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Thoroughly confused on self-inductance

  1. May 5, 2014 #1
    I am havingA LOT of trouble with self-inductance and have spent 10+ hours watching videos on youtube, and I'm still getting nowhere.

    From what I understand:

    self inductance = the ability of a thing (not sure what) to resist the change in voltage (and thereby causing current to change more steadily. (is this right?)


    In my textbook derivation of the self-inductance of a coaxial cable, they took a cross section like here:

    http://www.phys.nthu.edu.tw/~thschang/notes/GP32.pdf (page 8)

    I don't really get this at all intuitively, because can't you choose any cross section?? How do you know where to find the surface which you want to get the flux of?

    I guess the problem is that, I feel as if you need to multiply the result by 2pi*r to get the whole entire cylinder, however this is not true.

    Please help (and if you happen to know a good analogy for self-inductance, please share because i am very confused )
     
  2. jcsd
  3. May 5, 2014 #2

    Drakkith

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    Staff: Mentor

    Self inductance is the creation of a voltage in a conductor when the current through that conductor changes. It is important to understand that the basic principles of self induction are no different than mutual induction. A change in current through a conductor causes the magnetic field generated by that conductor to change. This change in the magnetic field generates an EMF in the conductor itself as well as any nearby conductors.

    The EMF generated in self induction always tries to resist the change in the current. That is, if the current is increasing, the EMF will be against the direction of current. It will oppose it. If instead the current is decreasing, the EMF generated will act to keep the current flowing.
     
  4. May 5, 2014 #3
    Thanks. Also, in the equation:


    ## emf = \dfrac{d\phi_B}{dt}##

    how do you know which surface to take the flux of?
     
  5. May 6, 2014 #4

    Jano L.

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    Gold Member

    You can take any surface and calculate flux through it. The result does not depend on the shape of the surface, only on the closed curve that is its boundary.
     
  6. May 6, 2014 #5
    As Jano L. said the law applies to any surface. When solving problems you should for a surface that simplifies the problem as much as possible often taking the symmetry of the physical situation into account.
     
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