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Thread: Derivative of magnetic flux

  1. #1
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    Question Derivative of magnetic flux

    Why is the derivative of magnetic flux like this defined by:

    $$ \frac{d\vec{\psi}}{dt} = \frac{d\psi e^{j\epsilon(t)}}{dt} = j\omega \vec{\psi}$$

    I don't understand why do we calculate the derivative of $\epsilon$, which is $\omega$.

    Thanks.
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    Forum Admin topsquark's Avatar
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    Re: Derivative of magnetic flux

    Quote Originally Posted by Nforce View Post
    Why is the derivative of magnetic flux like this defined by:

    $$ \frac{d\vec{\psi}}{dt} = \frac{d\psi e^{j\epsilon(t)}}{dt} = j\omega \vec{\psi}$$

    I don't understand why do we calculate the derivative of $\epsilon$, which is $\omega$.

    Thanks.
    This must be an Engineering equation. I've never seen this formula in my life. I presume "j" is the imaginary unit? And what then is $\displaystyle \epsilon (t)$? What is $\displaystyle \omega$?

    Also, you dropped the vector notation in the center part of the equation. I'm guessing it's still $\displaystyle \vec{ \psi }$?

    Lastly you have an apparent contradiction(?) How can $\displaystyle \dfrac{d \vec{ \psi }}{dt} \neq \dfrac{ d \left ( \vec{ \psi }e^{j \epsilon (t)} \right )}{dt} = \dfrac{ d \vec{ \psi }}{dt} e^{j \epsilon (t) } + \vec{ \psi } j e^{j \epsilon (t) } \dfrac{d \epsilon }{dt} \neq j \omega \vec{ \psi }$. I don't see how that works.

    I'm much more familiar with the Physics definition $\displaystyle \Phi _B = \oint _{\partial S} \vec{B} \cdot d \vec{s}$, where S is a closed, bounded surface.

    -Dan
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    Re: Derivative of magnetic flux

    j = imaginary unit
    $\epsilon$ = angular position
    $\omega$ = angular velocity

    Maybe $\psi$ isn't depended from time? We multiply $\psi$ with $e^{j\epsilon(t)}$ to rotate the vector. Does this make any sense?
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    Forum Admin topsquark's Avatar
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    Re: Derivative of magnetic flux

    Quote Originally Posted by Nforce View Post
    j = imaginary unit
    $\epsilon$ = angular position
    $\omega$ = angular velocity

    Maybe $\psi$ isn't depended from time? We multiply $\psi$ with $e^{j\epsilon(t)}$ to rotate the vector. Does this make any sense?
    Thanks. That helps. But I still have a problem with the original equation for the derivative of $\displaystyle \vec{ \psi }$. Note that if it has no time dependence we have $\displaystyle 0 = j \omega \vec{ \psi }$ in your original equation. So that's not right.

    It would help if you could explain to me where the $\displaystyle \vec{ \psi } e^{j \epsilon (t)}$ comes from? Why are we rotating $\displaystyle \vec{ \psi }$? That would mean that we have some field that is rotating the magnetic flux, which in turn should imply that the vector potentail $\displaystyle \vec{A}$ is changing over time. Is this part of an exercise or is it part of a derivation?

    -Dan
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    Re: Derivative of magnetic flux

    Quote Originally Posted by topsquark View Post
    Thanks. That helps. But I still have a problem with the original equation for the derivative of $\displaystyle \vec{ \psi }$. Note that if it has no time dependence we have $\displaystyle 0 = j \omega \vec{ \psi }$ in your original equation. So that's not right.

    It would help if you could explain to me where the $\displaystyle \vec{ \psi } e^{j \epsilon (t)}$ comes from? Why are we rotating $\displaystyle \vec{ \psi }$? That would mean that we have some field that is rotating the magnetic flux, which in turn should imply that the vector potentail $\displaystyle \vec{A}$ is changing over time. Is this part of an exercise or is it part of a derivation?

    -Dan
    terrible choice of notation using $\epsilon$ as anything but the permittivity constant when dealing with E/M

    the whole thing looks like phasors to me which was a shortcut notation for sinusoidal signals.
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    Re: Derivative of magnetic flux

    Yes, romsek it's a phasor. Sorry because I did not mention it at the start.

    But I still don't understand how do we get than $j\omega\psi$. By rules of math, it just does not compute. Can you explain to me?
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    Re: Derivative of magnetic flux

    Quote Originally Posted by Nforce View Post
    Yes, romsek it's a phasor. Sorry because I did not mention it at the start.

    But I still don't understand how do we get than $j\omega\psi$. By rules of math, it just does not compute. Can you explain to me?
    There is no way to explain it. Your original post states:
    $\displaystyle \dfrac{ d \vec{ \psi }}{dt} = \dfrac{ d \left ( \vec{ \psi } e^{j \epsilon (t)} \right ) }{dt}$
    Unless $\displaystyle \epsilon (t) = 1$ for all t then the two sides can't be equal.

    Now, if what you have written is part of a larger problem that you aren't sharing then I can make a guess. What happens if we rewrite $\displaystyle \vec{ \psi }$ as $\displaystyle \vec{ \psi } (t) = \vec{ \psi _0 }e^{j \epsilon (t) }$ where $\displaystyle \vec{ \psi _0}$ is a time independent vector quantity. Then we have
    $\displaystyle \dfrac{ \vec{ \psi }(t)}{dt} = \dfrac{ d \left ( \vec{ \psi _0 } e^{j \epsilon (t)} \right ) }{dt} = \dfrac{ \vec{ \psi _0 }}{dt} e^{j \epsilon (t)} + \vec{ \psi _0 }j e^{j \epsilon (t) } \dfrac{d \epsilon (t) }{dt} = \vec{ \psi (t) }j \dfrac{ d \epsilon (t)}{dt}$

    as you needed.

    -Dan
    Last edited by topsquark; Jun 4th 2019 at 06:07 PM.
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