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Computing for magnetic force. Please check my answer.

  1. Feb 7, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the magnitude and direction of the magnetic force on a charged particle with charge -4nC and velocity

    [itex]\vec{v}[/itex] = [itex] (2.5\times 10^{4}) \hat{i}[/itex] + [itex] (1.1 \times 10^{4}) \hat{j}[/itex] (m/s)

    if the magnetic field in the region is given by

    [itex]\vec{B}[/itex] = [itex](1.2\times 10^{-3}) \hat{i}[/itex] + [itex](5.6 \times 10^{-3}) \hat{j}[/itex] - [itex](3.2 \times 10^{-3}) \hat{k}[/itex] (T).


    2. Relevant equations

    [itex]\vec{F}[/itex] = q ([itex]\vec{v}[/itex] [itex]\times[/itex] [itex]\vec{B}[/itex])

    3. The attempt at a solution

    [itex]\vec{F}[/itex] = q ([itex]\vec{v}[/itex] [itex]\times[/itex] [itex]\vec{B}[/itex])

    [itex]\vec{F}[/itex] = ([itex] -4 \times 10^{-9} C [/itex]) [itex] \left[\left((2.5\times 10^{4}) \hat{i} + (1.1 \times 10^{4}) \hat{j}\right) \times \left((1.2\times 10^{-3}) \hat{i} + (5.6 \times 10^{-3}) \hat{j} - (3.2 \times 10^{-3}) \hat{k} \right) \right][/itex]

    [itex]\vec{F}[/itex] = ([itex] -4 \times 10^{-9} C [/itex])[itex] \left[\left[(1.1 \times 10^{4})(-3.2 \times 10^{-3})\right] \hat{i} - \left[(2.5 \times 10^{4})(-3.2 \times 10^{-3})\right] \hat{j}+ \left[(2.5 \times 10^{4})(5.6 \times 10^{-3}) - (1.1 \times 10^{4}) (1.2 \times 10^{-3})\right] \hat{k}\right] [/itex]

    [itex]\vec{F}[/itex] = ([itex] -4 \times 10^{-9} C [/itex]) ([itex]-35.2 \hat{i} + 80 \hat{j} + 126.8 \hat{k} [/itex])

    [itex]\vec{F}[/itex] = [itex] (1.4 \times 10^{-7}) \hat{i}[/itex] - [itex] (3.2 \times 10^{-7}) \hat{j} - (5.0 \times 10^{-7})\hat{k} [/itex] (N)

    Is this already the answer? Am I correct? Thanks in advance.
     
    Last edited: Feb 7, 2012
  2. jcsd
  3. Feb 7, 2012 #2

    tiny-tim

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    hi jhosamelly! :wink:

    looks ok so far :smile:

    now the question asks for the "magnitude and direction" :wink:
     
  4. Feb 7, 2012 #3
    Yah.., that's what I was thinking ..

    I know how to get the magnitude.. Its

    [itex]\left|\vec{F}\right|[/itex] = [itex]\sqrt{(A_x)^2 + (A_y)^2 + (A_z)^2 }[/itex]

    but the question also asks for the direction.. which I think is the i-hat, j-hat, and k-hat..

    so., I think my answer earlier is already the final answer.. If not. How can I find the direction?
     
  5. Feb 7, 2012 #4

    tiny-tim

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    hi jhosamelly! :smile:

    yes, that's the magnitude

    my guess is that they want the direction defined by the unit vector :wink:
     
  6. Feb 7, 2012 #5
    hmmm.. meaning the question is asking if the force is pointing along the + or - , x- y- or z- axis.....?? so i have three directions?? + x-axis, - y-axis and - z-axis? This is what confused me... I know this is not possible.. so, what is the direction of the force?
     
  7. Feb 7, 2012 #6

    tiny-tim

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    nooo …

    the is a unit vector in every direction

    every vector r is|r| (the magnitude) times the unit vector r^ :wink:
     
  8. Feb 7, 2012 #7

    ow yah! Right! Thanks for reminding me :))

    so I need to do

    [itex] \hat{F} [/itex] = [itex] \frac{\vec{F}}{\left|\vec{F}\right|} [/itex]
     
  9. Feb 7, 2012 #8

    tiny-tim

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  10. Feb 7, 2012 #9

    I already did it.

    [itex]\left|\vec{F}\right|[/itex] = [itex]\sqrt{(A_x)^2 + (A_y)^2 + (A_z)^2 }[/itex]

    [itex]\left|\vec{F}\right|[/itex] = [itex]\sqrt{(1.4 \times 10^{-7})^2 + (-3.2 \times 10^{-7})^2 + (-5.0 \times 10^{-7})^2 }[/itex]
    [itex]\left|\vec{F}\right|[/itex] = [itex] 6.18 \times 10^{-7} [/itex]


    [itex] \hat{F} [/itex] = [itex] \frac{\vec{F}}{\left|\vec{F}\right|} [/itex]

    [itex]\hat{F} [/itex] = [itex] \frac{(1.4 \times 10^{-7}) \hat{i} - (3.2 \times 10^{-7}) \hat{j} - (5.0 \times 10^{-7})\hat{k}}{6.18 \times 10^{-7}} [/itex]

    [itex] \hat{F} [/itex] = [itex] .23 \hat{i}[/itex] - [itex] .52 \hat{j} - .82 \hat{k} [/itex]

    so, again I still have 3 components.. Hmmmm... What's the direction then?
     
  11. Feb 7, 2012 #10

    tiny-tim

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    d'oh!! :rolleyes:

    that unit vector is the direction! :smile:
     
  12. Feb 7, 2012 #11
    Is that so?? hmmmm... I know that only happens if only one component is left?? O well, I was mistaken.. thanks for your help.
     
  13. Feb 7, 2012 #12

    tiny-tim

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    perhaps i should add …

    i j and k are the basis unit vectors​
     
  14. Feb 7, 2012 #13
    I see.. Big thanks :)) So the final answer is that unit vector and the magnitude. I see. Thanks :))))))))
     
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