Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Cross Product application problem

  1. Mar 15, 2012 #1
    Calculating magnitude and direction of an magnetic force on an electron

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

    Okay so the question says that the magnetic force (Vector)FM on a particle which is in a magnetic field is found by (Vector)FM = Vector I x Vector B.

    Vector I = charge multiplied by the velocity of a charged particle
    Vector B= the strength of the magnetic field (Measured in Tesla(T) ).

    An electron accelerated from REST to the RIGHT, in a horizontal directed electric field.
    The electron then leaves the electric field at a speed of 4.0 x 10^6, entering the magnetic field of magnitude 0.20 Tesla that is directed into the screen.

    From the given information calculate the magnitude and direction of the magnetic force on the electron given that the charge on an electron is q= 1.6 x 10^-19.

    2. Relevant equations

    All right, since we have to solve this in the cross product method i'm quite lost.
    I take it we want to end up with something like this: (ax, ay, az) and b = (bx, by, bz)
    giving us a x b = = (aybz − azby , azbx − axbz , axby − aybx)

    3. The attempt at a solution

    So far i analyzed that

    Vector I = charge of an electron (q = 1.6 x 10^-19) x velocity of a charged particle (4.06 x 10^6) (Since it says electron accelerates from rest to the right horizontally, i assume vector I is the x component?)

    and for

    Vector B = strength of the magnetic field in Tesla (0.20 T). (And since it says its directed into the screen, i assume its the z component? Right?)

    So now we have

    Vector I = 1.6x10^-19(4.0 x 10^6) = 6.4 x 10^-13
    Vector B = 0.20

    |6.4x10^-13| x |0.20| sin 90 = 1.28 x 10^-13

    So the magnitude is 1.28 x 10^-13 and the direction is in the y axis?
    I defenitley think i hugely messed up somewhere, please help
    Last edited: Mar 15, 2012
  2. jcsd
  3. Mar 15, 2012 #2


    User Avatar

    Staff: Mentor

    Your calculations look fine to me. But you should take care when dealing with vectors and vector components to make sure that you specify directions completely. "in the y-axis" doesn't indicate if the direction is towards +y or -y. You should be able to write each vector in component form: ## x \hat{i} + y \hat{j} + z \hat{k}## .
  4. Mar 15, 2012 #3
    I very much appreciate your reply, but since my answer is a positive value would it be correct in saying its towards the +y?

    So i would do
    |Vector Ix| = |vector I|cos 90°

    6.4x10^-13 cos 90

    0.20 cos 90


    Sorry im confused?
  5. Mar 15, 2012 #4


    User Avatar

    Staff: Mentor

    The sign of the result should depend upon your initial choice of axes, how the given values "lie" in that frame of reference, and any operations performed on those values. It's helpful to specify the coordinate system at the beginning and state the given values in terms of those coordinates. So for example, you might state that the +x-axis is chosen to coincide with the electron's direction of motion, and that the +z-axis is into the screen. The magnetic field is parallel to the z-axis and is directed towards +z. Or more concisely you could write:

    Let: ##\vec{v} = 4.0x10^6 \hat{i}~m/s~~~~~and~~~~~ \vec{B} = 0.20 \hat{k}~T ##
    If the result happened to have some odd direction that didn't coincide with any of the axes then you might want to write out the vector in ijk form. But if it coincides with an axis then it would be sufficient to specify which axis and which direction along that axis. For example, 1.28x10-13 N in the direction of the +y-axis.
  6. Mar 16, 2012 #5
    Thank you very very much, it helped a lot! The right hand rule could also be applied in this question!
    Kudos man!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook