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Quick question about acceleration caused by B field

  1. Mar 2, 2013 #1

    CAF123

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    I know that the magnetic force is calculated via ##\underline{F} = q( \underline{v} \times \underline{B})##. Consider a particle with a velocity at some angle to a constant B field, both v and B in the same plane. Then the force will be acting in/out of the page depending on their orientation.

    My question is: This particle will experience a net force perpendicular to the v vector throughout its entire motion, so the kinetic energy of the particle will not change. But since it is acted on my a net force, it will undergo an acceleration by NII. Is this acceleration simply calculated by F/m?

    So this means that the acceleration vector will be parallel to the force vector. Is it really sensible to talk about the acceleration of the particle here? (It doesn't seem so since it is not actually accelerating - it's speed and velocity are constant throughout). I thought I could clarify this myself via google, but there appears to be contradictory views

    Many thanks.
     
  2. jcsd
  3. Mar 2, 2013 #2

    tiny-tim

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    Hi CAF123! :smile:
    Correct. The particle will follow a helix whose axis lies along the B field.
    Yes, by good ol' Newton's second law, q(v x B) = ma.

    This a will be perpendicular to both v and B, and will be the centripetal acceleration perpendicularly towards the axis of the helix.
    The acceleration vector is always parallel to the force vector (if the mass is constant). :wink:
     
  4. Mar 2, 2013 #3

    CAF123

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    Hi tinytim,
    I understand that a helix path will be followed if v and B are not perpendicular, but here the v and B are in the same plane, so why would that still create a helix path? (if v and B are in same plane, then the force will be upwards so why would this tend to make the particle go in a helix shape?)
     
  5. Mar 2, 2013 #4

    tiny-tim

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    any two vectors are in the same plane!

    think about it! :wink:
     
  6. Mar 4, 2013 #5

    CAF123

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    If I take v and B lying in the xy plane, with v some angle from B, then F will point in z direction. Why does this mean the particle will follow a helical path?

    I can see why the helical path would be formed if v was at an angle to B in 3D since then there would be a component of the force tending to create a circle.

    Also, if I wanted to compute the kinetic energy of such a particle since it's speed does not change, can I just say T = 1/2 mv2 throughout the whole motion or would I need to consider the rotational kinetic energy?

    Many thanks.
     
  7. Mar 4, 2013 #6

    tiny-tim

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    Hi CAF123! :smile:
    Exactly!

    This is 3D …

    B is a uniform field, so it isn't only in the xy plane (ie, the plane z = 0), it's in every horizontal plane! :wink:
    That's correct, the energy stays at its original 1/2 mv2 value. :smile:

    (and the magnetic field doesn't affect the spin … at least, not until you start doing quantum theory … so you can forget about rotational kinetic energy :wink:)
     
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