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Magnetism, component values, and three dimensions

  1. Jul 31, 2007 #1

    exi

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    1. The problem statement, all variables and given/known data

    A magnetic field has these components:

    x: B = 0.056 T
    y: B = 0.046 T

    The particle has a charge of +6.5e-5 C, has a velocity of 2.6e3 m/s, and is moving along the z axis.

    Questions:

    1: What's the magnitude of the net magnetic force on the particle?

    2: What's the angle the net force makes with respect to the x axis?

    2. Relevant equations

    F = qvb(sinΘ)
    Component addition of the x- and y- values.

    3. The attempt at a solution

    I have the answer to #1 by adding the x- and y- components and getting a resulting magnitude of 0.0725 T at 39.4007° between those axes. (0.01225 N)

    On #2, I apparently incorrectly assumed that the net force was directed along the z-axis and is therefore 90° from the x-axis. Is the angle I found earlier (39.4007°) indeed the answer to this question? I'm unsure.
     
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  3. Jul 31, 2007 #2

    Gokul43201

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    I think you got lucky with the correct magnitude because x^2 + y^2 = y^2 + x^2. I suspect the values you have for Fx and Fy and in fact reversed.

    Use the Lorentz force equation in vector form (what do you know about the cross - or vector - product) to determine magnitudes and directions of each component.
     
  4. Jul 31, 2007 #3

    exi

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    The x- and y- magnitude components are correct; just double-checked them.

    Not sure what you're getting at as far as a Lorentz equation; that's new to me.
     
  5. Jul 31, 2007 #4

    Gokul43201

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    Okay, so did you use [itex]F_x = -qv_zB_y[/itex] and [itex]F_y = qv_zB_x[/itex]?

    This is simply the scalar breakup of the vector equation [itex]\vec{F} = q (\vec{v} \times \vec{B}) [/itex]. Have you seen this equation in this form? Have you dealt with cross products yet?
     
    Last edited: Jul 31, 2007
  6. Jul 31, 2007 #5

    exi

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    Whoa, for the first part? No, I did it the good ol' fashioned way, with [itex]|B| = \sqrt{0.056^2 + 0.046^2}[/itex].

    And yeah, that does look familiar, albeit in already-crossed [itex]F_B = qvBsin\theta[/itex] form. I think I'm having a bit of a hard time with the angle bit, since I have an angle from the x- and y- components and am not exactly sure what to do with the rest of what's here.
     
  7. Jul 31, 2007 #6

    Gokul43201

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    Okay, in that case, all you need to remember is that the force, F is always perpendicular to both the field, B, and the velocity v. So, it must be one of the 2 normals to B in the x-y plane. Which one, is determined by the right hand rule.
     
  8. Aug 1, 2007 #7

    exi

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    If I do what I think is the RHR for this question (with my thumb pointing upwards, towards the ceiling, for the particle moving along the z-axis, and my fingers extending in the direction of the x-y plane magnetic field at 39.4007° to the right), and if the force is in the direction indicated by my palm, it feels like that would put it at 129° to x. Am I going about this correctly?
     
  9. Aug 1, 2007 #8

    Gokul43201

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    Looks good.
     
  10. Aug 1, 2007 #9

    exi

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    I thought so too - but it was incorrect and my last submission on that question. Key won't be available until tomorrow am.

    Wonder where I went wrong...
     
  11. Aug 1, 2007 #10
    [tex]\displaystyle \vec{B} = (56\,\mathrm{mT})\hat{\i} + (46\,\mathrm{mT})\hat{\j}[/tex]

    [tex]\displaystyle \vec{v} = (2.6\,\mathrm{km/s})\hat{k}[/tex]

    When you have components, it is often wasier to calculate the cross product by finding the determinant of the matrix:

    [tex]\displaystyle \vec{v}\times\vec{B} =
    \begin{array}{|ccc|}
    \hat{\i} & \hat{\j} & \hat{k}\\
    0 & 0 & 2600\\
    0.056 & 0.046 & 0
    \end{array} = (-119.6)\hat{\i} + (145.6)\hat{\j}
    [/tex]

    Now you multiply this by q and you have the force:

    [tex]\displaystyle \vec{F} = (-q \cdot 119.6)\hat{\i} + (q\cdot 145.6)\hat{\j}
    [/tex].

    The magnitude of a vector [tex]\vec{F} = a\hat{\i} + b\hat{\j}[/tex] is, as you know, [tex]||F|| = \sqrt{a^2 + b^2}[/tex] and the angle of the vector with the x axis is [tex]\theta = \arctan{\frac b a}[/tex].
     
  12. Aug 1, 2007 #11

    exi

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    Thanks for the explanation. :smile:
     
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