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Finding acceleration of a particle due to the magnetic field

  1. Mar 26, 2015 #1
    1. The problem statement, all variables and given/known data
    A particle (mass 2.0 mg, charge –6.0 µC) moves in the positive direction along the x axis with a velocity of 3.0 km/s. It enters a magnetic field of (2.0i + 3.0j + 4.0k) mT. What is the acceleration of the particle?

    2. Relevant equations
    F = ma to find the acceleration
    F = qv x B to find the force of the magnetic field

    3. The attempt at a solution

    So I figured that since the velocity is in the positive x direction, and the x component of the magnetic field is also in a positive direction, the angle between them is 0 and so I can just not worry about the x component.

    To get Fj, I have Fj = qvBsin(90), since j is in the positive y direction and the velocity is in the positive x direction. This gives me a force of -5.4 x 10-5, which if I divide by the mass, I get a = -.027 m/s2. I do the same for the k component, which I think the angle separating the field and the velocity is also 90 degrees, so Fk = qvBsin(90), which I get to be -7.2 x 10-5, and if I divide by the mass, I get a = -0.036 m/s2.

    so my final answer would be a = (-.027j - 0.036k)m/s2. That is not one of my answer choices, which are as follows:

    However, the correct answer is (36j - 27k)m/s2. My order of magnitude is wrong but so are the vectors (my answer has 27j, the correct is 27k, and the same for 36). I'm really bad with dealing with vectors so I'm sure that's the problem, but can anyone guide me to the correct answer?
    Thank you
     
  2. jcsd
  3. Mar 26, 2015 #2

    Simon Bridge

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    ... why not check that by computing the determinant for the cross product?
     
  4. Mar 26, 2015 #3
    ..I do not have the slightest idea as to what that is.
     
  5. Mar 26, 2015 #4
    Okay, I just looked at my notes and see what you mean, but I don't understand why sometimes I can just do the cross product of A x B as ABsin(theta) and sometimes I have to do all of that. I have no clue what it means
     
  6. Mar 26, 2015 #5

    Simon Bridge

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    For two vectors given by:
    ##\vec r_1 = a\hat \imath + b\hat\jmath + c\hat k##
    ##\vec r_2 = d\hat \imath + e\hat\jmath + f\hat k##

    Then the cross product is given by:
    $$\vec r_1\times\vec r_2 = \begin{vmatrix} \hat\imath & \hat\jmath & \hat k\\ a & b & c \\ d & e & f \end{vmatrix}$$
     
  7. Mar 26, 2015 #6

    Simon Bridge

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    If you correctly guess the angle, you can use ##ABsin\theta## to get the magitude and the right-hand rule to get the direction.

    Go through the working for the determinant a step at a time and compare with your assumptions from before. i.e. can you safely ignore the x component of the magnetic field? Can you take the other components separately like you did? Did you absently switch two components over?
     
    Last edited: Mar 26, 2015
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