Magnetic force on capacitorI have this problem: I have a capacitor formed by two parallel square-shaped plates, of side a, and with distance d between them. They're asking me out of other things, the magnetic force as a function of some things. I calculate de magnetic force by calculating the gradient of the magnetic energy, given by: E = 1 2 ∭ V B ⋅ H d VI have calculated B and H, that form, during the charge of the capacitor, loops around the center proportional to r, being r the distance from the central axis. The problem is that I have to integrate that to a cubic volume: a 2 d, one of those variables is independent, but the other two are not, and this problems never involve complicated integrals, such that one: ∫ 0 a ∫ 0 a r d x d yIf the plates were circles it would be trivial, but being squares, I'm sure I'm missing something.

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Magnetic force on capacitorI have this problem: I have a capacitor formed by two parallel square-shaped plates, of side   a, and with distance   d between them. They&#039;re asking me out of other things, the magnetic force as a function of some things. I calculate de magnetic force by calculating the gradient of the magnetic energy, given by:  E  =      1    2        ∭    V    B  ⋅  H    d  VI have calculated   B and   H, that form, during the charge of the capacitor, loops around the center proportional to   r, being   r the distance from the central axis. The problem is that I have to integrate that to a cubic volume:       a    2    d, one of those variables is independent, but the other two are not, and this problems never involve complicated integrals, such that one:      ∫    0    a        ∫    0    a    r  d  x  d  yIf the plates were circles it would be trivial, but being squares, I&#039;m sure I&#039;m missing something.

priffEmipsy4i37m · Accepted Answer

You want to know      ∫    0    a        ∫    0    a    r  d  x  d  ywith  r  =            x      2        +          y      2      via substitution   x  =  α  a and   y  =  β  a, you can bring it into the form  a      ∫    0    1        ∫    0    1              α      2        +          β      2        d  α  d  β  .For the remaining intgral, Wolfram alpha is your friend. With some more work you can show that in the exact result is      ∫    0    a        ∫    0    a    r  d  x  d  y  =      a    3        [          2        +    log    ⁡          (      1      +              2            )        ]    .

Magnetic force on capacitor I have this problem: I have a capacitor formed by two parallel square-

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