac {d \vec v}{dt} = \vec v \times \vec \Omega[/tex]

where [itex]\vec \Omega = q \vec B / m_0[/itex]. There are a number of ways to proceed from here but it should be apparent that the same analysis you did in the classical case will work except that B is replaced by [itex]B / \gamma[/itex] from which your result follows.
 
  • #3
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still stuck

Still stuck since I don't really remember the classical case.
 
  • #4
Tide
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Homework Helper
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In that case, consider ...

[tex]\frac {d v_x} {dt} = \Omega v_y[/tex]

and

[tex]\frac {d v_y} {dt } = - \Omega v_x[/tex]

Differentiate, say, the first and substitute the second into the first:

[tex]\frac {d^2 v_x} {dt^2} = - \Omega^2 v_x[/tex]

from which it should be evident that the motion is sinusoidal with frequency [itex]\Omega[/itex].
 
Last edited:

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