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Kinetic Energy in B field (is it really conserved)

  1. Aug 27, 2012 #1
    Here is a quote from one of the MIT notes on magnetism :

    "Note that F(magnetic force) is perpendicular to v and B , and cannot change the particle’s speed v (and thus the kinetic energy). In other words, magnetic force cannot speed up or slow down a charged particle. "

    How is this possible, it kind of conflicts with the idea of acceleration and "change" in velocity.

    lets say the velocity of an electron is v in the direction of increasing x axis, while the direction of the B field is along the positive y axis. A force will act on the electron directed towards the negative z axis.

    The initial velocity was in the x-y plane only but after the force acts on the charge, the "z" component of its velocity changes.

    Would this not change the total Kinetic Energy possessed by the electron?
    Clearly the velocity is not the same as before !
    Thus the magnetic force HAS changed the velocity.

    Can anyone help clarify this ?
     
  2. jcsd
  3. Aug 27, 2012 #2

    gabbagabbahey

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    Kinetic energy ([itex]K=\frac{1}{2}mv^2[/itex] non-relativistically) depends on the speed of the particle. the direction of the velocity can change, but its magnitude, the speed of the particle cannot.
     
  4. Aug 27, 2012 #3
    Speed, v , initially = √(x^2+y^2)

    speed , v , final (after magnetic force acts on the electron) = √(x^2+y^2+z^2)

    Note that the x and y components of velocity remain constant thus the only difference between the final speed and the initial speed is the "z" component of velocity.

    As you said the the kinetic energy is same then :

    x^2+y^2 = x^2 + y^2 + z^2

    thus z=0

    Which is clearly another contradiction with the statement that the kinetic energy is conserved.
     
  5. Aug 27, 2012 #4
    No. As the z component increases, the x and/or y components decrease in a way that exactly compensates to keep the magnitude of the velocity constant.

    It may help to thing about a simpler example. Consider say, a ball on a string being whirled round in a circle at constant speed. Here the string's force always points perpendicular to the ball's current velocity, with the result that the ball's speed is constant.
     
  6. Aug 27, 2012 #5

    gabbagabbahey

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    No, initially [itex]\mathbf{v}(t=0)=v_0\mathbf{e}_x[/itex] and the force is [itex]\mathbf{F}(t=0)=v_0|\mathbf{B}|\mathbf{e}_z[/itex].

    An infinitesmal time later, the velocity is [itex]\mathbf{v}(t=0+dt) = v_0\mathbf{e}_x +\frac{v_0|\mathbf{B}|}{m}dt\mathbf{e}_z[/itex] and the force is [itex]\mathbf{F}(t=0 +dt)=v_0|\mathbf{B}|\mathbf{e}_z - \frac{v_0|\mathbf{B}|^2}{m}dt\mathbf{e}_y[/itex] (Notice the force has a negative y-component).

    But we don't measure at infinitesimal time intervals, only at finite time intervals. To do that, we integrate the acceleration:

    [tex]\mathbf{v}(t) = \int_0^{t} \frac{\mathbf{F}(t')}{m} dt' = \int_0^{t} \frac{q}{m}(\mathbf{v}(t') \times \mathbf{B}(t'))dt'[/tex]

    The easiest way to see that the speed is constant is to look at [itex]v^2 = \mathbf{v} \cdot \mathbf{v}[/itex] (if the square of the speed is constant, then so is the speed):

    [tex]\frac{d}{dt}v^2= \frac{d}{dt}\left[ \mathbf{v} \cdot \mathbf{v}\right] = 2\mathbf{v} \cdot \frac{d\mathbf{v}}{dt} = 2\mathbf{v} \cdot \left(\frac{q}{m} \mathbf{v} \times \mathbf{B} \right) = \frac{2q}{m}\mathbf{B} \cdot (\mathbf{v} \times \mathbf{v}) = 0 [/tex]

    where the second last step comes from the triple product identity [itex]\mathbf{A} \cdot ( \mathbf{B} \times \mathbf{C}) = \mathbf{B} \cdot ( \mathbf{C} \times \mathbf{A}) = \mathbf{C} \cdot ( \mathbf{A} \times \mathbf{B})[/itex].

    Evidently, [itex]v^2[/itex], and hence [itex]v[/itex] and [itex]K[/itex], is constant.
     
  7. Aug 28, 2012 #6
    I understood the method of Kinetic energy but i didn't really catch the first part of your post no.5

    Would you please explain/define what the variables represent .
     
  8. Aug 28, 2012 #7

    gabbagabbahey

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    I used [itex]v_0[/itex] to represent the initial speed of the particle, [itex]\mathbf{v}(t)[/itex] to represent the velocity vector of the particle at time [itex]t[/itex], [itex]\mathbf{B}[/itex] to represent the external magnetic field, and [itex]\mathbf{F}[/itex] to represent the force on the particle.
     
  9. Aug 29, 2012 #8
    Alright, but u missed one which is troubling me

    e (subscript x or y or z)
     
  10. Aug 29, 2012 #9
    what does e(x) represent
     
  11. Aug 29, 2012 #10

    gabbagabbahey

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    [itex]\mathbf{e}_x[/itex] is just one of many common ways to write the unit vector in the positive [itex]x[/itex]-direction. Other common ways to write it include [itex]\mathbf{i}[/itex], [itex]\mathbf{\hat{x}}[/itex], [itex]\mathbf{\hat{e}}_1[/itex], and [itex]\hat{i}[/itex] .
     
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