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Charge Acted By Magnetic Force AND Friction

  1. Jan 2, 2009 #1
    1. The problem statement, all variables and given/known data
    Part 1: A charge particle having charge q enters a region in which there is a friction force proportional to the speed of the particle. The constant of proportionality of friction is "b". The particle stops traveling a distance of 10m, in a straight line.

    Now, the experiment is repeated, but with magnetic field of unknown strength, in a direction perpendicular to entry velocity of the particle. The particle is found to stop at a point 6m from the entry point, then find the strength of magnetic field.

    Part 2:The magnetic field is now doubled in magnitude. How far from the point of entry the particle come to rest?

    2. Relevant equations
    v^2=u^2 +2as
    Perhaps something else as well??

    3. The attempt at a solution
    For the first case, the net force on the particle in the first case is simply the algebraic sum of the force due to the magnetic field and the friction i.e. F=quB-bu where B is magnetic field, u is the velocity, q= charge of the particle, now considering F=ma, we get a=(quB-bu)/m (where m is the mass of the particle), then we can use v^2=u^2+2as (a is retardation so V^2=u^2-2as)

    so 0=u -2(qB-b)10/m (as we are given tht s=10m)
    so mu=20 (qB-b) ------------->1

    When the field is made perpendicular to the particle, the magnetic field will, at every point acts tangentially to the circular path of the particle, and the frictional force acts in a dirction exactly opposite to this .......atleast thts what I think

    Tht's all I've been able to get upto .......haven't even started off on the second part!!

    Respected sirs, I'm a 17 yr old student, so please go easy on me if you feel that this problem is too simple or if my method is completely wrong

    Thanks for your help
    Last edited: Jan 2, 2009
  2. jcsd
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