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Homework Help: Radius of path of electron in a magnetic field

  1. Jul 25, 2006 #1
    # Electrons are moving in a uniform magnetic field of 50 Oersted having a velocity of 8.8 x 10^6 cm/sec. What is the radius of the circular path they follow?
    I solved it in the following way:
    In C.G.S system 1 Gauss = 1 Oersted (In vacuum)
    So, B = 50 Gauss
    V = 8.8 x 10^6 cm/sec
    Let e be the charge of the electron, r be radius of its circular path and m be its mass.
    e = 1.6 x 10^(-19) Coulomb = 4.8 x 10^(-10) Stat Coulomb
    m = 9.1 x 10^(-28) grams
    In a perpendicular magnetic field,
    BeV = {m(V^2)}/r
    r = (mV)/(Be)
    Solving I get,
    r = 3.3 x 10^(-13) cm
    I think somewhere I have gone wrong as the radius is too small. Please help.
  2. jcsd
  3. Jul 25, 2006 #2
    In cgs units the formula for the Lorentz force is (qv/c) x B. The final formula should read r = (mVc/Be). This gaves a much more reasonable answer.
  4. Jul 27, 2006 #3
    I have learnt the expression for Lorentz force in C.G.S system.Thanks.
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