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Heat dissipated by a cylinder in a magnetic field

  1. Jan 11, 2012 #1
    1. The problem statement, all variables and given/known data

    We have a cylinder (radius r, height h, conductivity γ) and it is immersed in a variable magnetic field B(t) = B0* e^(-t/T). The field is parallel to the axis of the cylinder.

    The problem then asks to find the heat dissipated by the cylinder after an infinitely long time.

    2. Relevant equations

    W = [itex]\int R i^2 dt[/itex]

    [itex] i = (-1/R) * \pi r^2 dB/dt [/itex]

    3. The attempt at a solution


    I consider the cylinder as a succession of circular loops.

    Let's consider one loop:

    The heat dissipated is: [itex] W = \int R i^2 dt[/itex]

    where

    [itex] R = \rho \frac{l}{S} = \frac{2\pi r}{γ*S}[/itex]

    [itex] i^2 = (E/R)^2 = (1/R^2) (\frac{d\varphi(B)}{dt})^2 = (1/R^2) \pi^2 r^4 (dB/dt)^2 [/itex]

    where E is the Electromotive force.

    So:

    [itex] (dB/dt)^2 = (B0/T)^2 * e ^ {-2t/T} [/itex]

    [itex] W = \int R i^2 dt = R/R^2 \pi^2 r^4 (B0/T)^2 \int_0^\infty e^{-2t/T} dt =(\pi^2 r^4 (B0/T)^2)*T/2R [/itex]

    Since there are h loops, the result is [itex] \frac{h*\pi^2 *r^4 (B_0)^2}{2RT} [/itex]

    Is this right?

    Thank you in advance.
     
  2. jcsd
  3. Jan 11, 2012 #2

    rude man

    User Avatar
    Homework Helper
    Gold Member

    You're on the right track, but realize that h → ∞ !
    And so of course the current di in each loop must be infitesmally small, also that the resistance of a loop will vary with r, 0 < r < radius. (I am making r variable. Dumb to use r for radius, but can't use R either since R is resistance).

    In other words, consider an annulus of differential width dr and get the corresponding resistance R(r), then di = emf(r)/R(r), then integrate dP = (di^2)*R from 0 to r, then integrate dP over time t from 0 to ∞ (energy = ∫power*dt).

    Good problem!
     
  4. Jan 17, 2012 #3
    Hey :) Thanks a lot for your help.

    Let's give it a second try step by step.

    1. why does h → ∞ ? it isn't an infinitely long cylinder!

    2. The cross section of the wire is A (constant), so

    dR(r) = rho * dL/A = (rho * 2pi * dr)/A, right?

    Now, di = emf(r)/R(r) with

    d emf(r) = 2pi*rdr dB/dt

    but now di = (dB/dt)*A*r since dr disappeared. I think it can't be right. :\
     
  5. Jan 17, 2012 #4

    rude man

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    Mea culpa. I got confused when you spoke of "h loops".

    OK, think of an elemental annulus, height = h, thickness = dr, radius = r. The annuli comprising the cylinder range in radius fro 0 to R. (I will call italicized R the radius of your wire/cylinder.) In other words, think of the cylinder as kind of a cylindrical onion, with the annuli represented by the onion's layers.

    There will be an emf generated around this elemental annulus. The conductance of this annulus will be dS = γ*h*dr/2πr Siemens. You need to visualize the annulus as being unfolded to a slab of metal with dimensions length = 2πr width = h, thickness = dr.

    The emf around this annulus' contour will be area * dB/dt (ignoring sign). Area = πr2.

    The current di will then be emf*dS, and the power dissipation will be dP = (emf)2*dS.

    To get the total power you now need to integrate dP from r = 0 to r = R.

    Finally you need to integrate P from t = 0 to t = ∞ to get the total energy, which of course is the heat you're looking for.





    2. The cross section of the wire is A (constant), so

    dR(r) = rho * dL/A = (rho * 2pi * dr)/A, right?

    Now, di = emf(r)/R(r) with

    d emf(r) = 2pi*rdr dB/dt

    but now di = (dB/dt)*A*r since dr disappeared. I think it can't be right. :\[/QUOTE]

    The current does not flow along the wire. The current flows in circles inside the wire. Other than that I don't follow your math. Where did 2pi*dr come from?
     
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