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Magnetic energy of a conductor with a hole

  1. Nov 20, 2016 #1
    1. The problem statement, all variables and given/known data
    A cylindrical conductor of radius ##a## has a hole of radius ##b## and there is a current ##i## uniformly spread inside the conductor.
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    Calculate the magnetic energy of the conductor.

    2. Relevant equations
    $$U_{m}=\int_{Volume} \frac{B^2}{2 \mu_0} dV$$

    3. The attempt at a solution

    I know that this system is equivalent to a integer cylindrical conductor with uniformly spread current ##i_{+}##, plus another conductor in the place of the hole, with uniformly spread current ##i_{-}##, where

    $$i_{+}=\frac{i \, \pi a^2}{\pi (a^2-b^2)} \,\,\,\, \mathrm{and} \,\,\,\, i_{-}\frac{i \, \pi b^2}{\pi (a^2-b^2)}$$

    I see that on my textbook this exercise is solved by saying that the total energy would be the sum of the energy due to ##i_{+}## and the energy due to ##i_{-}##, that is (doing the integral above):

    $$U_{m}=\frac{\mu_0}{16 \pi} (i_{+}^2-i_{-}^2)$$

    But I do not agree with that. The superposition principle is valid in the calculation of the magnetic field but not in the calculation of the total energy.

    Roughly said, the sum of squares is not the square of the sum, so shouldn't I instead firstly calculate the total magnetic field il all the conductor (summing vectorially the magnetic field due to ##i_{+}## and the one due to ##i_{-}## in every point) and then calculate

    $$U_{m}=\int \int \int_{Volume} \frac{B_{total}(x,y,z)^2}{2 \mu_0} dx dy dz$$

    ?

    This process looks more correct because the superposition principle is allowed while calculating fields.

    Did I get this right is the process done on my textbook valid to calculate the total energy?
     

    Attached Files:

  2. jcsd
  3. Nov 21, 2016 #2

    Simon Bridge

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    Do you get a different answer?

    If two systems are physically equivalent, then they will follow equivalent rules.
     
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