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An electron moving through a uniform magnetic field

  1. Apr 7, 2015 #1
    1. The problem statement, all variables and given/known data

    An electron moves through a uniform magnetic field given by B = Bx i + (3.68 Bx) j . At a particular instant, the electron has velocity v = (1.88 i +4.86 j) m/s and the magnetic force acting on it is (2.43 × 10-19) N. Find Bx.

    Givens

    B = [Bx i + (3.68 Bx)]

    v = (1.88 + 4.86)

    Fb = (2.43 × 10-19) N. Find Bx.

    2. Relevant equations


    Fb = q(B x V)


    3. The attempt at a solution

    (2.43 x 10^-19)k = (1.602 x 10^-19) [(Bx i + (3.68Bx) j) x (1.88 i + 4.86 j)

    cross product resultant:

    (2.43 x 10^-19)k = (1.602 x 10^-19) (-2.06*Bx)k

    (2.43 x 10^-19)k = (-3.30 x 10^-19*Bx) k

    I don't understand how to solve for the Bx, we haven't learned how to divide vectors, and I can't really pull the Bx out of the k vector as a scalar. How do I solve for the Bx???
     
  2. jcsd
  3. Apr 7, 2015 #2

    haruspex

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    Since the two unit vectors are identical, you can cancel them. But if you want to do it by pure methods, there are two ways: consider magnitudes; take the dot product of each side with something convenient.
     
  4. Apr 7, 2015 #3

    BvU

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    BX is not a vector, it is a number with a dimension. The vectors are ##\vec v## and ##\vec B = B_X \hat\imath + (3.68\;B_X) \hat\jmath\ ##.
    So check the dimensions (they should be fine) and find the value as you intended to do: by dividing.
     
  5. Apr 7, 2015 #4

    haruspex

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    I believe Ryan is referring to the two k vectors.
     
  6. Apr 7, 2015 #5

    collinsmark

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    Hello RyanTAsher,

    Before I go through the math (privately on my own), could you help clear up a few ambiguities?

    I'm assuming that the x in Bx is a subscript and not a variable. (In other words, I'm assuming that [itex] \vec B [/itex] is uniform and not a function of x.) Is that correct?

    Are you sure about [itex] \vec F = q \left( \vec B \times \vec v \right) [/itex]? Is that correct or is something backwards?

    (Also, don't forget that the charge of an electron is negative. i.e., [itex] q = -e [/itex])

    In the initial problem statement, the force was given as (2.43 × 10-19) N with no direction specified. Later, you tacked on a [itex] \hat k [/itex]. I'm just wondering if the original problem statement had the [itex] \hat k [/itex] in it, or if the (2.43 × 10-19) N figure was meant to be magnitude only without specifying a direction. [Edit: Going through the math, I realize that the force must be along the z axis. But ultimately, one still needs to determine if the force is pointing in the positive or negative direction of the axis.]
     
    Last edited: Apr 7, 2015
  7. Apr 7, 2015 #6

    BvU

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    @Haru: of course, naive of me to overlook that; thanks!
    @CM: well pointed out. But now the wrong expression is appearing twice already in this thread (shudder) ....

    @RTA: you're in good hands ! Bedtime for me. Good luck. And dividing out vectors is like dividing out Yuzzamatuzzes: if Zfluff Yuzzamatuzzes = 2.5 Yuzzamatuzzes, then you can be sure Zfluff = 2.5. It's a property of multiplication, not a property of Yuzzamatuzzes. There is only one snag to beware of: things go awry if Yuzzamatuzz = 0, because then you can't "divide them out" any more. (With thanks to dr. Seuss)
     
  8. Apr 8, 2015 #7
    Yes, the Bx is a B with subscript x, Bx, and the force value did have the [itex] \hat k [/itex] with it. Sorry for the confusion. I am pretty positive about the [itex] \vec F = q \left( \vec B \times \vec v \right) [/itex] as that is what it says in my textbook.
     
  9. Apr 8, 2015 #8

    BvU

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    The Lorentz force is definitely $$
    \vec F = q\, \left (\vec E + \vec v \times \vec B \right )
    $$no matter what your textbook says.
     
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