1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Force on a charge on magnetic field

  1. Apr 14, 2012 #1
    Hi guys,

    I am solving some magnetism questions,but somewhat confused about right hand rule.

    Determine the magnitude and direction of the force on an electron traveling 8.75 * 10^5 m/s horizontally to the east in vertically upward magnetic field of strnegth 0.75T?

    We know since this is perpindcular to the magnetic field it will have maximum since sin(90) would be 1,so we can calculate the the force easily.

    F = qVBsin;
    F = 1.05;

    What I don't understand is the direction first I did normally right hand rule,but since that charge is an electron it should be the opposite to what right hand rule gives us.

    So I got the direction as South,but my book says west that should be the direction on a given proton not electron right ?
  2. jcsd
  3. Apr 14, 2012 #2


    User Avatar
    Homework Helper

    You can visualize the direction of the force if you imagine that you rotate v into B (by the angle less than pi). The force points to that direction from where the rotation looks anti-clockwise. To the South for a proton, opposite for the electron, too North. You did something wrong with the right-hand rule. F=q vxB (vector product).


    Attached Files:

  4. Apr 14, 2012 #3
    I see I just rechecked I orientated by hand wrong that's why I got wrong answer :S.
    Is their a way to check if I got my answer correctly mathematically ?
  5. Apr 14, 2012 #4


    User Avatar
    Homework Helper

    You can find the force from the components of v and B. Do you know the determinant method for calculating a vector product?

  6. Apr 14, 2012 #5
    No I don't know about it :S could you suggest a tutorial to read about this method?
  7. Apr 14, 2012 #6


    User Avatar
    Homework Helper

    The determinant only makes easier to memorize cross product. (See for example http://mathworld.wolfram.com/CrossProduct.html).

    You certainly know that

    [itex]\vec a\times \vec b=(a_yb_z-a_zb_y) \vec i+(a_zb_x-a_xb_z) \vec j+(a_xb_y-a_yb_x) \vec k [/itex].

    The force is [itex]\vec F= q\left[\vec v \times \vec B\right][/itex]. If the x axis points to East, the y axis points to North and the z axis point upward, [itex]\vec v=v \vec i[/itex] and [itex]\vec B=B \vec k[/itex], and [itex]\vec F=q v \vec i\times B \vec k=-qvB \vec j[/itex].

  8. Apr 14, 2012 #7
    I see thanks alot :).
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Threads - Force charge magnetic Date
Radial force on charged particle in beam of positive ions Oct 29, 2017
Electrons move, causing magnetic force, find v? Jun 13, 2017