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F=qvB components

  1. Aug 25, 2010 #1
    1. The problem statement, all variables and given/known data
    2 A particle with charge 94.5 nC is moving in a region where there is a uniform magnetic field of 0.450 T in the positive x-direction. At a particular instant of time, the velocity of the particle has components:
    vx = - 1.68 x 104 ms-1, vy = - 3.11 x 104 ms-1 and vz = 5.85 x 104 ms-1.
    What are the components of the force on the particle at this time?



    2. Relevant equations

    F=qvBsin(theta)

    3. The attempt at a solution
    1st how do i get Greek letters into a post????
    Fx=qvBsin0 = 0N
    Fy=94.5*10-9*-3.11*104*.45sin90 = -1.322*10-3N
    Fz=94.5*10-9*5.85*104*.45sin90 = 2.49*10-3N

    The reason Im posting this is because it seemed way to easy so I think I may have totally missed/screwed something up????
    Thanks for any help.
     
  2. jcsd
  3. Aug 25, 2010 #2

    kuruman

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    It is best to start from
    F = q v x B
    F = q(vxi + vyj + vzk) x Bi = q B(vx ixi + vy jxi + vz kxi)

    calculate the three cross products separately and see what you get.

    For special symbols, save the items below somewhere, then cut and paste from there.

    α β γ δ ε θ λ μ ν π ρ σ τ η φ χ ψ ω Γ Δ Θ Λ Π Σ Φ Ψ Ω
    ∂ ∏ ∑ ← → ↓ ↑ ↔ ⇐⇑⇒⇓⇔
    ± − ÷ √ ∫ ½ ∞∴ ~ ≈ ≠ ≡ ≤ ≥ ° ∇∝
     
  4. Aug 25, 2010 #3
    Hey Kuruman, thanks for the reply.
    I chucked your equation in my 89 an came up with -7.144*10-4i2-1.3225*10-3ij+2.4877*10-3ik
    the only inconsistent value is the 1st 1 which i hope is simply because of the lack of sin(theta) since the force in the i plane would be zero????
    also unsure of what "calculate the three cross products separately and see what you get." means? I cant see how to do three of them, is what I did what you meant for me to do????
    Thanks
     
  5. Aug 25, 2010 #4
    There are three cross products in Kuruman's last expression (i x i, j x i, and k x i)

    i x i is not equal to i^2 (i is a vector)
     
  6. Aug 26, 2010 #5

    kuruman

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    Listen to JaWiB. These cross products are vectors and your 89 appears not to be able to handle them correctly. You need to simplify cross products of unit vectors like ixj and write them as single unit vectors. Look up "cross product" and do it correctly.
     
  7. Aug 26, 2010 #6
    89 can do cross product, I told it to solve that equation because I've never done them outside of our maths course where the questions are usually just "find the cross product of a and b" so admittedly i was still a little confused about what to do??? I was thinking that the cross required two separate vectors but I only had one with the velocity broken up into component form. Now I have done the cross product of the magnetic field and charge*velocity resulting in:
    {0,2.48771E-3xz,1.32253E-3xy} x y z is i j k respectively
    Im still confused about what exactly this tells me? Do I need to do this twice more with something???sorry about being a bit dense about this and thanks for your help.
     
  8. Aug 26, 2010 #7

    kuruman

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    How do you interpret this? What are the three elements separated by commas and what do xz and xy stand for?
     
  9. Aug 26, 2010 #8
    I think the coefficients are the magnitude of the force components? what the xz and xy mean im not sure. if I had to guess i would say some sort of direction but i think i might remember something about cross product that says not(maybe dot- I cant remember at all)??
     
  10. Aug 26, 2010 #9

    kuruman

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    OK, here is a simple question. What is the cross product i x j. Figure this out without using your 89 because it seems you do not understand what it's spitting back at you.
     
  11. Aug 26, 2010 #10
    Downside of having a halfway decent calculator is that I forget the process of solving things like this almost immediately after the exam!

    Reading the PF "cross product page" i get ixj=ijsin90=ij

    I also found that "The cross product of two vectors A and B is a third vector (strictly, a pseudovector or axial vector) perpendicular to both of the original vectors, with magnitude equal to the product of their magnitudes times the (positive) sine of the angle between them, and in the direction determined by the right-hand rule."
    This leaves me more confused about what my xy and xz's are???????
     
  12. Aug 26, 2010 #11

    kuruman

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    So the cross product is a vector. To specify it, you need to specify its magnitude and direction. First the magnitude. What is the product of the magnitude of i times the magnitude of j times sin90o? Don't forget that i and j are unit vectors.

    Now for the direction. What does the right hand rule say about it?
     
  13. Aug 26, 2010 #12
    again trying to remember 11c, a unit vector is just a vector of length 1unit??? so the magnitude of ixj must be 1????
    I could be applying the rhr to the wrong section of this problem but for Fy the direction i get the force to be is to the right and for Fz the force is straight down(using conventional planes)
    I am using http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/magfor.html just replacing i with q. Just to try and let you know how Im thinking I used the values that were in the question(could have worked out direction without doing any calculations???).
     
  14. Aug 26, 2010 #13

    kuruman

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    Yes and yes. Now for the direction. It must be perpendicular to the xy plane, but there are two perpendiculars. By definition the +z direction (unit vector k) is in the ixj direction. So the cross product between two unit vectors is another unit vector in a direction perpendicular to the two (or zero). Back to my first post. Find the following cross products

    ixi = ________

    jxi = _______

    kxi = ________
     
  15. Aug 26, 2010 #14
  16. Aug 26, 2010 #15
    i found the cross product of v x B at once,

    the vector i found perpendicular to both v and B is 0i -26585j - 13995k
    Thats before multiplying through the charge, so after i do that it SHOULD give the magnitude of the force in component form.

    does that sound right?

    I used determinants to find cross product, all by hand.
     
  17. Aug 27, 2010 #16

    kuruman

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    I didn't check the numbers, but yes, determinants is another way of doing this.
     
  18. Aug 27, 2010 #17
    i got F= 0i - 0.025j - 0.013k ??

    got anything like this pat?
     
  19. Aug 27, 2010 #18
    very 1st post on this thread, exactly what i got-that should be right then(hopefully).
     
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