1. Not finding help here? Sign up for a free 30min 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!

Acceleration of a Particle Due to Magnetic Field

  1. Apr 15, 2016 #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 (2i + 3j + 4k) mT. What is the acceleration of the particle in component form?

    2. Relevant equations
    F = q(vXB)
    F = ma

    3. The attempt at a solution
    First convert all values to SI units:
    mass = 0.002 kg
    q = -6x10^-6 C
    v = 3000 i m/s
    B = (0.002i + 0.003j + 0.004k) T

    Find vXB:
    vXB = -12j + 9k

    Plug vXB and q into F = q(vXB) and find a:

    F = (-6x10^-6)(-12j + 9k) = ma
    a = (7.2x10^-5 j - 5.4X10^-5 k) / 0.002
    a = 0.036j - 0.027k m/s^2

    However the correct answer provided to me is:
    (36j - 27k) m/s^2

    I think i made a mistake converting the units for B from mT to T. i'm not sure how to convert units for vector components.


     
    Last edited by a moderator: Apr 15, 2016
  2. jcsd
  3. Apr 15, 2016 #2
    Never-mind I found my mistake. I didn't convert the mass correctly.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted