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Probability that a magnetic dipole is oriented with theta

  1. Jan 28, 2016 #1
    the problem goes like this :
    The energy of interaction of a classical magnetic dipole with the magnetic field B is given by
    E = −μ·B.
    The sum over microstates becomes an integral over all directions of μ. The direction of μ
    in three dimensions is given by the angles θ and φ of a spherical coordinate system
    The integral is over the solid angle element
    = sin θdθdφ. In this coordinate system
    the energy of the dipole is given by E = −μB cos θ.
    Choose spherical coordinates and show that the probability p(θ, φ)dθdφ that the dipole is
    between the angles θ and + dθ and φ and φ + dφ is given by

    p(θ, φ)dθdφ = (e^(μB cos θ) sin(θ) dθ dφ)/z

    2. Relevant equations

    z = ∫∫ e^(μB cos θ) sin(θ) dθ dφ .

    3. The attempt at a solution
    i have no idea what to do , and i tried all i know
    i know that the boltzmann distribution gives you the probability that a particle has an energy is :
    μB cos θ)/∫e^(μB cos θ) , but how do i integrate the spherical coordinates i don t know . please help me and thank you .
    Last edited: Jan 28, 2016
  2. jcsd
  3. Jan 28, 2016 #2
    this is the best that i could do
    the probability that the dipole between x and dx is
    dp(x) = (1 / Z ) e^(μ(x)B cos θ) dx = dx because we assume that B is parallel to z so μ(x) . B = 0
    dp(y) =
    (1 / Z ) e^(μ(y)B cos θ) dy = dy
    dp(z) = (1 / Z )e^(μ(z)B cos θ) dz
    d^3 p( x,y,z)= (1 / Z )e^(μ(z)B cos ) dz dy dx = (1 / Z )[B][B][B][B]e^(μ(z)B cos [B][B][B][B]θ[/B][/B][/B][/B]) dv[/B][/B][/B][/B]
    d^3 p( r,
    = (1 / Z )e^(μ(z)B cos θ) rd²r d[B][B][B][B]θ dφ[/B][/B][/B][/B]

    is this true ? or am i making horrible mistakes ?
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